Steven Shreve: Stochastic Calculus and Finance P RASAD C HALASANI Carnegie Mellon University
[email protected]
S OMESH J HA Carnegie Mellon University
[email protected]
THIS IS A DRAFT: PLEASE DO NOT DISTRIBUTE c Copyright; Steven E. Shreve, 1996 July 25, 1997
Contents 1 Introduction to Probability Theory
11
1.1
The Binomial Asset Pricing Model . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.2
Finite Probability Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
1.3
Lebesgue Measure and the Lebesgue Integral . . . . . . . . . . . . . . . . . . . .
22
1.4
General Probability Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
1.5
Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
1.5.1
Independence of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
1.5.2
Independence of algebras . . . . . . . . . . . . . . . . . . . . . . . . .
41
1.5.3
Independence of random variables . . . . . . . . . . . . . . . . . . . . . .
42
1.5.4
Correlation and independence . . . . . . . . . . . . . . . . . . . . . . . .
44
1.5.5
Independence and conditional expectation. . . . . . . . . . . . . . . . . .
45
1.5.6
Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
1.5.7
Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
2 Conditional Expectation
49
2.1
A Binomial Model for Stock Price Dynamics . . . . . . . . . . . . . . . . . . . .
49
2.2
Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
2.3
Conditional Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
2.3.1
An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
2.3.2
Definition of Conditional Expectation . . . . . . . . . . . . . . . . . . . .
53
2.3.3
Further discussion of Partial Averaging . . . . . . . . . . . . . . . . . . .
54
2.3.4
Properties of Conditional Expectation . . . . . . . . . . . . . . . . . . . .
55
2.3.5
Examples from the Binomial Model . . . . . . . . . . . . . . . . . . . . .
57
Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
2.4
1
2 3 Arbitrage Pricing
59
3.1
Binomial Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3.2
General onestep APT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.3
RiskNeutral Probability Measure . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.3.1
Portfolio Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
3.3.2
Selffinancing Value of a Portfolio Process . . . . . . . . . . . . . . . .
62
3.4
Simple European Derivative Securities . . . . . . . . . . . . . . . . . . . . . . . .
63
3.5
The Binomial Model is Complete . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
4 The Markov Property
67
4.1
Binomial Model Pricing and Hedging . . . . . . . . . . . . . . . . . . . . . . . .
67
4.2
Computational Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
4.3
Markov Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
4.3.1
Different ways to write the Markov property . . . . . . . . . . . . . . . .
70
4.4
Showing that a process is Markov . . . . . . . . . . . . . . . . . . . . . . . . . .
73
4.5
Application to Exotic Options . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
5 Stopping Times and American Options
77
5.1
American Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
5.2
Value of Portfolio Hedging an American Option . . . . . . . . . . . . . . . . . . .
79
5.3
Information up to a Stopping Time . . . . . . . . . . . . . . . . . . . . . . . . . .
81
6 Properties of American Derivative Securities
85
6.1
The properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
6.2
Proofs of the Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
6.3
Compound European Derivative Securities . . . . . . . . . . . . . . . . . . . . . .
88
6.4
Optimal Exercise of American Derivative Security . . . . . . . . . . . . . . . . . .
89
7 Jensen’s Inequality
91
7.1
Jensen’s Inequality for Conditional Expectations . . . . . . . . . . . . . . . . . . .
91
7.2
Optimal Exercise of an American Call . . . . . . . . . . . . . . . . . . . . . . . .
92
7.3
Stopped Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
8 Random Walks 8.1
First Passage Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97 97
3 8.2 8.3
is almost surely finite
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
The moment generating function for . . . . . . . . . . . . . . . . . . . . . . . .
99
8.4
Expectation of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
8.5
The Strong Markov Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
8.6
General First Passage Times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
8.7
Example: Perpetual American Put . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.8
Difference Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8.9
Distribution of First Passage Times . . . . . . . . . . . . . . . . . . . . . . . . . . 107
8.10 The Reflection Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9 Pricing in terms of Market Probabilities: The RadonNikodym Theorem.
111
9.1
RadonNikodym Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
9.2
RadonNikodym Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
9.3
The State Price Density Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
9.4
Stochastic Volatility Binomial Model . . . . . . . . . . . . . . . . . . . . . . . . . 116
9.5
Another Applicaton of the RadonNikodym Theorem . . . . . . . . . . . . . . . . 118
10 Capital Asset Pricing
119
10.1 An Optimization Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 11 General Random Variables
123
11.1 Law of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 11.2 Density of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 11.3 Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 11.4 Two random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 11.5 Marginal Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 11.6 Conditional Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 11.7 Conditional Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 11.8 Multivariate Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 11.9 Bivariate normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 11.10MGF of jointly normal random variables . . . . . . . . . . . . . . . . . . . . . . . 130 12 SemiContinuous Models
131
12.1 Discretetime Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
4 12.2 The Stock Price Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 12.3 Remainder of the Market . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.4 RiskNeutral Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.5 RiskNeutral Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 12.6 Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 12.7 Stalking the RiskNeutral Measure . . . . . . . . . . . . . . . . . . . . . . . . . . 135 12.8 Pricing a European Call . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 13 Brownian Motion
139
13.1 Symmetric Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 13.2 The Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 13.3 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 13.4 Brownian Motion as a Limit of Random Walks . . . . . . . . . . . . . . . . . . . 141 13.5 Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 13.6 Covariance of Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 13.7 FiniteDimensional Distributions of Brownian Motion . . . . . . . . . . . . . . . . 144 13.8 Filtration generated by a Brownian Motion . . . . . . . . . . . . . . . . . . . . . . 144 13.9 Martingale Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 13.10The Limit of a Binomial Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 13.11Starting at Points Other Than 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 13.12Markov Property for Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . 147 13.13Transition Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 13.14First Passage Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 14 The Itˆo Integral
153
14.1 Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 14.2 First Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 14.3 Quadratic Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 14.4 Quadratic Variation as Absolute Volatility . . . . . . . . . . . . . . . . . . . . . . 157 14.5 Construction of the Itˆo Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 14.6 Itˆo integral of an elementary integrand . . . . . . . . . . . . . . . . . . . . . . . . 158 14.7 Properties of the Itˆo integral of an elementary process . . . . . . . . . . . . . . . . 159 14.8 Itˆo integral of a general integrand . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
5 14.9 Properties of the (general) Itˆo integral . . . . . . . . . . . . . . . . . . . . . . . . 163 14.10Quadratic variation of an Itˆo integral . . . . . . . . . . . . . . . . . . . . . . . . . 165 15 Itˆo’s Formula
167
15.1 Itˆo’s formula for one Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . 167 15.2 Derivation of Itˆo’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 15.3 Geometric Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 15.4 Quadratic variation of geometric Brownian motion . . . . . . . . . . . . . . . . . 170 15.5 Volatility of Geometric Brownian motion . . . . . . . . . . . . . . . . . . . . . . 170 15.6 First derivation of the BlackScholes formula . . . . . . . . . . . . . . . . . . . . 170 15.7 Mean and variance of the CoxIngersollRoss process . . . . . . . . . . . . . . . . 172 15.8 Multidimensional Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . 173 15.9 Crossvariations of Brownian motions . . . . . . . . . . . . . . . . . . . . . . . . 174 15.10Multidimensional Itˆo formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 16 Markov processes and the Kolmogorov equations
177
16.1 Stochastic Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 16.2 Markov Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 16.3 Transition density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 16.4 The Kolmogorov Backward Equation . . . . . . . . . . . . . . . . . . . . . . . . 180 16.5 Connection between stochastic calculus and KBE . . . . . . . . . . . . . . . . . . 181 16.6 BlackScholes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 16.7 BlackScholes with pricedependent volatility . . . . . . . . . . . . . . . . . . . . 186 17 Girsanov’s theorem and the riskneutral measure
189
P . . . . . . . . . . . . . . . . . . . . . . . . . . 191 17.1 Conditional expectations under If 17.2 Riskneutral measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
18 Martingale Representation Theorem
197
18.1 Martingale Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 197 18.2 A hedging application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 18.3 18.4
ddimensional Girsanov Theorem . . . . . . . . . ddimensional Martingale Representation Theorem
. . . . . . . . . . . . . . . . . 199 . . . . . . . . . . . . . . . . . 200
18.5 Multidimensional market model . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
6 19 A twodimensional market model
203
19.1 Hedging when ,1 < < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 19.2 Hedging when = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
20 Pricing Exotic Options
209
20.1 Reflection principle for Brownian motion . . . . . . . . . . . . . . . . . . . . . . 209 20.2 Up and out European call. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 20.3 A practical issue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 21 Asian Options
219
21.1 FeynmanKac Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 21.2 Constructing the hedge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 21.3 Partial average payoff Asian option . . . . . . . . . . . . . . . . . . . . . . . . . . 221 22 Summary of Arbitrage Pricing Theory
223
22.1 Binomial model, Hedging Portfolio . . . . . . . . . . . . . . . . . . . . . . . . . 223 22.2 Setting up the continuous model . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 22.3 Riskneutral pricing and hedging . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 22.4 Implementation of riskneutral pricing and hedging . . . . . . . . . . . . . . . . . 229 23 Recognizing a Brownian Motion
233
23.1 Identifying volatility and correlation . . . . . . . . . . . . . . . . . . . . . . . . . 235 23.2 Reversing the process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 24 An outside barrier option
239
24.1 Computing the option value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 24.2 The PDE for the outside barrier option . . . . . . . . . . . . . . . . . . . . . . . . 243 24.3 The hedge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 25 American Options
247
25.1 Preview of perpetual American put . . . . . . . . . . . . . . . . . . . . . . . . . . 247 25.2 First passage times for Brownian motion: first method . . . . . . . . . . . . . . . . 247 25.3 Drift adjustment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 25.4 Driftadjusted Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 25.5 First passage times: Second method . . . . . . . . . . . . . . . . . . . . . . . . . 251
7 25.6 Perpetual American put . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 25.7 Value of the perpetual American put . . . . . . . . . . . . . . . . . . . . . . . . . 256 25.8 Hedging the put . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 25.9 Perpetual American contingent claim . . . . . . . . . . . . . . . . . . . . . . . . . 259 25.10Perpetual American call . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 25.11Put with expiration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 25.12American contingent claim with expiration . . . . . . . . . . . . . . . . . . . . . 261 26 Options on dividendpaying stocks
263
26.1 American option with convex payoff function . . . . . . . . . . . . . . . . . . . . 263 26.2 Dividend paying stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 26.3 Hedging at time t1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
27 Bonds, forward contracts and futures
267
27.1 Forward contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 27.2 Hedging a forward contract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 27.3 Future contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 27.4 Cash flow from a future contract . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.5 Forwardfuture spread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 27.6 Backwardation and contango . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 28 Termstructure models
275
28.1 Computing arbitragefree bond prices: first method . . . . . . . . . . . . . . . . . 276 28.2 Some interestrate dependent assets . . . . . . . . . . . . . . . . . . . . . . . . . 276 28.3 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 28.4 Forward rate agreement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 28.5 Recovering the interest r(t) from the forward rate . . . . . . . . . . . . . . . . . . 278 28.6 Computing arbitragefree bond prices: HeathJarrowMorton method . . . . . . . . 279 28.7 Checking for absence of arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . 280 28.8 Implementation of the HeathJarrowMorton model . . . . . . . . . . . . . . . . . 281 29 Gaussian processes
285
29.1 An example: Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 30 Hull and White model
293
8 30.1 Fiddling with the formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 30.2 Dynamics of the bond price . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 30.3 Calibration of the Hull & White model . . . . . . . . . . . . . . . . . . . . . . . . 297 30.4 Option on a bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 31 CoxIngersollRoss model
303
31.1 Equilibrium distribution of r(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 31.2 Kolmogorov forward equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 31.3 CoxIngersollRoss equilibrium density . . . . . . . . . . . . . . . . . . . . . . . 309 31.4 Bond prices in the CIR model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 31.5 Option on a bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 31.6 Deterministic time change of CIR model . . . . . . . . . . . . . . . . . . . . . . . 313 31.7 Calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
31.8 Tracking down '0(0) in the time change of the CIR model . . . . . . . . . . . . . 316 32 A twofactor model (Duffie & Kan)
319
32.1 Nonnegativity of Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 32.2 Zerocoupon bond prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 32.3 Calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 33 Change of num´eraire
325
33.1 Bond price as num´eraire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 33.2 Stock price as num´eraire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 33.3 Merton option pricing formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 34 BraceGatarekMusiela model
335
34.1 Review of HJM under riskneutral IP . . . . . . . . . . . . . . . . . . . . . . . . . 335 34.2 BraceGatarekMusiela model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 34.3 LIBOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 34.4 Forward LIBOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 34.5 The dynamics of L(t; ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 34.6 Implementation of BGM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 34.7 Bond prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 34.8 Forward LIBOR under more forward measure . . . . . . . . . . . . . . . . . . . . 343
9 34.9 Pricing an interest rate caplet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 34.10Pricing an interest rate cap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 34.11Calibration of BGM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 34.12Long rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 34.13Pricing a swap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
10
Chapter 1
Introduction to Probability Theory 1.1 The Binomial Asset Pricing Model The binomial asset pricing model provides a powerful tool to understand arbitrage pricing theory and probability theory. In this course, we shall use it for both these purposes. In the binomial asset pricing model, we model stock prices in discrete time, assuming that at each step, the stock price will change to one of two possible values. Let us begin with an initial positive stock price S0. There are two positive numbers, d and u, with
0 < d < u;
(1.1)
such that at the next period, the stock price will be either dS0 or uS0. Typically, we take d and u to satisfy 0 < d < 1 < u, so change of the stock price from S0 to dS0 represents a downward movement, and change of the stock price from S0 to uS0 represents an upward movement. It is common to also have d = u1 , and this will be the case in many of our examples. However, strictly speaking, for what we are about to do we need to assume only (1.1) and (1.2) below. Of course, stock price movements are much more complicated than indicated by the binomial asset pricing model. We consider this simple model for three reasons. First of all, within this model the concept of arbitrage pricing and its relation to riskneutral pricing is clearly illuminated. Secondly, the model is used in practice because with a sufficient number of steps, it provides a good, computationally tractable approximation to continuoustime models. Thirdly, within the binomial model we can develop the theory of conditional expectations and martingales which lies at the heart of continuoustime models. With this third motivation in mind, we develop notation for the binomial model which is a bit different from that normally found in practice. Let us imagine that we are tossing a coin, and when we get a “Head,” the stock price moves up, but when we get a “Tail,” the price moves down. We denote the price at time 1 by S1 (H ) = uS0 if the toss results in head (H), and by S1 (T ) = dS0 if it 11
12 S2 (HH) = 16
S (H) = 8 1 S2 (HT) = 4 S =4 0
S2 (TH) = 4 S1 (T) = 2
S2 (TT) = 1
Figure 1.1: Binomial tree of stock prices with S0
= 4, u = 1=d = 2.
results in tail (T). After the second toss, the price will be one of:
S2 (HH ) = uS1(H ) = u2 S0; S2 (HT ) = dS1(H ) = duS0; S2 (TH ) = uS1 (T ) = udS0; S2 (TT ) = dS1(T ) = d2S0 : After three tosses, there are eight possible coin sequences, although not all of them result in different stock prices at time 3. For the moment, let us assume that the third toss is the last one and denote by
= fHHH; HHT; HTH; HTT; THH; THT; TTH; TTT g the set of all possible outcomes of the three tosses. The set of all possible outcomes of a random experiment is called the sample space for the experiment, and the elements ! of are called sample points. In this case, each sample point ! is a sequence of length three. We denote the kth component of ! by !k . For example, when ! = HTH , we have !1 = H , !2 = T and !3 = H .
The stock price Sk at time k depends on the coin tosses. To emphasize this, we often write Sk (! ). Actually, this notation does not quite tell the whole story, for while S3 depends on all of ! , S2 depends on only the first two components of ! , S1 depends on only the first component of ! , and S0 does not depend on ! at all. Sometimes we will use notation such S2(!1 ; !2) just to record more explicitly how S2 depends on ! = (!1; !2; !3 ).
Example 1.1 Set S0 = 4, u = 2 and d = 21 . We have then the binomial “tree” of possible stock prices shown in Fig. 1.1. Each sample point ! = (!1; !2 ; !3) represents a path through the tree. Thus, we can think of the sample space as either the set of all possible outcomes from three coin tosses or as the set of all possible paths through the tree. To complete our binomial asset pricing model, we introduce a money market with interest rate r; $1 invested in the money market becomes $(1 + r) in the next period. We take r to be the interest
CHAPTER 1. Introduction to Probability Theory
13
rate for both borrowing and lending. (This is not as ridiculous as it first seems, because in a many applications of the model, an agent is either borrowing or lending (not both) and knows in advance which she will be doing; in such an application, she should take r to be the rate of interest for her activity.) We assume that
d < 1 + r < u:
(1.2)
The model would not make sense if we did not have this condition. For example, if 1 + r u, then the rate of return on the money market is always at least as great as and sometimes greater than the return on the stock, and no one would invest in the stock. The inequality d 1 + r cannot happen unless either r is negative (which never happens, except maybe once upon a time in Switzerland) or d 1. In the latter case, the stock does not really go “down” if we get a tail; it just goes up less than if we had gotten a head. One should borrow money at interest rate r and invest in the stock, since even in the worst case, the stock price rises at least as fast as the debt used to buy it. With the stock as the underlying asset, let us consider a European call option with strike price
K > 0 and expiration time 1. This option confers the right to buy the stock at time 1 for K dollars, and so is worth S1 , K at time 1 if S1 , K is positive and is otherwise worth zero. We denote by V1(!) = (S1(! ) , K )+ = maxfS1(! ) , K; 0g the value (payoff) of this option at expiration. Of course, V1(! ) actually depends only on !1 , and we can and do sometimes write V1 (!1) rather than V1(! ). Our first task is to compute the arbitrage price of this option at time zero. Suppose at time zero you sell the call for V0 dollars, where V0 is still to be determined. You now have an obligation to pay off (uS0 , K )+ if !1 = H and to pay off (dS0 , K )+ if !1 = T . At the time you sell the option, you don’t yet know which value !1 will take. You hedge your short position in the option by buying 0 shares of stock, where 0 is still to be determined. You can use the proceeds V0 of the sale of the option for this purpose, and then borrow if necessary at interest rate r to complete the purchase. If V0 is more than necessary to buy the 0 shares of stock, you invest the residual money at interest rate r. In either case, you will have V0 , 0S0 dollars invested in the money market, where this quantity might be negative. You will also own 0 shares of stock. If the stock goes up, the value of your portfolio (excluding the short position in the option) is
0 S1(H ) + (1 + r)(V0 , 0 S0 ); and you need to have V1(H ). Thus, you want to choose V0 and 0 so that
V1(H ) = 0S1 (H ) + (1 + r)(V0 , 0 S0):
(1.3)
If the stock goes down, the value of your portfolio is
0 S1 (T ) + (1 + r)(V0 , 0 S0); and you need to have V1(T ). Thus, you want to choose V0 and 0 to also have
V1(T ) = 0S1 (T ) + (1 + r)(V0 , 0S0 ):
(1.4)
14 These are two equations in two unknowns, and we solve them below Subtracting (1.4) from (1.3), we obtain
V1 (H ) , V1(T ) = 0 (S1(H ) , S1 (T ));
(1.5)
) , V1(T ) 0 = SV1((H H ) , S (T ) :
(1.6)
so that
1
1
This is a discretetime version of the famous “deltahedging” formula for derivative securities, according to which the number of shares of an underlying asset a hedge should hold is the derivative (in the sense of calculus) of the value of the derivative security with respect to the price of the underlying asset. This formula is so pervasive the when a practitioner says “delta”, she means the derivative (in the sense of calculus) just described. Note, however, that my definition of 0 is the number of shares of stock one holds at time zero, and (1.6) is a consequence of this definition, not the definition of 0 itself. Depending on how uncertainty enters the model, there can be cases in which the number of shares of stock a hedge should hold is not the (calculus) derivative of the derivative security with respect to the price of the underlying asset. To complete the solution of (1.3) and (1.4), we substitute (1.6) into either (1.3) or (1.4) and solve for V0 . After some simplification, this leads to the formula
1 + r , d 1 u , (1 + r ) V0 = 1 + r u , d V1(H ) + u , d V1 (T ) :
(1.7)
This is the arbitrage price for the European call option with payoff V1 at time 1. To simplify this formula, we define
p~ = 1 +u ,r ,d d ; q~ = u ,u(1, +d r) = 1 , p~;
(1.8)
V0 = 1 +1 r [~pV1(H ) + q~V1(T )]:
(1.9)
so that (1.7) becomes
Because we have taken d < u, both p~ and q~ are defined,i.e., the denominator in (1.8) is not zero. Because of (1.2), both p~ and q~ are in the interval (0; 1), and because they sum to 1, we can regard them as probabilities of H and T , respectively. They are the riskneutral probabilites. They appeared when we solved the two equations (1.3) and (1.4), and have nothing to do with the actual probabilities of getting H or T on the coin tosses. In fact, at this point, they are nothing more than a convenient tool for writing (1.7) as (1.9). We now consider a European call which pays off K dollars at time 2. At expiration, the payoff of this option is V2 = (S2 , K )+ , where V2 and S2 depend on !1 and !2 , the first and second coin tosses. We want to determine the arbitrage price for this option at time zero. Suppose an agent sells the option at time zero for V0 dollars, where V0 is still to be determined. She then buys 0 shares
CHAPTER 1. Introduction to Probability Theory
15
of stock, investing V0 , 0S0 dollars in the money market to finance this. At time 1, the agent has a portfolio (excluding the short position in the option) valued at
X1 = 0S1 + (1 + r)(V0 , 0 S0):
(1.10)
Although we do not indicate it in the notation, S1 and therefore X1 depend on !1 , the outcome of the first coin toss. Thus, there are really two equations implicit in (1.10):
X1(H ) = 0S1 (H ) + (1 + r)(V0 , 0 S0); X1(T ) = 0S1 (T ) + (1 + r)(V0 , 0S0): After the first coin toss, the agent has X1 dollars and can readjust her hedge. Suppose she decides to now hold 1 shares of stock, where 1 is allowed to depend on !1 because the agent knows what value !1 has taken. She invests the remainder of her wealth, X1 , 1S1 in the money market. In the next period, her wealth will be given by the righthand side of the following equation, and she wants it to be V2. Therefore, she wants to have
V2 = 1 S2 + (1 + r)(X1 , 1 S1):
(1.11)
Although we do not indicate it in the notation, S2 and V2 depend on !1 and !2 , the outcomes of the first two coin tosses. Considering all four possible outcomes, we can write (1.11) as four equations:
V2(HH ) V2(HT ) V2(TH ) V2(TT )
= = = =
1(H )S2(HH ) + (1 + r)(X1(H ) , 1 (H )S1(H )); 1(H )S2(HT ) + (1 + r)(X1(H ) , 1(H )S1(H )); 1(T )S2(TH ) + (1 + r)(X1(T ) , 1(T )S1(T )); 1(T )S2(TT ) + (1 + r)(X1(T ) , 1 (T )S1(T )):
We now have six equations, the two represented by (1.10) and the four represented by (1.11), in the six unknowns V0 , 0 , 1(H ), 1 (T ), X1 (H ), and X1 (T ). To solve these equations, and thereby determine the arbitrage price V0 at time zero of the option and the hedging portfolio 0 , 1(H ) and 1 (T ), we begin with the last two
V2(TH ) = 1(T )S2(TH ) + (1 + r)(X1(T ) , 1(T )S1(T )); V2(TT ) = 1(T )S2(TT ) + (1 + r)(X1(T ) , 1 (T )S1(T )): Subtracting one of these from the other and solving for mula”
1(T ), we obtain the “deltahedging for
) , V2(TT ) ; 1(T ) = SV2 ((TH TH ) , S (TT ) 2
2
(1.12)
and substituting this into either equation, we can solve for
X1 (T ) = 1 +1 r [~pV2(TH ) + q~V2 (TT )]:
(1.13)
16 Equation (1.13), gives the value the hedging portfolio should have at time 1 if the stock goes down between times 0 and 1. We define this quantity to be the arbitrage value of the option at time 1 if !1 = T , and we denote it by V1(T ). We have just shown that
V1(T ) = 1 +1 r [~pV2 (TH ) + q~V2(TT )]:
(1.14)
The hedger should choose her portfolio so that her wealth X1 (T ) if !1 = T agrees with V1(T ) defined by (1.14). This formula is analgous to formula (1.9), but postponed by one step. The first two equations implicit in (1.11) lead in a similar way to the formulas
1 (H ) = V2 (HH ) , V2(HT ) S2(HH ) , S2(HT ) and X1(H ) = V1(H ), where V1(H ) is the value of the option at time 1 if !1
(1.15)
= H , defined by
V1(H ) = 1 +1 r [~pV2(HH ) + q~V2(HT )]:
(1.16)
This is again analgous to formula (1.9), postponed by one step. Finally, we plug the values X1(H ) = V1(H ) and X1(T ) = V1(T ) into the two equations implicit in (1.10). The solution of these equations for 0 and V0 is the same as the solution of (1.3) and (1.4), and results again in (1.6) and (1.9). The pattern emerging here persists, regardless of the number of periods. If Vk denotes the value at time k of a derivative security, and this depends on the first k coin tosses !1 ; : : :; !k , then at time k , 1, after the first k , 1 tosses !1 ; : : :; !k,1 are known, the portfolio to hedge a short position should hold k,1 (!1 ; : : :; !k,1) shares of stock, where
H ) , Vk (!1; : : :; !k,1 ; T ) ; k,1 (!1; : : :; !k,1 ) = SVk ((!!1 ;; :: :: :;:; !!k,1 ;; H ) , S (! ; : : :; ! ; T ) k 1
k,1
k 1
k,1
(1.17)
and the value at time k , 1 of the derivative security, when the first k , 1 coin tosses result in the outcomes !1 ; : : :; !k,1 , is given by
Vk,1(!1; : : :; !k,1) = 1 +1 r [~pVk (!1; : : :; !k,1 ; H ) + q~Vk (!1; : : :; !k,1; T )]
(1.18)
1.2 Finite Probability Spaces Let be a set with finitely many elements. An example to keep in mind is
= fHHH; HHT; HTH; HTT; THH; THT; TTH; TTT g
(2.1)
of all possible outcomes of three coin tosses. Let F be the set of all subsets of . Some sets in F are ;, fHHH; HHT; HTH; HTT g, fTTT g, and itself. How many sets are there in F ?
CHAPTER 1. Introduction to Probability Theory
IP
Definition 1.1 A probability measure properties:
17
is a function mapping
F into [0; 1] with the following
IP ( ) = 1, (ii) If A1 ; A2; : : : is a sequence of disjoint sets in F , then (i)
IP
1 ! X 1 [
k=1
Ak =
k=1
IP (Ak ):
Probability measures have the following interpretation. Let A be a subset of F . Imagine that is the set of all possible outcomes of some random experiment. There is a certain probability, between 0 and 1, that when that experiment is performed, the outcome will lie in the set A. We think of IP (A) as this probability. Example 1.2 Suppose a coin has probability 13 for H and 23 for T . For the individual elements of
in (2.1), define
3
2
IP fHHH g = 13 ; 2 IP fHTH g = 13 23 ; 2 IP fTHH g = 13 13 ; 2 IP fTTH g = 13 23 ;
IP fHHT g = 31 23 ; 2 IP fHTT g = 13 23 ; 2 IP fTHT g = 13 23 ; 3 IP fTTT g = 23 :
For A 2 F , we define
IP (A) = For example,
X !2A
1 3
IP fHHH; HHT; HTH; HTT g = 3
IP f!g:
(2.2)
1 2 2 1 2 2 1 +2 + = ; 3
3
3
3
3
which is another way of saying that the probability of H on the first toss is 13 . As in the above example, it is generally the case that we specify a probability measure on only some of the subsets of and then use property (ii) of Definition 1.1 to determine IP (A) for the remaining sets A 2 F . In the above example, we specified the probability measure only for the sets containing a single element, and then used Definition 1.1(ii) in the form (2.2) (see Problem 1.4(ii)) to determine IP for all the other sets in F . Definition 1.2 Let be a nonempty set. A algebra is a collection following three properties: (i)
; 2 G,
G of subsets of with the
18 (ii) If A 2 G , then its complement Ac
2 G,
(iii) If A1; A2; A3; : : : is a sequence of sets in G , then [1 k=1 Ak is also in G . Here are some important algebras of subsets of the set in Example 1.2:
F0 = F1 = F2 =
(
)
;; ;
(
)
;; ; fHHH; HHT; HTH; HTT g; fTHH; THT; TTH; TTT g ;
(
;; ; fHHH; HHT g; fHTH; HTT g; fTHH; THT g; fTTH; TTT g; ) and all sets which can be built by taking unions of these
;
F3 = F = The set of all subsets of : To simplify notation a bit, let us define
AH = fHHH; HHT; HTH; HTT g = fH on the first tossg; AT = fTHH; THT; TTH; TTT g = fT on the first tossg; so that
F1 = f;; ; AH; AT g;
and let us define
AHH = fHHH; HHT g = fHH on the first two tossesg; AHT = fHTH; HTT g = fHT on the first two tossesg; ATH = fTHH; THT g = fTH on the first two tossesg; ATT = fTTH; TTT g = fTT on the first two tossesg; so that
F2 = f;; ; AHH; AHT ; ATH ; ATT ; AH ; AT ; AHH [ ATH ; AHH [ ATT ; AHT [ ATH ; AHT [ ATT ; AcHH ; AcHT ; AcTH ; AcTT g: We interpret algebras as a record of information. Suppose the coin is tossed three times, and you are not told the outcome, but you are told, for every set in F1 whether or not the outcome is in that set. For example, you would be told that the outcome is not in ; and is in . Moreover, you might be told that the outcome is not in AH but is in AT . In effect, you have been told that the first toss was a T , and nothing more. The algebra F1 is said to contain the “information of the first toss”, which is usually called the “information up to time 1”. Similarly, F2 contains the “information of
CHAPTER 1. Introduction to Probability Theory
19
the first two tosses,” which is the “information up to time 2.” The algebra F3 = F contains “full information” about the outcome of all three tosses. The socalled “trivial” algebra F0 contains no information. Knowing whether the outcome ! of the three tosses is in ; (it is not) and whether it is in (it is) tells you nothing about ! Definition 1.3 Let be a nonempty finite set. A filtration is a sequence of algebras F0 ; F1; F2; : : :; Fn such that each algebra in the sequence contains all the sets contained by the previous algebra. Definition 1.4 Let be a nonempty finite set and let random variable is a function mapping into IR.
F be the algebra of all subsets of .
A
Example 1.3 Let be given by (2.1) and consider the binomial asset pricing Example 1.1, where S0 = 4, u = 2 and d = 21 . Then S0, S1 , S2 and S3 are all random variables. For example, S2(HHT ) = u2 S0 = 16. The “random variable” S0 is really not random, since S0(!) = 4 for all ! 2 . Nonetheless, it is a function mapping into IR, and thus technically a random variable, albeit a degenerate one. A random variable maps into IR, and we can look at the preimage under the random variable of sets in IR. Consider, for example, the random variable S2 of Example 1.1. We have
S2(HHH ) = S2(HHT ) = 16; S2(HTH ) = S2 (HTT ) = S2 (THH ) = S2 (THT ) = 4; S2(TTH ) = S2(TTT ) = 1: Let us consider the interval [4; 27]. The preimage under S2 of this interval is defined to be
f! 2 ; S2(!) 2 [4; 27]g = f! 2 ; 4 S2 27g = AcTT : The complete list of subsets of we can get as preimages of sets in IR is:
;; ; AHH; AHT [ ATH ; ATT ; and sets which can be built by taking unions of these. This collection of sets is a algebra, called the algebra generated by the random variable S2, and is denoted by (S2). The information content of this algebra is exactly the information learned by observing S2 . More specifically, suppose the coin is tossed three times and you do not know the outcome ! , but someone is willing to tell you, for each set in (S2), whether ! is in the set. You might be told, for example, that ! is not in AHH , is in AHT [ ATH , and is not in ATT . Then you know that in the first two tosses, there was a head and a tail, and you know nothing more. This information is the same you would have gotten by being told that the value of S2(! ) is 4.
Note that F2 defined earlier contains all the sets which are in (S2), and even more. This means that the information in the first two tosses is greater than the information in S2 . In particular, if you see the first two tosses, you can distinguish AHT from ATH , but you cannot make this distinction from knowing the value of S2 alone.
20 Definition 1.5 Let be a nonemtpy finite set and let F be the algebra of all subsets of . Let X be a random variable on ( ; F ). The algebra (X ) generated by X is defined to be the collection of all sets of the form f! 2 ; X (! ) 2 Ag, where A is a subset of IR. Let G be a sub algebra of F . We say that X is G measurable if every set in (X ) is also in G . Note: We normally write simply fX
2 Ag rather than f! 2 ; X (!) 2 Ag. Definition 1.6 Let be a nonempty, finite set, let F be the algebra of all subsets of , let IP be a probabilty measure on ( ; F ), and let X be a random variable on . Given any set A IR, we
define the induced measure of A to be
LX (A) = IP fX 2 Ag: In other words, the induced measure of a set A tells us the probability that X takes a value in A. In the case of S2 above with the probability measure of Example 1.2, some sets in IR and their induced measures are:
LS (;) = IP (;) = 0; LS (IR) = IP ( ) = 1; LS [0; 1) = IP ( ) = 1; 2 LS [0; 3] = IP fS2 = 1g = IP (ATT ) = 23 : 2 In fact, the induced measure of S2 places a mass of size 13 = 19 at the number 16, a mass of size 2 2 4 4 2 2 2
2
9 at the number 4, and a mass of size 3
=
9 at the number 1. A common way to record this information is to give the cumulative distribution function FS2 (x) of S2 , defined by
8 > 0; > < 4 FS (x) = IP (S2 x) = > 98 ;; > : 19; 2
By the distribution of a random variable
if x < 1; if 1 x < 4; if 4 x < 16; if 16 x:
(2.3)
X , we mean any of the several ways of characterizing
LX . If X is discrete, as in the case of S2 above, we can either tell where the masses are and how
large they are, or tell what the cumulative distribution function is. (Later we will consider random variables X which have densities, in which case the induced measure of a set A IR is the integral of the density over the set A.) Important Note. In order to work through the concept of a riskneutral measure, we set up the definitions to make a clear distinction between random variables and their distributions. A random variable is a mapping from to IR, nothing more. It has an existence quite apart from discussion of probabilities. For example, in the discussion above, S2 (TTH ) = S2(TTT ) = 1, regardless of whether the probability for H is 13 or 12 .
CHAPTER 1. Introduction to Probability Theory
21
The distribution of a random variable is a measure LX on IR, i.e., a way of assigning probabilities to sets in IR. It depends on the random variable X and the probability measure IP we use in . If we set the probability of H to be 13 , then LS2 assigns mass 19 to the number 16. If we set the probability of H to be 12 , then LS2 assigns mass 14 to the number 16. The distribution of S2 has changed, but the random variable has not. It is still defined by
S2(HHH ) = S2(HHT ) = 16; S2(HTH ) = S2 (HTT ) = S2 (THH ) = S2 (THT ) = 4; S2(TTH ) = S2(TTT ) = 1: Thus, a random variable can have more than one distribution (a “market” or “objective” distribution, and a “riskneutral” distribution). In a similar vein, two different random variables can have the same distribution. Suppose in the binomial model of Example 1.1, the probability of H and the probability of T is 12 . Consider a European call with strike price 14 expiring at time 2. The payoff of the call at time 2 is the random variable (S2 , 14)+, which takes the value 2 if ! = HHH or ! = HHT , and takes the value 0 in every other case. The probability the payoff is 2 is 14 , and the probability it is zero is 34 . Consider also a European put with strike price 3 expiring at time 2. The payoff of the put at time 2 is (3 , S2)+ , which takes the value 2 if ! = TTH or ! = TTT . Like the payoff of the call, the payoff of the put is 2 with probability 14 and 0 with probability 34 . The payoffs of the call and the put are different random variables having the same distribution. Definition 1.7 Let be a nonempty, finite set, let F be the algebra of all subsets of , let IP be a probabilty measure on ( ; F ), and let X be a random variable on . The expected value of X is defined to be
IEX =
X
! 2
X (!)IP f!g:
(2.4)
Notice that the expected value in (2.4) is defined to be a sum over the sample space . Since is a finite set, X can take only finitely many values, which we label x1; : : :; xn . We can partition into the subsets fX1 = x1 g; : : :; fXn = xn g, and then rewrite (2.4) as
IEX = = = = =
X
!2
n X
X (!)IP f! g
X
X (!)IP f! g
k=1 !2fXk =xk g n X X xk IP f!g k=1 !2fXk =xk g n X xk IP fXk = xk g k=1 n X xk LX fxk g: k=1
22 Thus, although the expected value is defined as a sum over the sample space , we can also write it as a sum over IR.
To make the above set of equations absolutely clear, we consider S2 with the distribution given by (2.3). The definition of IES2 is
IES2 = S2(HHH )IP fHHH g + S2(HHT )IP fHHT g +S2 (HTH )IP fHTH g + S2(HTT )IP fHTT g +S2 (THH )IP fTHH g + S2(THT )IP fTHT g +S2 (TTH )IP fTTH g + S2 (TTT )IP fTTT g = 16 IP (AHH ) + 4 IP (AHT [ ATH ) + 1 IP (ATT ) = 16 IP fS2 = 16g + 4 IP fS2 = 4g + 1 IP fS2 = 1g = 16 LS f16g + 4 LS f4g + 1 LS f1g = 16 19 + 4 49 + 4 49 = 48 : 9 2
2
2
Definition 1.8 Let be a nonempty, finite set, let F be the algebra of all subsets of , let IP be a probabilty measure on ( ; F ), and let X be a random variable on . The variance of X is defined to be the expected value of (X , IEX )2, i.e.,
Var(X ) =
X
(X (! ) , IEX )2IP f! g:
!2
(2.5)
One again, we can rewrite (2.5) as a sum over IR rather than over . Indeed, if X takes the values
x1 ; : : :; xn, then
Var(X ) =
n X k=1
(xk , IEX )2IP fX = xk g =
n X
(xk , IEX )2LX (xk ):
k=1
1.3 Lebesgue Measure and the Lebesgue Integral In this section, we consider the set of real numbers IR, which is uncountably infinite. We define the Lebesgue measure of intervals in IR to be their length. This definition and the properties of measure determine the Lebesgue measure of many, but not all, subsets of IR. The collection of subsets of IR we consider, and for which Lebesgue measure is defined, is the collection of Borel sets defined below. We use Lebesgue measure to construct the Lebesgue integral, a generalization of the Riemann integral. We need this integral because, unlike the Riemann integral, it can be defined on abstract spaces, such as the space of infinite sequences of coin tosses or the space of paths of Brownian motion. This section concerns the Lebesgue integral on the space IR only; the generalization to other spaces will be given later.
CHAPTER 1. Introduction to Probability Theory
23
Definition 1.9 The Borel algebra, denoted B(IR), is the smallest algebra containing all open intervals in IR. The sets in B(IR) are called Borel sets. Every set which can be written down and just about every set imaginable is in B(IR). The following discussion of this fact uses the algebra properties developed in Problem 1.3. By definition, every open interval (a; b) is in B(IR), where a and b are real numbers. Since B(IR) is a algebra, every union of open intervals is also in B(IR). For example, for every real number a, the open halfline
1 [
(a; 1) =
n=1
is a Borel set, as is
(a; a + n)
1 [
(,1; a) =
(a , n; a):
n=1
For real numbers a and b, the union
(,1; a) [ (b; 1) is Borel. Since B(IR) is a algebra, every complement of a Borel set is Borel, so B(IR) contains
c
[a; b] = (,1; a) [ (b; 1) : This shows that every closed interval is Borel. In addition, the closed halflines
1 [
[a; 1) = and
[a; a + n]
n=1
(,1; a] =
1 [
[a , n; a]
n=1
are Borel. Halfopen and halfclosed intervals are also Borel, since they can be written as intersections of open halflines and closed halflines. For example,
(a; b] = (,1; b] \ (a; 1): Every set which contains only one real number is Borel. Indeed, if a is a real number, then
fag =
1 \
n=1
1 1 a, ;a+ : n
n
This means that every set containing finitely many real numbers is Borel; if A then
A=
[n
k=1
fak g:
= fa1; a2; : : :; an g,
24 In fact, every set containing countably infinitely many numbers is Borel; if A = fa1; a2; : : : g, then
A=
[n
k=1
fak g:
This means that the set of rational numbers is Borel, as is its complement, the set of irrational numbers. There are, however, sets which are not Borel. We have just seen that any nonBorel set must have uncountably many points. Example 1.4 (The Cantor set.) This example gives a hint of how complicated a Borel set can be. We use it later when we discuss the sample space for an infinite sequence of coin tosses. Consider the unit interval [0; 1], and remove the middle half, i.e., remove the open interval
A1 = 14 ; 43 :
1 3 C1 = 0; 4 [ 4 ; 1 has two pieces. From each of these pieces, remove the middle half, i.e., remove the open set 1 3 [ 13 15 The remaining set
A2 = 16 ; 16
The remaining set
;
16 16
:
1 [ 3 1 [ 3 13 [ 15 C2 = 0; 16 16 ; 4 4 ; 16 16 ; 1 :
has four pieces. Continue this process, so at stage k, the set Ck has 2k pieces, and each piece has length 41k . The Cantor set
C =
1 \
k=1
Ck
is defined to be the set of points not removed at any stage of this nonterminating process. Note that the length of A1 , the first set removed, is 12 . The “length” of A2 , the second set removed, 1 = 1 , and in general, the length of the is 18 + 18 = 41 . The “length” of the next set removed is 4 32 8 , k kth set removed is 2 . Thus, the total length removed is
1 1 X 2k = 1;
k=1
and so the Cantor set, the set of points not removed, has zero “length.” Despite the fact that the Cantor set has no “length,” there are lots of points in this set. In particular, none of the endpoints of the pieces of the sets C1; C2; : : : is ever removed. Thus, the points
1 ; 3 ; 13 ; 15 ; 1 ; : : : 0; 14 ; 43 ; 1; 16 16 16 16 64
are all in C . This is a countably infinite set of points. We shall see eventually that the Cantor set has uncountably many points.
CHAPTER 1. Introduction to Probability Theory
25
Definition 1.10 Let B(IR) be the algebra of Borel subsets of IR. A measure on (IR; B(IR)) is a function mapping B into [0; 1] with the following properties:
(;) = 0, (ii) If A1 ; A2; : : : is a sequence of disjoint sets in B(IR), then (i)
! X 1
1 [ k=1
Ak =
k=1
(Ak ):
Lebesgue measure is defined to be the measure on (IR; B(IR)) which assigns the measure of each interval to be its length. Following Williams’s book, we denote Lebesgue measure by 0 . A measure has all the properties of a probability measure given in Problem 1.4, except that the total measure of the space is not necessarily 1 (in fact, 0 (IR) = 1), one no longer has the equation
(Ac ) = 1 , (A) in Problem 1.4(iii), and property (v) in Problem 1.4 needs to be modified to say: (v) If A1 ; A2; : : : is a sequence of sets in B(IR) with A1
1 \ k=1
!
A2 and (A1) < 1, then
Ak = nlim !1 (An ):
To see that the additional requirment (A1 ) < 1 is needed in (v), consider
A1 = [1; 1); A2 = [2; 1); A3 = [3; 1); : : :: Then \1 k=1 Ak
= ;, so 0 (\1 k=1 Ak ) = 0, but limn!1 0 (An ) = 1.
We specify that the Lebesgue measure of each interval is its length, and that determines the Lebesgue measure of all other Borel sets. For example, the Lebesgue measure of the Cantor set in Example 1.4 must be zero, because of the “length” computation given at the end of that example. The Lebesgue measure of a set containing only one point must be zero. In fact, since
fag a , n1 ; a + n1
for every positive integer n, we must have
2 1 1 0 0 fag 0 a , ; a + = :
Letting n ! 1, we obtain
n
0 fag = 0:
n
n
26 The Lebesgue measure of a set containing countably many points must also be zero. Indeed, if A = fa1; a2; : : : g, then
0 (A) =
1 X
k=1
0 fak g =
1 X
k=1
0 = 0:
The Lebesgue measure of a set containing uncountably many points can be either zero, positive and finite, or infinite. We may not compute the Lebesgue measure of an uncountable set by adding up the Lebesgue measure of its individual members, because there is no way to add up uncountably many numbers. The integral was invented to get around this problem. In order to think about Lebesgue integrals, we must first consider the functions to be integrated. Definition 1.11 Let
f be a function from IR to IR.
We say that
f
is Borelmeasurable if the set
fx 2 IR; f (x) 2 Ag is in B(IR) whenever A 2 B(IR). In the language of Section 2, we want the algebra generated by f to be contained in B(IR).
Definition 3.4 is purely technical and has nothing to do with keeping track of information. It is difficult to conceive of a function which is not Borelmeasurable, and we shall pretend such functions don’t exist. Hencefore, “function mapping IR to IR” will mean “Borelmeasurable function mapping IR to IR” and “subset of IR” will mean “Borel subset of IR”. Definition 1.12 An indicator function g from IR to IR is a function which takes only the values 0 and 1. We call
A = fx 2 IR; g (x) = 1g the set indicated by g . We define the Lebesgue integral of g to be
Z
IR
g d0 = 0 (A):
A simple function h from IR to IR is a linear combination of indicators, i.e., a function of the form
h(x) = where each gk is of the form
gk (x) =
(
n X
k=1
1; 0;
ck gk (x); if x 2 Ak ; if x 2 = Ak ;
and each ck is a real number. We define the Lebesgue integral of h to be
Z
R
h d0 =
Z n X
k=1
ck
Let f be a nonnegative function defined on define the Lebesgue integral of f to be
Z
IR
f d0
sup =
Z
IR
IR
gk d0 =
n X
k=1
ck 0(Ak ):
IR, possibly taking the value 1 at some points.
h d0; h is simple and h(x) f (x) for every x 2 IR :
We
CHAPTER 1. Introduction to Probability Theory
27
It is possible that this integral is infinite. If it is finite, we say that f is integrable.
Finally, let f be a function defined on IR, possibly taking the value 1 at some points and the value ,1 at other points. We define the positive and negative parts of f to be
f + (x) = maxff (x); 0g; f ,(x) = maxf,f (x); 0g; respectively, and we define the Lebesgue integral of f to be
Z
IR
f d0
=
Z
IR
0,,
f + d
Z
IR
f , d0;
R
R
provided the righthand side is not of the form 1 , 1. If both IR f + d0 and IR f , d0 are finite R (or equivalently, IR jf j d0 < 1, since jf j = f + + f , ), we say that f is integrable.
Let f be a function defined on IR, possibly taking the value 1 at some points and the value ,1 at other points. Let A be a subset of IR. We define
Z
A where
lIA
f d0 =
(x) =
(
is the indicator function of A.
Z
IR
1; 0;
lIAf d0 ; if x 2 A; if x 2 = A;
The Lebesgue integralRjust defined is related to the Riemann integral in Rone very important way: if the Riemann integral ab f (x)dx is defined, then the Lebesgue integral [a;b] f d0 agrees with the Riemann integral. The Lebesgue integral has two important advantages over the Riemann integral. The first is that the Lebesgue integral is defined for more functions, as we show in the following examples.
Example 1.5 Let Q be the set of rational numbers in [0; 1], and consider f = lIQ . Being a countable set, Q has Lebesgue measure zero, and so the Lebesgue integral of f over [0; 1] is
Z
R
[0;1]
f d0 = 0:
To compute the Riemann integral 01 f (x)dx, we choose partition points 0 = x0 < x1 < < xn = 1 and divide the interval [0; 1] into subintervals [x0; x1]; [x1; x2]; : : :; [xn,1 ; xn]. In each subinterval [xk,1 ; xk ] there is a rational point qk , where f (qk ) = 1, and there is also an irrational point rk , where f (rk ) = 0. We approximate the Riemann integral from above by the upper sum
n X
k=1
f (qk )(xk , xk,1 ) =
n X
k=1
1 (xk , xk,1 ) = 1;
and we also approximate it from below by the lower sum
n X k=1
f (rk )(xk , xk,1 ) =
n X k=1
0 (xk , xk,1 ) = 0:
28 No matter how fine we take the partition of [0; 1], the upper sum is always 1 and the lower sum is always 0. Since these two do not converge to a common value as the partition becomes finer, the Riemann integral is not defined. Example 1.6 Consider the function
f (x) =
(
1;
if x = 0; if x 6= 0:
0;
This is not a simple function because simple function cannot take the value function which lies between 0 and f is of the form
h(x) = for some y
(
y; 0;
1.
Every simple
if x = 0; if x 6= 0;
2 [0; 1), and thus has Lebesgue integral Z IR
It follows that
Z
IR
f d0 = sup
h d0 = y0 f0g = 0:
Z IR
h d0; h is simple and h(x) f (x) for every x 2 IR = 0:
R
1 f (x) dx, which for this function f is the same as the Now consider the Riemann integral ,1 R Riemann integral ,11 f (x) dx. When we partition [,1; 1] into subintervals, one of these will contain R the point 0, and when we compute the upper approximating sum for ,11 f (x) dx, this point will contribute 1 times the length of the subinterval containing it. Thus the upper approximating sum is 1. On the other hand, the lower approximating sum is 0, and again the Riemann integral does not exist. The Lebesgue integral has all linearity and comparison properties one would expect of an integral. In particular, for any two functions f and g and any real constant c,
Z
IR
Z
(f + g ) d0 =
Z
IR
cf d0 = c
IR Z
and whenever f (x) g (x) for all x 2 IR, we have
Z
IR
f d0
Finally, if A and B are disjoint sets, then
Z
A[B
f d0 =
Z A
Z
f d0 +
IR
Z IR
f d0;
IR
g d0 ;
gd d0:
f d0 +
Z B
f d0:
CHAPTER 1. Introduction to Probability Theory
29
There are three convergence theorems satisfied by the Lebesgue integral. In each of these the situation is that there is a sequence of functions fn ; n = 1; 2; : : : converging pointwise to a limiting function f . Pointwise convergence just means that
nlim !1 fn (x) = f (x) for every x 2 IR: There are no such theorems for the Riemann integral, because the Riemann integral of the limiting function f is too often not defined. Before we state the theorems, we given two examples of pointwise convergence which arise in probability theory. Example 1.7 Consider a sequence of normal densities, each with variance mean n: 1 , (x,n)2
fn (x) = p e 2
2
1 and the nth having
:
These converge pointwise to the function
f (x) = 0 for every x 2 IR: R R R We have IR fn d0 = 1 for every n, so lim n!1 IR fn d0 = 1, but IR f d0 = 0.
Example 1.8 Consider a sequence of normal densities, each with mean 0 and the nth having variance n1 : r 2 fn (x) = n e, x2n :
2
These converge pointwise to the function
f (x) =
(
1; 0;
R
if x = 0; if x 6= 0:
R
R
We have again IR fn d0 = 1 for every n, so limn!1 IR fn d0 = 1, but IR f d0 = 0. The function f is not the Dirac delta; the Lebesgue integral of this function was already seen in Example 1.6 to be zero. Theorem 3.1 (Fatou’s Lemma) Let fn ; n verging pointwise to a function f . Then
Z
R
IR
= 1; 2; : : : be a sequence of nonnegative functions con
f d0 lim inf n!1
Z IR
fn d0 :
If limn!1 IR fn d0 is defined, then Fatou’s Lemma has the simpler conclusion
Z
IR
f d0 nlim !1
Z
IR
fn d0 :
This is the case in Examples 1.7 and 1.8, where
Z
nlim !1 IR fn d0 = 1;
30
R
while IR f d0 = 0. We could modify either Example 1.7 or 1.8 Rby setting gn = fn if n is even, R but gn = 2Rfn if n is odd. Now IR gn d0 = 1 if n is even, but IR gn d0 = 2 if n is odd. The sequence f IR gn d0 g1 n=1 has two cluster points, 1 and 2. R By definition, the smaller one, 1, is R lim inf n!1 IR gn d0 and the larger one, 2, is lim sup n!1 IR gn d0 . Fatou’s Lemma guarantees that even the smaller cluster point will be greater than or equal to the integral of the limiting function. The key assumption in Fatou’s Lemma is that all the functions take only nonnegative values. Fatou’s Lemma does not assume much but it is is not very satisfying because it does not conclude that
Z
IR
f d0 = nlim !1
Z
IR
fn d0 :
There are two sets of assumptions which permit this stronger conclusion. Theorem 3.2 (Monotone Convergence Theorem) Let fn ; n converging pointwise to a function f . Assume that
= 1; 2; : : : be a sequence of functions
0 f1 (x) f2 (x) f3 (x) for every x 2 IR:
Z
Then
IR
where both sides are allowed to be 1.
f d0 = nlim !1
Z IR
fn d0 ;
Theorem 3.3 (Dominated Convergence Theorem) Let fn ; n = 1; 2; : : : be a sequence of functions, which may take either positive or negative values, converging pointwise to a function f . Assume R that there is a nonnegative integrable function g (i.e., IR g d0 < 1) such that
jfn(x)j g(x) for every x 2 IR for every n: Z Z
Then
IR
f d0 = nlim !1
IR
fn d0 ;
and both sides will be finite.
1.4 General Probability Spaces Definition 1.13 A probability space ( ; F ; IP ) consists of three objects: (i)
, a nonempty set, called the sample space, which contains all possible outcomes of some random experiment;
(ii) (iii)
F , a algebra of subsets of ;
IP , a probability measure on ( ; F ), i.e., a function which assigns to each set A 2 F a number IP (A) 2 [0; 1], which represents the probability that the outcome of the random experiment lies in the set A.
CHAPTER 1. Introduction to Probability Theory
31
Remark 1.1 We recall from Homework Problem 1.4 that a probability measure IP has the following properties: (a)
IP (;) = 0.
(b) (Countable additivity) If A1 ; A2; : : : is a sequence of disjoint sets in F , then
IP
1 ! X 1 [
k=1
Ak =
k=1
IP (Ak ):
(c) (Finite additivity) If n is a positive integer and A1 ; : : :; An are disjoint sets in F , then
IP (A1 [ [ An ) = IP (A1 ) + + IP (An ):
(d) If A and B are sets in F and A B , then
IP (B ) = IP (A) + IP (B n A):
In particular,
IP (B ) IP (A):
(d) (Continuity from below.) If A1 ; A2; : : : is a sequence of sets in F with A1
IP
1 ! [
k=1
Ak = nlim !1 IP (An ):
(d) (Continuity from above.) If A1 ; A2; : : : is a sequence of sets in F with A1
IP
1 ! \
k=1
A2 , then A2 , then
Ak = nlim !1 IP (An ):
We have already seen some examples of finite probability spaces. We repeat these and give some examples of infinite probability spaces as well. Example 1.9 Finite coin toss space. Toss a coin n times, so that is the set of all sequences of H and T which have n components. We will use this space quite a bit, and so give it a name: n . Let F be the collection of all subsets of n . Suppose the probability of H on each toss is p, a number between zero and one. Then the probability of T is q = 1 , p. For each ! = (!1 ; !2; : : :; !n ) in n , we define
IP f!g = pNumber of H in ! qNumber of T in ! :
For each A 2 F , we define
IP (A) =
X !2A
IP f!g:
(4.1)
We can define IP (A) this way because A has only finitely many elements, and so only finitely many terms appear in the sum on the righthand side of (4.1).
32 Example 1.10 Infinite coin toss space. Toss a coin repeatedly without stopping, so that is the set of all nonterminating sequences of H and T . We call this space 1 . This is an uncountably infinite space, and we need to exercise some care in the construction of the algebra we will use here.
For each positive integer n, we define Fn to be the algebra determined by the first n tosses. For example, F2 contains four basic sets,
AHH = f! = (!1; !2 ; !3; : : : ); !1 = H; !2 = H g = The set of all sequences which begin with HH; AHT = f! = (!1; !2 ; !3; : : : ); !1 = H; !2 = T g = The set of all sequences which begin with HT; ATH = f! = (!1; !2 ; !3; : : : ); !1 = T; !2 = H g = The set of all sequences which begin with TH; ATT = f! = (!1; !2 ; !3; : : : ); !1 = T; !2 = T g = The set of all sequences which begin with TT: Because sets.
F2 is a algebra, we must also put into it the sets ;, , and all unions of the four basic
In the algebra F , we put every set in every algebra Fn , where n ranges over the positive integers. We also put in every other set which is required to make F be a algebra. For example, the set containing the single sequence
fHHHHH g = fH on every tossg is not in any of the Fn algebras, because it depends on all the components of the sequence and not just the first n components. However, for each positive integer n, the set
fH on the first n tossesg is in Fn and hence in F . Therefore,
fH on every tossg = is also in F .
1 \
fH on the first n tossesg
n=1
We next construct the probability measure IP on ( 1 ; F ) which corresponds to probability p 2 [0; 1] for H and probability q = 1 , p for T . Let A 2 F be given. If there is a positive integer n such that A 2 Fn , then the description of A depends on only the first n tosses, and it is clear how to define IP (A). For example, suppose A = AHH [ ATH , where these sets were defined earlier. Then A is in F2. We set IP (AHH ) = p2 and IP (ATH ) = qp, and then we have
IP (A) = IP (AHH [ ATH ) = p2 + qp = (p + q )p = p: In other words, the probability of a H on the second toss is p.
CHAPTER 1. Introduction to Probability Theory
33
Let us now consider a set A 2 F for which there is no positive integer n such that A 2 F . Such is the case for the set fH on every tossg. To determine the probability of these sets, we write them in terms of sets which are in Fn for positive integers n, and then use the properties of probability measures listed in Remark 1.1. For example,
fH on the first tossg fH on the first two tossesg fH on the first three tossesg ; and
1 \ n=1
fH on the first n tossesg = fH on every tossg:
According to Remark 1.1(d) (continuity from above),
n IP fH on every tossg = nlim !1 IP fH on the first n tossesg = nlim !1 p : If p = 1, then IP fH on every tossg = 1; otherwise, IP fH on every tossg = 0.
A similar argument shows that if 0 < p < 1 so that 0 < q < 1, then every set in 1 which contains only one element (nonterminating sequence of H and T ) has probability zero, and hence very set which contains countably many elements also has probabiliy zero. We are in a case very similar to Lebesgue measure: every point has measure zero, but sets can have positive measure. Of course, the only sets which can have positive probabilty in 1 are those which contain uncountably many elements. In the infinite coin toss space, we define a sequence of random variables Y1 ; Y2; : : : by
Yk
(! ) =
(
1 0
if !k if !k
= H; = T;
and we also define the random variable
X (! ) =
n Y (! ) X k k=1
2k
:
Since each Yk is either zero or one, X takes values in the interval [0; 1]. Indeed, X (TTTT ) = 0, X (HHHH ) = 1 and the other values of X lie in between. We define a “dyadic rational number” to be a number of the form 2mk , where k and m are integers. For example, 34 is a dyadic rational. Every dyadic rational in (0,1) corresponds to two sequences ! 2 1 . For example,
X (HHTTTTT ) = X (HTHHHHH ) = 43 : The numbers in (0,1) which are not dyadic rationals correspond to a single ! have a unique binary expansion.
2 1; these numbers
34 Whenever we place a probability measure IP on ( ; F ), we have a corresponding induced measure LX on [0; 1]. For example, if we set p = q = 12 in the construction of this example, then we have
1 LX 0; 2 = IP fFirst toss is T g = 21 ; 1 LX 2 ; 1 = IP fFirst toss is H g = 21 ; 1 LX 0; 4 = IP fFirst two tosses are TT g = 14 ; 1 1 LX 4 ; 2 = IP fFirst two tosses are TH g = 14 ; 1 3 LX 2 ; 4 = IP fFirst two tosses are HT g = 14 ; 3 LX 4 ; 1 = IP fFirst two tosses are HH g = 14 :
Continuing this process, we can verify that for any positive integers k and m satisfying
0 m2,k 1 < 2mk 1; we have
m , 1 m 1 LX 2k ; 2k = 2k :
In other words, the LX measure of all intervals in [0; 1] whose endpoints are dyadic rationals is the same as the Lebesgue measure of these intervals. The only way this can be is for LX to be Lebesgue measure.
It is interesing to consider what LX would look like if we take a value of p other than 12 when we construct the probability measure IP on .
We conclude this example with another look at the Cantor set of Example 3.2. Let pairs be the subset of in which every evennumbered toss is the same as the oddnumbered toss immediately preceding it. For example, HHTTTTHH is the beginning of a sequence in pairs , but HT is not. Consider now the set of real numbers
C 0 = fX (!); ! 2 pairs g: The numbers between ( 41 ; 12 ) can be written as X (! ), but the sequence ! must begin with either TH or HT . Therefore, none of these numbers is in C 0. Similarly, the numbers between ( 161 ; 163 ) can be written as X (! ), but the sequence ! must begin with TTTH or TTHT , so none of these numbers is in C 0. Continuing this process, we see that C 0 will not contain any of the numbers which were removed in the construction of the Cantor set C in Example 3.2. In other words, C 0 C . With a bit more work, one can convince onself that in fact C 0 = C , i.e., by requiring consecutive coin tosses to be paired, we are removing exactly those points in [0; 1] which were removed in the Cantor set construction of Example 3.2.
CHAPTER 1. Introduction to Probability Theory
35
In addition to tossing a coin, another common random experiment is to pick a number, perhaps using a random number generator. Here are some probability spaces which correspond to different ways of picking a number at random. Example 1.11 Suppose we choose a number from IR in such a way that we are sure to get either 1, 4 or 16. Furthermore, we construct the experiment so that the probability of getting 1 is 49 , the probability of getting 4 is 49 and the probability of getting 16 is 19 . We describe this random experiment by taking
to be IR, F to be B(IR), and setting up the probability measure so that
IP f1g = 94 ; IP f4g = 49 ; IP f16g = 19 : This determines IP (A) for every set A 2 B(IR). For example, the probability of the interval (0; 5] is 89 , because this interval contains the numbers 1 and 4, but not the number 16.
The probability measure described in this example is LS2 , the measure induced by the stock price S2 , when the initial stock price S0 = 4 and the probability of H is 13 . This distribution was discussed immediately following Definition 2.8. Example 1.12 Uniform distribution on [0; 1]. Let = [0; 1] and let F = B([0; 1]), the collection of all Borel subsets containined in [0; 1]. For each Borel set A [0; 1], we define IP (A) = 0 (A) to be the Lebesgue measure of the set. Because 0[0; 1] = 1, this gives us a probability measure.
This probability space corresponds to the random experiment of choosing a number from [0; 1] so that every number is “equally likely” to be chosen. Since there are infinitely mean numbers in [0; 1], this requires that every number have probabilty zero of being chosen. Nonetheless, we can speak of the probability that the number chosen lies in a particular set, and if the set has uncountably many points, then this probability can be positive. I know of no way to design a physical experiment which corresponds to choosing a number at random from [0; 1] so that each number is equally likely to be chosen, just as I know of no way to toss a coin infinitely many times. Nonetheless, both Examples 1.10 and 1.12 provide probability spaces which are often useful approximations to reality. Example 1.13 Standard normal distribution. Define the standard normal density
x2
, '(x) = p1 e 2 : 2 Let = IR, F
= B(IR) and for every Borel set A IR, define
IP (A) =
Z
A
' d0:
(4.2)
36 If A in (4.2) is an interval [a; b], then we can write (4.2) as the less mysterious Riemann integral:
IP [a; b] =
Z b 1 , x2 p e 2 dx: 2 a
This corresponds to choosing a point at random on the real line, and every single point has probability zero of being chosen, but if a set A is given, then the probability the point is in that set is given by (4.2). The construction of the integral in a general probability space follows the same steps as the construction of Lebesgue integral. We repeat this construction below. Definition 1.14 Let ( ; F ; IP ) be a probability space, and let X be a random variable on this space, i.e., a mapping from to IR, possibly also taking the values 1.
If X is an indicator, i.e, X (! ) = lIA (!) = for some set A 2 F , we define
Z
If X is a simple function, i.e,
(
if ! if !
1 0
2 A; 2 Ac ;
X dIP = IP (A):
X (! ) =
n X k=1
ck lIAk (!);
where each ck is a real number and each Ak is a set in F , we define
Z
X dIP =
Z n X
k=1
ck
lIAk dIP =
n X
k=1
ck IP (Ak ):
If X is nonnegative but otherwise general, we define Z X dIP
=
sup
Z
Y dIP ; Y
is simple and Y (! ) X (! ) for every !
2 :
In fact, we can always construct a sequence of simple functions Yn ; n = 1; 2; : : : such that
0 Y1 (! ) Y2 (! ) Y3 (! ) : : : for every ! 2 ; and Y (! ) = limn!1 Yn (! ) for every !
Z
2 . With this sequence, we can define Z
X dIP = nlim !1
Yn dIP:
CHAPTER 1. Introduction to Probability Theory
If X is integrable, i.e,
Z
Z
X + dIP < 1;
where
37
X , dIP < 1;
X +(!) = maxfX (!); 0g; X ,(!) = maxf,X (!); 0g;
Z
then we define
X dIP
=
Z
Z
, , X , dIP:
X + dIP
If A is a set in F and X is a random variable, we define
Z
A
Z
X dIP =
lIA X dIP:
The expectation of a random variable X is defined to be
IEX =
Z
X dIP:
The above integral has all the linearity and comparison properties one would expect. In particular, if X and Y are random variables and c is a real constant, then
Z
If X (! ) Y (! ) for every !
Z
(X + Y ) dIP =
Z
Z
X dIP +
cX dIP = c X dP;
Z
Y dIP;
2 , then Z
X dIP
Z
Y dIP:
In fact, we don’t need to have X (! ) Y (! ) for every ! 2 in order to reach this conclusion; it is enough if the set of ! for which X (! ) Y (! ) has probability one. When a condition holds with probability one, we say it holds almost surely. Finally, if A and B are disjoint subsets of and X is a random variable, then
Z
A[B
X dIP =
Z
A
X dIP +
Z
B
X dIP:
We restate the Lebesgue integral convergence theorem in this more general context. We acknowledge in these statements that conditions don’t need to hold for every ! ; almost surely is enough. Theorem 4.4 (Fatou’s Lemma) Let Xn ; n = 1; 2; : : : be a sequence of almost surely nonnegative random variables converging almost surely to a random variable X . Then
Z
or equivalently,
X dIP lim inf n!1
Z
Xn dIP;
IEX lim inf IEXn : n!1
38 Theorem 4.5 (Monotone Convergence Theorem) Let Xn ; n = 1; 2; : : : be a sequence of random variables converging almost surely to a random variable X . Assume that
0 X1 X2 X3 almost surely: Then
Z
Z
X dIP = nlim !1 Xn dIP;
or equivalently,
IEX = nlim !1 IEXn :
Theorem 4.6 (Dominated Convergence Theorem) Let Xn ; n = 1; 2; : : : be a sequence of random variables, converging almost surely to a random variable X . Assume that there exists a random variable Y such that jXnj Y almost surely for every n: Then
Z
X dIP = nlim !1
or equivalently,
Z
Xn dIP;
IEX = nlim !1 IEXn :
In Example 1.13, we constructed a probability measure on (IR; B(IR)) by integrating R the standard normal density. In fact, whenever ' is a nonnegative function defined on R satisfying IR ' d0 = 1, we call ' a density and we can define an associated probability measure by
IP (A) =
Z
A
' d0 for every A 2 B(IR):
(4.3)
We shall often have a situation in which two measure are related by an equation like (4.3). In fact, the market measure and the riskneutral measures in financial markets are related this way. We say that ' in (4.3) is the RadonNikodym derivative of dIP with respect to 0 , and we write
dIP : ' = d 0
(4.4)
The probability measure IP weights different parts of the real line according to the density '. Now suppose f is a function on (R; B(IR); IP ). Definition 1.14 gives us a value for the abstract integral
Z
Z
IR
We can also evaluate
IR
f dIP:
f' d0;
which is an integral with respec to Lebesgue measure over the real line. We want to show that
Z
IR
f dIP =
Z
IR
f' d0;
(4.5)
CHAPTER 1. Introduction to Probability Theory
39
dIP for ' in (4.5) and an equation which is suggested by the notation introduced in (4.4) (substitute d 0 “cancel” the d0 ). We include a proof of this because it allows us to illustrate the concept of the standard machine explained in Williams’s book in Section 5.12, page 5. The standard machine argument proceeds in four steps. Step 1. Assume that f is an indicator function, i.e., f (x) that case, (4.5) becomes Z
= lIA (x) for some Borel set A IR. In
IP (A) = ' d0: A This is true because it is the definition of IP (A). Step 2. Now that we know that (4.5) holds when f is an indicator function, assume that f
is a
simple function, i.e., a linear combination of indicator functions. In other words,
n X
f (x) =
k=1
ck hk (x);
where each ck is a real number and each hk is an indicator function. Then
Z
IR
f dIP =
Z "X n
#
=
ck hk dIP IR k=1 Z n X ck hk dIP IR k=1 Z n X ck hk ' d0 k=1 " IR # Z X n ck hk ' d0 IR k=1 Z
= = =
IR
f' d0:
Step 3. Now that we know that (4.5) holds when f is a simple function, we consider a general nonnegative function f . We can always construct a sequence of nonnegative simple functions fn; n = 1; 2; : : : such that
0 f1 (x) f2 (x) f3 (x) : : : for every x 2 IR; and f (x) = limn!1 fn (x) for every x 2 IR. We have already proved that
Z
IR
fn dIP =
Z
IR
fn ' d0 for every n:
We let n ! 1 and use the Monotone Convergence Theorem on both sides of this equality to get Z Z
IR
f dIP =
IR
f' d0:
40 Step 4. In the last step, we consider an integrable function f , which can take both positive and negative values. By integrable, we mean that
Z
IR ¿From Step 3, we have
f + dIP
Z ZIR IR
< 1;
f + dIP = f , dIP =
Z
IR
f , dIP < 1:
Z ZIR
f + ' d0; f , ' d0:
IR
Subtracting these two equations, we obtain the desired result:
Z
IR
f dIP = = =
Z
ZIR ZIR R
f + dIP ,
Z
0,
f + ' d
IRZ
f , dIP
f' d0:
IR
f , ' d0
1.5 Independence In this section, we define and discuss the notion of independence in a general probability space ( ; F ; IP ), although most of the examples we give will be for coin toss space.
1.5.1
Independence of sets
Definition 1.15 We say that two sets A 2 F and B
2 F are independent if
IP (A \ B) = IP (A)IP (B):
Suppose a random experiment is conducted, and ! is the outcome. The probability that ! 2 A is IP (A). Suppose you are not told !, but you are told that ! 2 B . Conditional on this information, the probability that ! 2 A is
(A \ B ) : IP (AjB) = IP IP (B )
The sets A and B are independent if and only if this conditional probability is the uncondidtional probability IP (A), i.e., knowing that ! 2 B does not change the probability you assign to A. This discussion is symmetric with respect to A and B ; if A and B are independent and you know that ! 2 A, the conditional probability you assign to B is still the unconditional probability IP (B ).
Whether two sets are independent depends on the probability measure IP . For example, suppose we toss a coin twice, with probability p for H and probability q = 1 , p for T on each toss. To avoid trivialities, we assume that 0 < p < 1. Then
IP fHH g = p2 ; IP fHT g = IP fTH g = pq; IP fTT g = q 2:
(5.1)
CHAPTER 1. Introduction to Probability Theory
41
Let A = fHH; HT g and B = fHT; TH g. In words, A is the set “H on the first toss” and B is the set “one H and one T .” Then A \ B = fHT g. We compute
IP (A) = p2 + pq = p; IP (B ) = 2pq; IP (A)IP (B ) = 2p2q; IP (A \ B) = pq: These sets are independent if and only if 2p2 q
= pq , which is the case if and only if p = 21 .
If p = 21 , then IP (B ), the probability of one head and one tail, is 12 . If you are told that the coin tosses resulted in a head on the first toss, the probability of B , which is now the probability of a T on the second toss, is still 12 .
Suppose however that p = 0:01. By far the most likely outcome of the two coin tosses is TT , and the probability of one head and one tail is quite small; in fact, IP (B ) = 0:0198. However, if you are told that the first toss resulted in H , it becomes very likely that the two tosses result in one head and one tail. In fact, conditioned on getting a H on the first toss, the probability of one H and one T is the probability of a T on the second toss, which is 0:99.
1.5.2
Independence of algebras
Definition 1.16 Let G and H be sub algebras of F . We say that G and H are independent if every set in G is independent of every set in H, i.e,
IP (A \ B) = IP (A)IP (B) for every A 2 H; B 2 G : Example 1.14 Toss a coin twice, and let IP be given by (5.1). Let determined by the first toss: G contains the sets
G = F1 be the algebra
;; ; fHH; HT g; fTH; TT g: Let H be the albegra determined by the second toss: H contains the sets
;; ; fHH; TH g; fHT; TT g: These two algebras are independent. For example, if we choose the set fHH; HT g from G and the set fHH; TH g from H, then we have
IP fHH; HT gIP fHH; TH g =(p2 + pq )(p2 + pq ) = p2 ; IP fHH; HT g\ fHH; TH g = IP fHH g = p2 : No matter which set we choose in G and which set we choose in H, we will find that the product of the probabilties is the probability of the intersection.
42 Example 1.14 illustrates the general principle that when the probability for a sequence of tosses is defined to be the product of the probabilities for the individual tosses of the sequence, then every set depending on a particular toss will be independent of every set depending on a different toss. We say that the different tosses are independent when we construct probabilities this way. It is also possible to construct probabilities such that the different tosses are not independent, as shown by the following example. Example 1.15 Define IP for the individual elements of = fHH; HT; TH; TT g to be
IP fHH g = 91 ; IP fHT g = 29 ; IP fTH g = 13 ; IP fTT g = 31 ;
and for every set A , define IP (A) to be the sum of the probabilities of the elements in A. Then IP ( ) = 1, so IP is a probability measure. Note that the sets fH on first tossg = fHH; HT g and fH on second tossg = fHH; TH g have probabilities IP fHH; HT g = 13 and IP fHH; TH g = 4 4 of fHH; HT g 9 , so the product of the probabilities is 27 . On the other hand, the intersection and fHH; TH g contains the single element fHH g, which has probability 19 . These sets are not independent.
1.5.3
Independence of random variables
Definition 1.17 We say that two random variables X and Y are independent if the algebras they generate (X ) and (Y ) are independent. In the probability space of three independent coin tosses, the price S2 of the stock at time 2 is independent of SS32 . This is because S2 depends on only the first two coin tosses, whereas SS32 is either u or d, depending on whether the third coin toss is H or T . Definition 1.17 says that for independent random variables X and Y , every set defined in terms of X is independent of every set defined in terms of Y . In the case of S2 and SS32 just considered, for example, the sets fS2 = are indepedent sets.
udS0g = fHTH; HTT g and
nS
o
S = u = fHHH; HTH; THH; TTH g 3 2
Suppose X and Y are independent random variables. We defined earlier the measure induced by X on IR to be
LX (A) = IP fX 2 Ag; A IR:
Similarly, the measure induced by Y is
LY (B) = IP fY 2 Bg; B IR:
Now the pair (X; Y ) takes values in the plane IR2 , and we can define the measure induced by the pair
LX;Y (C ) = IP f(X; Y ) 2 C g; C IR2:
The set C in this last equation is a subset of the plane IR2 . In particular, C could be a “rectangle”, i.e, a set of the form A B , where A IR and B IR. In this case,
f(X; Y ) 2 A Bg = fX 2 Ag \ fY 2 Bg;
CHAPTER 1. Introduction to Probability Theory
43
and X and Y are independent if and only if
LX;Y (A B) = IP fX 2 Ag \ fY 2 Bg = IP fX 2 AgIP fY 2 B g = LX (A)LY (B ):
(5.2)
In other words, for independent random variables X and Y , the joint distribution represented by the measure LX;Y factors into the product of the marginal distributions represented by the measures LX and LY . A joint density for (X; Y ) is a nonnegative function fX;Y (x; y ) such that
Z Z
LX;Y (A B) =
A B
fX;Y (x; y ) dx dy:
Not every pair of random variables (X; Y ) has a joint density, but if a pair does, then the random variables X and Y have marginal densities defined by
fX (x) =
Z1
,1
fX;Y (x; ) d; fY (y )
These have the properties
LX (A) = LY (B) =
Z ZA B
Z1
,1
fX;Y (; y ) d:
fX (x) dx; A IR; fY (y ) dy; B IR:
Suppose X and Y have a joint density. Then X and Y are independent variables if and only if the joint density is the product of the marginal densities. This follows from the fact that (5.2) is equivalent to independence of X and Y . Take A = (,1; x] and B = (,1; y ], write (5.1) in terms of densities, and differentiate with respect to both x and y . Theorem 5.7 Suppose X and Y are independent random variables. Let g and h be functions from IR to IR. Then g(X ) and h(Y ) are also independent random variables. P ROOF : Let us denote W = g (X ) and Z a typical set in (W ) is of the form
= h(Y ). We must consider sets in (W ) and (Z ). But
f!; W (!) 2 Ag = f! : g(X (!)) 2 Ag; which is defined in terms of the random variable X . Therefore, this set is in (X ). (In general, we have that every set in (W ) is also in (X ), which means that X contains at least as much information as W . In fact, X can contain strictly more information than W , which means that (X ) will contain all the sets in (W ) and others besides; this is the case, for example, if W = X 2.) In the same way that we just argued that every set in (W ) is also in (X ), we can show that every set in (Z ) is also in (Y ). Since every set in (X ) is independent of every set in (Y ), we conclude that every set in (W ) is independent of every set in (Z ).
44 Definition 1.18 Let X1; X2; : : : be a sequence of random variables. We say that these random variables are independent if for every sequence of sets A1 2 (X1); A2 2 (X2); : : : and for every positive integer n,
IP (A1 \ A2 \ An) = IP (A1 )IP (A2 ) IP (An ):
1.5.4
Correlation and independence
Theorem 5.8 If two random variables X and Y are independent, and if g and h are functions from IR to IR, then
IE [g(X )h(Y )] = IEg (X ) IEh(Y );
provided all the expectations are defined. P ROOF : Let g (x) = lIA (x) and trying to prove becomes
h(y) = lIB (y ) be indicator functions.
Then the equation we are
IP fX 2 Ag \ fY 2 Bg = IP fX 2 AgIP fY 2 Bg; which is true because X and Y are independent. Now use the standard machine to get the result for general functions g and h. The variance of a random variable X is defined to be
Var(X ) = IE [X , IEX ]2: The covariance of two random variables X and Y is defined to be Cov(X; Y )
h
i
IE (X , IEX )(Y , IEY ) = = IE [XY ] , IEX IEY:
According to Theorem 5.8, for independent random variables, the covariance is zero. If X and Y both have positive variances, we define their correlation coefficient
(X; Y ) = p Cov(X; Y ) : Var(X )Var(Y ) For independent random variables, the correlation coefficient is zero. Unfortunately, two random variables can have zero correlation and still not be independent. Consider the following example. Example 1.16 Let X be a standard normal random variable, let Z be independent of X and have the distribution IP fZ = 1g = IP fZ = ,1g = 0. Define Y = XZ . We show that Y is also a standard normal random variable, X and Y are uncorrelated, but X and Y are not independent. The last claim is easy to see. If X 2 = Y 2 almost surely.
X and Y
were independent, so would be X 2 and Y 2 , but in fact,
CHAPTER 1. Introduction to Probability Theory
45
2 IR, we have IP fY y and Z = 1g + IP fY y and Z = ,1g IP fX y and Z = 1g + IP f,X y and Z = ,1g IP fX y gIP fZ = 1g + IP f,X y gIP fZ = ,1g 1 IP fX y g + 1 IP f,X y g: 2 2
We next check that Y is standard normal. For y
IP fY yg = = = =
Since X is standard normal, IP fX y g = IP fX which shows that Y is also standard normal.
,yg, and we have IP fY yg = IP fX yg,
Being standard normal, both X and Y have expected value zero. Therefore, Cov(X; Y ) = IE [XY ] = IE [X 2Z ] = IEX 2 IEZ
= 1 0 = 0:
Where in IR2 does the measure LX;Y put its mass, i.e., what is the distribution of (X; Y )? We conclude this section with the observation that for independent random variables, the variance of their sum is the sum of their variances. Indeed, if X and Y are independent and Z = X + Y , then Var(Z )
h
i
IE (Z , IEZ )2 = i = IE X + Y , IEX , IEY )2 h i = IE (X , IEX )2 + 2(X , IEX )(Y , IEY ) + (Y , IEY )2 = Var(X ) + 2IE [X , IEX ]IE [Y , IEY ] + Var(Y ) = Var(X ) + Var(Y ):
This argument extends to any finite number of random variables. If we are given independent random variables X1; X2; : : :; Xn , then Var(X1 + X2 + + Xn ) = Var(X1) + Var(X2) + + Var(Xn ):
1.5.5
(5.3)
Independence and conditional expectation.
We now return to property (k) for conditional expectations, presented in the lecture dated October 19, 1995. The property as stated there is taken from Williams’s book, page 88; we shall need only the second assertion of the property: (k) If a random variable X is independent of a algebra H, then
IE [X jH] = IEX: The point of this statement is that if X is independent of H, then the best estimate of X based on the information in H is IEX , the same as the best estimate of X based on no information.
46 To show this equality, we observe first that IEX is Hmeasurable, since it is not random. We must also check the partial averaging property
Z
A
IEX dIP =
Z
A
X dIP for every A 2 H:
If X is an indicator of some set B , which by assumption must be independent of H, then the partial averaging equation we must check is
Z
A
IP (B) dIP =
Z
A
lIB dIP:
The lefthand side of this equation is IP (A)IP (B ), and the right hand side is
Z
lIAlIB dIP =
Z
lIA\B dIP = IP (A \ B):
The partial averaging equation holds because A and B are independent. The partial averaging equation for general X independent of H follows by the standard machine.
1.5.6
Law of Large Numbers
There are two fundamental theorems about sequences of independent random variables. Here is the first one. Theorem 5.9 (Law of Large Numbers) Let X1 ; X2; : : : be a sequence of independent, identically distributed random variables, each with expected value and variance 2. Define the sequence of averages
Yn = X1 + X2 +n + Xn ; n = 1; 2; : : :: Then Yn converges to almost surely as n ! 1.
We are not going to give the proof of this theorem, but here is an argument which makes it plausible. We will use this argument later when developing stochastic calculus. The argument proceeds in two steps. We first check that IEYn = for every n. We next check that Var(Yn ) ! 0 as n ! 0. In other words, the random variables Yn are increasingly tightly distributed around as n ! 1. For the first step, we simply compute
IEYn = n1 [IEX1 + IEX2 + + IEXn] = n1 [ + +{z + }] = : n times
For the second step, we first recall from (5.3) that the variance of the sum of independent random variables is the sum of their variances. Therefore, Var(Yn ) = As n ! 1, we have Var(Yn ) ! 0.
n X k=1
Var
Xk X n 2 n
=
2
: = 2 n k=1 n
CHAPTER 1. Introduction to Probability Theory
1.5.7
47
Central Limit Theorem
The Law of Large Numbers is a bit boring because the limit is nonrandom. This is because the denominator in the definition of Yn is so large that the variance of Yn converges to zero. If we want p to prevent this, we should divide by n rather than n. In particular, if we again have a sequence of independent, identically distributed random variables, each with expected value and variance 2 , but now we set
Zn = (X1 , ) + (X2 ,pn) + + (Xn , ) ;
then each Zn has expected value zero and Var(Zn ) =
n X
k=1
Var
Xk , X n 2 pn = n = 2: k=1
As n ! 1, the distributions of all the random variables Zn have the same degree of tightness, as measured by their variance, around their expected value 0. The Central Limit Theorem asserts that as n ! 1, the distribution of Zn approaches that of a normal random variable with mean (expected value) zero and variance 2. In other words, for every set A IR,
1 Z e, x dx: p lim IP f Z 2 A g = n n!1 2 A 2 2 2
48
Chapter 2
Conditional Expectation Please see Hull’s book (Section 9.6.)
2.1 A Binomial Model for Stock Price Dynamics Stock prices are assumed to follow this simple binomial model: The initial stock price during the period under study is denoted S0. At each time step, the stock price either goes up by a factor of u or down by a factor of d. It will be useful to visualize tossing a coin at each time step, and say that
the stock price moves up by a factor of u if the coin comes out heads (H ), and down by a factor of d if it comes out tails (T ).
Note that we are not specifying the probability of heads here. Consider a sequence of 3 tosses of the coin (See Fig. 2.1) The collection of all possible outcomes (i.e. sequences of tosses of length 3) is
= fHHH; HHT; HTH; HTT; THH; THH; THT; TTH; TTT g: A typical sequence of will be denoted ! , and ! k will denote the kth element in the sequence ! . We write Sk (! ) to denote the stock price at “time” k (i.e. after k tosses) under the outcome !. Note that Sk (! ) depends only on ! 1 ; !2; : : : ; !k . Thus in the 3cointoss example we write for instance,
S1 (!) =4 S1(!1 ; !2; !3) =4 S1(! 1); S2 (!) =4 S2(! 1 ; !2; !3) =4 S2(! 1; !2): Each Sk is a random variable defined on the set . More precisely, let F = P ( ). Then F is a algebra and ( ; F ) is a measurable space. Each Sk is an F measurable function !IR, that is, Sk,1 is a function B!F where B is the Borel algebra on IR. We will see later that Sk is in fact 49
50 ω = Η 3
3 S3 (HHH) = u S0
2 S2 (HH) =u S0 ω =Η 2
ω = Τ 3
S1 (H) = uS0 ω1= Η
ω =Τ 2 S2 (HT) = ud S0
S0 ω =Τ 1
S2 (TH) = ud S 0
ω = Η 3 ω = Τ 3
ω2 = Η
S1 (T) = dS0 ω2 = Τ
S3 (HHT) = u2 d S0 2 S3 (HTH) = u d S0 2 S3 (THH) = u d S0
ω = Η 3
2 S3 (HTT) = d u 2 S3 (THT) = d u 2 S3 (TTH) = d u
S0 S0 S0
S2 (TT) = d 2S0 ω3 = Τ
S3 (TTT) = d 3 S0
Figure 2.1: A three coin period binomial model. measurable under a subalgebra of F . Recall that the Borel algebra B is the algebra generated by the open intervals of IR. In this course we will always deal with subsets of R I that belong to B. For any random variable X defined on a sample space and any y
2 IR, we will use the notation:
fX yg =4 f! 2 ; X (!) yg: The sets fX < y g; fX y g; fX = y g; etc, are defined similarly. Similarly for any subset B of IR, we define
Assumption 2.1
fX 2 Bg =4 f! 2 ; X (!) 2 Bg: u > d > 0.
2.2 Information Definition 2.1 (Sets determined by the first k tosses.) We say that a set A is determined by the first k coin tosses if, knowing only the outcome of the first k tosses, we can decide whether the outcome of all tosses is in A. In general we denote the collection of sets determined by the first k tosses by F k . It is easy to check that F k is a algebra. Note that the random variable Sk is F k measurable, for each k = 1; 2; : : : ; n. Example 2.1 In the 3 cointoss example, the collection F 1 of sets determined by the first toss consists of:
CHAPTER 2. Conditional Expectation 1. 2. 3. 4.
51
4 fHHH; HHT; HTH; HT T g, AH = AT =4 fTHH; THT; TTH; TTT g, ,
.
The collection F 2 of sets determined by the first two tosses consists of: 1. 2. 3.
AHH =4 fHHH; HHT g, 4 fHTH; HTT g, AHT = 4 fTHH; THT g, AT H = 4 fTTH; TTT g, AT T =
4. 5. The complements of the above sets, 6. Any union of the above sets (including the complements), 7. and .
Definition 2.2 (Information carried by a random variable.) Let X be a random variable !IR. We say that a set A is determined by the random variable X if, knowing only the value X (!) of the random variable, we can decide whether or not ! 2 A. Another way of saying this is that for every y 2 IR, either X ,1(y ) A or X ,1 (y ) \ A = . The collection of susbets of determined by X is a algebra, which we call the algebra generated by X , and denote by (X ). If the random variable X takes finitely many different values, then (X ) is generated by the collection of sets
fX ,1(X (!))j! 2 g;
these sets are called the atoms of the algebra (X ).
In general, if X is a random variable !IR, then (X ) is given by
(X ) = fX ,1(B); B 2 Bg: Example 2.2 (Sets determined by S2 ) The algebra generated by S2 consists of the following sets: 1. 2. 3.
AHH = fHHH; HHT g = f! 2 ; S2 (!) = u2S0 g, AT T = fTTH; TTT g = fS2 = d2S0 g; AHT [ AT H = fS2 = udS0g;
4. Complements of the above sets, 5. Any union of the above sets, 6. = fS2 (!) 2 g, 7. = fS2 (!) 2 IRg.
52
2.3 Conditional Expectation In order to talk about conditional expectation, we need to introduce a probability measure on our cointoss sample space . Let us define
p 2 (0; 1) is the probability of H , q =4 (1 , p) is the probability of T , the coin tosses are independent, so that, e.g., IP (HHT ) = p2q; etc. IP (A) =4 P!2A IP (!), 8A . Definition 2.3 (Expectation.)
IEX =4 If A then
and
X ! 2
4
IA (!) = IE (IA X ) =
Z A
(
X (!)IP (! ):
1 0
2A 62 A X
if ! if !
XdIP =
!2A
X (!)IP (! ):
We can think of IE (IA X ) as a partial average of X over the set A.
2.3.1
An example
Let us estimate S1, given S2. Denote the estimate by IE (S1jS2). From elementary probability, IE (S1jS2) is a random variable Y whose value at ! is defined by
Y (!) = IE (S1jS2 = y ); where y = S2 (!). Properties of IE (S1jS2):
IE (S1jS2) should depend on !, i.e., it is a random variable. If the value of S2 is known, then the value of IE (S1jS2) should also be known. In particular, – If ! = HHH or ! = HHT , then S2 (! ) = u2 S0. If we know that S2(! ) = u2 S0 , then even without knowing ! , we know that S1 (! ) = uS0. We define
–
IE (S1jS2)(HHH ) = IE (S1jS2)(HHT ) = uS0: If ! = TTT or ! = TTH , then S2(! ) = d2S0 . If we know that S2(! ) = d2S0 , then even without knowing ! , we know that S1 (! ) = dS0. We define IE (S1jS2)(TTT ) = IE (S1jS2 )(TTH ) = dS0:
CHAPTER 2. Conditional Expectation – If !
53
2 A = fHTH; HTT; THH; THT g, then S2(!) = udS0. If we know S2(!) =
udS0, then we do not know whether S1 average:
= uS0 or S1 = dS0. We then take a weighted
IP (A) = p2 q + pq2 + p2 q + pq 2 = 2pq:
Furthermore,
Z A
For !
S1 dIP = p2 quS0 + pq2uS0 + p2qdS0 + pq 2dS0 = pq (u + d)S0
2 A we define
R S dIP IE (S1jS2)(!) = AIP (1A) = 12 (u + d)S0: Z
Then
A In conclusion, we can write
IE (S1jS2)dIP =
A
S1dIP:
IE (S1jS2)(!) = g (S2(! ));
8 > < uS0 g(x) = > 21 (u + d)S0 : dS0
where
Z
if x = u2 S0 if x = udS0 if x = d2 S0
In other words, IE (S1jS2) is random only through dependence on S2 . We also write
IE (S1jS2 = x) = g (x); where g is the function defined above.
The random variable IE (S1jS2) has two fundamental properties:
IE (S1jS2) is (S2)measurable. For every set A 2 (S2), Z A 2.3.2
IE (S1jS2)dIP =
Z A
S1dIP:
Definition of Conditional Expectation
Please see Williams, p.83.
Let ( ; F ; IP ) be a probability space, and let G be a subalgebra of F . Let X be a random variable on ( ; F ; IP ). Then IE (X jG) is defined to be any random variable Y that satisfies: (a)
Y is G measurable,
54 (b) For every set A 2 G , we have the “partial averaging property”
Z
A
Y dIP =
Z
A
XdIP:
Existence. There is always a random variable Y satisfying the above properties (provided that IE jX j < 1), i.e., conditional expectations always exist. Uniqueness. There can be more than one random variable Y satisfying the above properties, but if Y 0 is another one, then Y = Y 0 almost surely, i.e., IP f! 2 ; Y (!) = Y 0 (!)g = 1: Notation 2.1 For random variables X; Y , it is standard notation to write
IE (X jY ) =4 IE (X j(Y )): Here are some useful ways to think about IE (X jG):
2.3.3
A random experiment is performed, i.e., an element ! of is selected. The value of ! is partially but not fully revealed to us, and thus we cannot compute the exact value of X (!). Based on what we know about ! , we compute an estimate of X (!). Because this estimate depends on the partial information we have about ! , it depends on !, i.e., IE [X jY ](!) is a function of ! , although the dependence on ! is often not shown explicitly.
If the algebra G contains finitely many sets, there will be a “smallest” set A in G containing ! , which is the intersection of all sets in G containing !. The way ! is partially revealed to us is that we are told it is in A, but not told which element of A it is. We then define IE [X jY ](!) to be the average (with respect to IP ) value of X over this set A. Thus, for all ! in this set A, IE [X jY ](!) will be the same.
Further discussion of Partial Averaging
The partial averaging property is
Z
A
IE (X jG)dIP =
Z A
XdIP; 8A 2 G :
(3.1)
We can rewrite this as
IE [IA:IE (X jG)] = IE [IA :X ]:
(3.2)
Note that IA is a G measurable random variable. In fact the following holds: Lemma 3.10 If V is any G measurable random variable, then provided IE jV:IE (X jG)j < 1,
IE [V:IE (X jG)] = IE [V:X ]:
(3.3)
CHAPTER 2. Conditional Expectation
55
Proof: To see this, first use (3.2) and linearity of expectations to prove (3.3) when V is a simple G measurable random variable, i.e., V is of the form V = Pnk=1 ck IAK , where each Ak is in G and each ck is constant. Next consider the case that V is a nonnegative G measurable random variable, but is not necessarily simple. Such a V can be written as the limit of an increasing sequence of simple random variables Vn ; we write (3.3) for each Vn and then pass to the limit, using the Monotone Convergence Theorem (See Williams), to obtain (3.3) for V . Finally, the general G measurable random variable V can be written as the difference of two nonnegative randomvariables V = V + , V , , and since (3.3) holds for V + and V , it must hold for V as well. Williams calls this argument the “standard machine” (p. 56). Based on this lemma, we can replace the second condition in the definition of a conditional expectation (Section 2.3.2) by: (b’) For every G measurable randomvariable V , we have
IE [V:IE (X jG)] = IE [V:X ]: 2.3.4
(3.4)
Properties of Conditional Expectation
Please see Willams p. 88. Proof sketches of some of the properties are provided below. (a)
(b)
IE (IE (X jG)) = IE (X ): Proof: Just take A in the partial averaging property to be . The conditional expectation of X is thus an unbiased estimator of the random variable X . If X is G measurable, then IE (X jG) = X: Proof: The partial averaging property holds trivially when Y is replaced by X . And since X is G measurable, X satisfies the requirement (a) of a conditional expectation as well. If the information content of G is sufficient to determine X , then the best estimate of X based on G is X itself.
(c) (Linearity) (d) (Positivity) If X
IE (a1X1 + a2 X2jG ) = a1IE (X1jG) + a2IE (X2jG ):
0 almost surely, then IE (X jG) 0:
Proof: Take A = f! 2 ; RIE (X jG)(! ) < 0g. This R set is in G since IE (X jG) is G measurable. Partial averaging implies A IE (X jG)dIP = A XdIP . The righthand side is greater than or equal to zero, and the lefthand side is strictly negative, unless IP (A) = 0. Therefore, IP (A) = 0.
56 (h) (Jensen’s Inequality) If : R!R is convex and IE j(X )j < 1, then
IE ((X )jG) (IE (X jG)): Recall the usual Jensen’s Inequality: IE(X ) (IE (X )): (i) (Tower Property) If H is a subalgebra of G , then
IE (IE (X jG)jH) = IE (X jH):
(j)
H is a subalgebra of G means that G contains more information than H. If we estimate X based on the information in G , and then estimate the estimator based on the smaller amount of information in H, then we get the same result as if we had estimated X directly based on the information in H. (Taking out what is known) If Z is G measurable, then IE (ZX jG) = Z:IE (X jG): When conditioning on G , the G measurable random variable Z acts like a constant. Proof: Let Z be a G measurable random variable. A random variable Y is IE (ZX jG) if and only if (a) Y is G measurable; R R (b) A Y dIP = A ZXdIP; 8A 2 G .
Take Y = Z:IE (X jG). Then Y satisfies (a) (a product of G measurable random variables is Gmeasurable). Y also satisfies property (b), as we can check below:
Z
A
Y dIP = IE (IA:Y ) = IE [IAZIE (X jG)] = ZIE [IAZ:X ] ((b’) with V = IA Z = ZXdIP: A
(k) (Role of Independence) If H is independent of ((X ); G), then
IE (X j(G ; H)) = IE (X jG): In particular, if X is independent of H, then
IE (X jH) = IE (X ): If H is independent of X and G , then nothing is gained by including the information content of H in the estimation of X .
CHAPTER 2. Conditional Expectation
2.3.5
57
Examples from the Binomial Model
Recall that F 1
= f; AH ; AT ; g. Notice that IE (S2jF 1 ) must be constant on AH and AT . Now since IE (S2jF 1 ) must satisfy the partial averaging property,
Z
AH
Z
AT We compute
Z AH
AH Therefore,
IE (S2jF 1)dIP =
Z
AH
Z
AT
S2dIP; S2dIP:
IE (S2jF 1 )dIP = IP (AH ):IE (S2jF 1)(!) = pIE (S2jF 1)(! ); 8! 2 AH :
Z
On the other hand,
IE (S2jF 1)dIP =
S2 dIP = p2u2 S0 + pqudS0:
IE (S2jF 1 )(!) = pu2 S0 + qudS0; 8! 2 AH :
We can also write
IE (S2jF 1)(! ) = pu2 S0 + qudS0 = (pu + qd)uS0 = (pu + qd)S1(!); 8! 2 AH Similarly,
IE (S2jF 1 )(!) = (pu + qd)S1(!); 8! 2 AT :
Thus in both cases we have
IE (S2jF 1 )(!) = (pu + qd)S1(!); 8! 2 : A similar argument one time step later shows that
IE (S3jF 2)(!) = (pu + qd)S2(!): We leave the verification of this equality as an exercise. We can verify the Tower Property, for instance, from the previous equations we have
IE [IE (S3jF 2)jF 1] = IE [(pu + qd)S2jF 2] = (pu + qd)IE (S2jF 1) = (pu + qd)2S1: This final expression is IE (S3jF 1).
(linearity)
58
2.4 Martingales The ingredients are:
A probability space ( ; F ; IP ).
A sequence of random variables M0; M1; : : : ; Mn . This is called a stochastic process.
A sequence of algebras F 0; F 1 ; : : : ; F n , with the property that F 0 F . Such a sequence of algebras is called a filtration.
F1 : : : Fn
Conditions for a martingale: 1. Each Mk is F k measurable. If you know the information in F k , then you know the value of Mk . We say that the process fMk g is adapted to the filtration fF k g. 2. For each k, IE (Mk+1 jF k ) = Mk . Martingales tend to go neither up nor down.
A supermartingale tends to go down, i.e. the second condition above is replaced by IE (Mk+1 jF k )
Mk ; a submartingale tends to go up, i.e. IE (Mk+1jF k ) Mk . Example 2.3 (Example from the binomial model.) For k
= 1; 2 we already showed that
IE(Sk+1 jF k ) = (pu + qd)Sk : For k = 0, we set F 0 = f; g, the “trivial algebra”. This algebra contains no information, and any F 0 measurable random variable must be constant (nonrandom). Therefore, by definition, IE(S1 jF 0 ) is that constant which satisfies the averaging property
Z
IE(S1 jF 0 )dIP =
Z
The right hand side is IES1 = (pu + qd)S0 , and so we have
S1 dIP:
IE(S1 jF 0) = (pu + qd)S0 : In conclusion,
If (pu + qd) = 1 then fSk ; F k ; k = 0; 1; 2; 3g is a martingale. If (pu + qd) 1 then fSk ; F k ; k = 0; 1; 2; 3g is a submartingale. If (pu + qd) 1 then fSk ; F k ; k = 0; 1; 2; 3g is a supermartingale.
Chapter 3
Arbitrage Pricing 3.1 Binomial Pricing Return to the binomial pricing model Please see:
Cox, Ross and Rubinstein, J. Financial Economics, 7(1979), 229–263, and Cox and Rubinstein (1985), Options Markets, PrenticeHall.
Example 3.1 (Pricing a Call Option) Suppose u = 2; d = 0:5; r = 25%(interest rate), S0 = 50. (In this and all examples, the interest rate quoted is per unit time, and the stock prices S0 ; S1; : : : are indexed by the same time periods). We know that
S1 (!) =
100 25
if !1 if !1
=H =T
Find the value at time zero of a call option to buy one share of stock at time 1 for $50 (i.e. the strike price is $50). The value of the call at time 1 is
V1(!) = (S1 (!) , 50)+ =
50 0
if !1 if !1
Suppose the option sells for $20 at time 0. Let us construct a portfolio: 1. Sell 3 options for $20 each. Cash outlay is ,$60: 2. Buy 2 shares of stock for $50 each. Cash outlay is $100. 3. Borrow $40. Cash outlay is ,$40:
59
=H =T
60 This portfolio thus requires no initial investment. For this portfolio, the cash outlay at time 1 is: Pay off option Sell stock Pay off debt
!1 = H $150 ,$200 $50
!1 = T $0 ,$50 $50
,,,,, ,,,,, $0
$0 The arbitrage pricing theory (APT) value of the option at time 0 is V0 = 20. Assumptions underlying APT:
Unlimited short selling of stock. Unlimited borrowing. No transaction costs. Agent is a “small investor”, i.e., his/her trading does not move the market.
Important Observation: The APT value of the option does not depend on the probabilities of H and T .
3.2 General onestep APT Suppose a derivative security pays off the amount V1 at time 1, where V1 is an F 1 measurable random variable. (This measurability condition is important; this is why it does not make sense to use some stock unrelated to the derivative security in valuing it, at least in the straightforward method described below).
Sell the security for V0 at time 0. (V0 is to be determined later).
Buy 0 shares of stock at time 0. (0 is also to be determined later)
Invest V0 , negative).
0S0 in the money market, at riskfree interest rate r. (V0 , 0S0 might be
Then wealth at time 1 is
X1 =4 0 S1 + (1 + r)(V0 , 0S0) = (1 + r)V0 + 0 (S1 , (1 + r)S0): We want to choose V0 and 0 so that
X1 = V1 regardless of whether the stock goes up or down.
CHAPTER 3. Arbitrage Pricing
61
The last condition above can be expressed by two equations (which is fortunate since there are two unknowns):
(1 + r)V0 + 0(S1 (H ) , (1 + r)S0) = V1 (H )
(2.1)
(1 + r)V0 + 0 (S1(T ) , (1 + r)S0) = V1 (T )
(2.2)
Note that this is where we use the fact that the derivative security value Vk is a function of Sk , i.e., when Sk is known for a given ! , Vk is known (and therefore nonrandom) at that ! as well. Subtracting the second equation above from the first gives
) , V1(T ) : 0 = SV1((H H ) , S (T ) 1
(2.3)
1
Plug the formula (2.3) for 0 into (2.1):
(1 + r)V0 = V1 (H ) , 0 (S1(H ) , (1 + r)S0) = V1 (H ) , V1((Hu ),,d)VS1(T ) (u , 1 , r)S0 0 1 = [(u , d)V (H ) , (V (H ) , V (T ))(u , 1 , r)] 1 1 1 u,d 1,r = 1 +u ,r ,d d V1(H ) + u , u , d V1(T ):
We have already assumed u > d > 0. We now also assume d 1 + r be an arbitrage opportunity). Define
u (otherwise there would
p~ =4 1 +u ,r ,d d ; q~ =4 u ,u ,1 ,d r :
Then p~ > 0 and q~ > 0. Since p~ + q~ = 1, we have 0 < p~ < 1 and q~ = 1 , p~. Thus, p~; q~ are like probabilities. We will return to this later. Thus the price of the call at time 0 is given by
V0 = 1 +1 r [~pV1(H ) + q~V1(T )]:
(2.4)
3.3 RiskNeutral Probability Measure Let be the set of possible outcomes from n coin tosses. Construct a probability measure If P on
by the formula If P (!1; !2 ; : : : ; !n) =4 p~#fj ;! j =H gq~#fj;!j =T g
If P is called the riskneutral probability measure. We denote by If E the expectation under If P . Equation 2.4 says
V0 = If E
1 1 + r V1 :
62 Theorem 3.11 Under If P , the discounted stock price process f(1+ r),k Sk ; F k gnk=0 is a martingale. Proof:
If E [(1 + r),(k+1) Sk+1 jF k ] = (1 + r),(k+1)(~pu + q~d)Sk = (1 + r),(k+1) u(1u+,r d, d) + d(uu,,1 d, r) Sk + du , d , dr S = (1 + r),(k+1) u + ur , udu , k d = (1 + r),(k+1) (u , d)(1 + r) Sk u,d , k = (1 + r) Sk :
3.3.1
Portfolio Process
The portfolio process is = (0 ; 1; : : : ; n,1), where
k is the number of shares of stock held between times k and k + 1. Each k is F k measurable. (No insider trading). 3.3.2
Selffinancing Value of a Portfolio Process
Start with nonrandom initial wealth X0 , which need not be 0. Define recursively
Xk+1 = k Sk+1 + (1 + r)(Xk , k Sk ) = (1 + r)Xk + k (Sk+1 , (1 + r)Sk ):
(3.1) (3.2)
Then each Xk is F k measurable.
Theorem 3.12 Under If P , the discounted selffinancing portfolio process value f(1 + r),k Xk ; F k gnk=0 is a martingale. Proof: We have
(1 + r),(k+1)Xk+1 = (1 + r),k Xk + k (1 + r),(k+1) Sk+1 , (1 + r),k Sk :
CHAPTER 3. Arbitrage Pricing Therefore,
63
If E [(1 + r),(k+1) Xk+1 jF k ] = If E [(1 + r),k Xk jF k ] +If E [(1 + r),(k+1)k Sk+1jF k ] ,If E [(1 + r),k k Sk jF k ] = (1 + r),k Xk (requirement (b) of conditional exp.) +k If E[(1 + r),(k+1)Sk+1jF k ] (taking out what is known) ,(1 + r),k k Sk (property (b)) = (1 + r),k Xk (Theorem 3.11)
3.4 Simple European Derivative Securities Definition 3.1 () A simple European derivative security with expiration time m is an F m measurable random variable Vm . (Here, m is less than or equal to n, the number of periods/cointosses in the model). Definition 3.2 () A simple European derivative security Vm is said to be hedgeable if there exists a constant X0 and a portfolio process = (0; : : : ; m,1 ) such that the selffinancing value process X0; X1; : : : ; Xm given by (3.2) satisfies
Xm(!) = Vm(!); 8! 2 : In this case, for k = 0; 1; : : : ; m, we call Xk the APT value at time k of Vm .
Theorem 4.13 (Corollary to Theorem 3.12) If a simple European security Vm is hedgeable, then for each k = 0; 1; : : : ; m, the APT value at time k of Vm is
Vk =4 (1 + r)k If E [(1 + r),m Vm jF k ]:
Proof: We first observe that if martingale property
(4.1)
fMk ; F k ; k = 0; 1; : : : ; mg is a martingale, i.e., satisfies the If E [Mk+1 jF k ] = Mk
= 0; 1; : : : ; m , 1, then we also have If E [Mm jF k ] = Mk ; k = 0; 1; : : : ; m , 1: (4.2) When k = m , 1, the equation (4.2) follows directly from the martingale property. For k = m , 2,
for each k
we use the tower property to write
If E [MmjF m,2 ] = If E [If E[MmjF m,1 ]jF m,2 ] f = IE [Mm,1 jF m,2 ] = Mm,2 :
64 We can continue by induction to obtain (4.2). If the simple European security Vm is hedgeable, then there is a portfolio process whose selffinancing value process X0; X1; : : : ; Xm satisfies Xm = Vm . By definition, Xk is the APT value at time k of Vm . Theorem 3.12 says that
X0; (1 + r),1 X1; : : : ; (1 + r),m Xm is a martingale, and so for each k, (1 + r),k Xk = If E[(1 + r),m XmjF k ] = If E [(1 + r),mVm jF k ]: Therefore,
Xk = (1 + r)k If E [(1 + r),mVm jF k ]:
3.5 The Binomial Model is Complete Can a simple European derivative security always be hedged? It depends on the model. If the answer is “yes”, the model is said to be complete. If the answer is “no”, the model is called incomplete. Theorem 5.14 The binomial model is complete. In particular, let Vm be a simple European derivative security, and set
Vk (!1 ; : : : ; !k ) = (1 + r)k If E [(1 + r),mVmjF k ](!1; : : : ; !k );
(5.1)
H ) , Vk+1 (!1; : : : ; !k ; T ) : k (!1 ; : : : ; !k ) = SVk+1 ((!!1;; :: :: :: ;; !!k ;; H ) , S (! ; : : : ; ! ; T )
(5.2)
k+1 1
k
k+1 1
k
E[(1 + r),m Vm], the selffinancing value of the portfolio process Starting with initial wealth V0 = If 0 ; 1; : : : ; m,1 is the process V0 ; V1; : : : ; Vm. Proof: Let V0 ; : : : ; Vm,1 and 0 ; : : : ; m,1 be defined by (5.1) and (5.2). Set X0 = V0 and define the selffinancing value of the portfolio process 0 ; : : : ; m,1 by the recursive formula 3.2:
Xk+1 = k Sk+1 + (1 + r)(Xk , k Sk ): We need to show that
Xk = Vk ; 8k 2 f0; 1; : : : ; mg:
(5.3)
We proceed by induction. For k = 0, (5.3) holds by definition of X0. Assume that (5.3) holds for some value of k, i.e., for each fixed (!1 ; : : : ; !k ), we have
Xk (!1 ; : : : ; !k ) = Vk(!1 ; : : : ; !k ):
CHAPTER 3. Arbitrage Pricing
65
We need to show that
Xk+1 (!1; : : : ; !k ; H ) = Vk+1(!1; : : : ; !k ; H ); Xk+1 (!1 ; : : : ; !k ; T ) = Vk+1(!1; : : : ; !k ; T ): We prove the first equality; the second can be shown similarly. Note first that
If E [(1 + r),(k+1)Vk+1jF k ] = If E [If E[(1 + r),m VmjF k+1 ]jF k ] = If E [(1 + r),m VmjF k ] = (1 + r),k Vk
In other words, f(1 + r),k Vk gnk=0 is a martingale under If P . In particular,
Vk (!1 ; : : : ; !k ) = If E [(1 + r),1 Vk+1 jF k ](!1; : : : ; !k ) = 1 +1 r (~pVk+1 (!1 ; : : : ; !k ; H ) + q~Vk+1 (!1 ; : : : ; !k ; T )) :
Since (!1; : : : ; !k ) will be fixed for the rest of the proof, we simplify notation by suppressing these symbols. For example, we write the last equation as
Vk = 1 +1 r (~pVk+1 (H ) + q~Vk+1(T )) : We compute
Xk+1 (H ) = k Sk+1 (H ) + (1 + r)(Xk , k Sk ) = k (Sk+1 (H ) , (1 + r)Sk ) + (1 + r)Vk H ) , Vk+1(T ) (S (H ) , (1 + r)S ) = SVk+1 ((H k+1 k k+1 ) , Sk+1 (T ) +~pVk+1 (H ) + q~Vk+1 (T ) (H ) , Vk+1 (T ) (uS , (1 + r)S ) = Vk+1uS k k k , dSk +~pVk+1 (H ) + q~Vk+1(T ) = (Vk+1 (H ) , Vk+1 (T )) u , 1 , r + p~Vk+1 (H ) + q~Vk+1 (T ) u,d = (Vk+1 (H ) , Vk+1 (T )) q~ + p~Vk+1 (H ) + q~Vk+1 (T ) = Vk+1 (H ):
66
Chapter 4
The Markov Property 4.1 Binomial Model Pricing and Hedging Recall that Vm is the given simple European derivative security, and the value and portfolio processes are given by:
Vk = (1 + r)k If E [(1 + r),m Vm jF k ]; k = 0; 1; : : : ; m , 1: ) , Vk+1(!1 ; : : : ; !k ; T ) ; k = 0; 1; : : : ; m , 1: k (!1 ; : : : ; !k ) = SVk+1 ((!!1;; :: :: :: ;; !!k ;; H H k+1 1 k ) , Sk+1 (!1 ; : : : ; !k ; T ) r,d Example 4.1 (Lookback Option) u = 2; d = 0:5; r = 0:25; S0 = 4; p~ = 1+ u,d = 0:5; q~ = 1 , p~ = Consider a simple European derivative security with expiration 2, with payoff given by (See Fig. 4.1):
V2 = 0max (S , 5)+ : k2 k Notice that
V2 (HH) = 11; V2 (HT) = 3 6= V2 (TH) = 0; V2 (TT ) = 0:
The payoff is thus “path dependent”. Working backward in time, we have:
V1 (H) = 1 +1 r [~pV2 (HH) + q~V2 (HT)] = 45 [0:5 11 + 0:5 3] = 5:60; V1 (T) = 45 [0:5 0 + 0:5 0] = 0; V0 = 54 [0:5 5:60 + 0:5 0] = 2:24:
Using these values, we can now compute:
, V1 (T) = 0:93; 0 = SV1(H) (H) , S (T) 1
1
, V2 (HT) 1(H) = SV2 (HH) (HH) , S (HT) = 0:67; 2
2
67
0:5:
68 S2 (HH) = 16
S (H) = 8 1 S2 (HT) = 4 S =4 0
S2 (TH) = 4 S1 (T) = 2
S2 (TT) = 1
Figure 4.1: Stock price underlying the lookback option.
, V2 (TT ) = 0: 1(T) = SV2 (TH) (TH) , S (TT ) 2
2
Working forward in time, we can check that
X1 (H) = 0S1 (H) + (1 + r)(X0 , 0S0 ) = 5:59; V1 (H) = 5:60; X1 (T) = 0S1 (T ) + (1 + r)(X0 , 0S0 ) = 0:01; V1 (T) = 0; X1 (HH) = 1(H)S1 (HH) + (1 + r)(X1 (H) , 1(H)S1 (H)) = 11:01; V1 (HH) = 11; etc.
Example 4.2 (European Call) Let u = 2; d = with expiration time 2 and payoff function
1 2
; r = 14 ; S0 = 4; p~ = q~ =
1 2
, and consider a European call
V2 = (S2 , 5)+ : Note that
V2 (HH) = 11; V2 (HT) = V2 (TH) = 0; V2 (TT) = 0; V1 (H) = 45 [ 21 :11 + 12 :0] = 4:40 V1(T) = 45 [ 21 :0 + 12 :0] = 0 V0 = 45 [ 21 4:40 + 12 0] = 1:76: Define vk (x) to be the value of the call at time k when Sk = x. Then v2 (x) = (x , 5)+ v1(x) = 45 [ 12 v2 (2x) + 12 v2 (x=2)]; v0(x) = 45 [ 12 v1 (2x) + 12 v1 (x=2)]:
CHAPTER 4. The Markov Property
69
In particular,
v2(16) = 11; v2 (4) = 0; v2 (1) = 0; v1(8) = 45 [ 21 :11 + 12 :0] = 4:40; v1(2) = 45 [ 12 :0 + 21 :0] = 0; v0 = 45 [ 21 4:40 + 12 0] = 1:76: Let k (x) be the number of shares in the hedging portfolio at time k when Sk = x. Then , vk+1 (x=2) ; k = 0; 1: k (x) = vk+1 (2x) 2x , x=2
4.2 Computational Issues For a model with n periods (coin tosses), equations of the form
has 2n
elements. For period k, we must solve 2k
Vk (!1 ; : : : ; !k ) = 1 +1 r [~pVk+1(!1; : : : ; !k ; H ) + q~Vk+1 (!1; : : : ; !k ; T )]: For example, a threemonth option has 66 trading days. If each day is taken to be one period, then n = 66 and 266 7 1019. There are three possible ways to deal with this problem: 1. Simulation. We have, for example, that
V0 = (1 + r),n If EVn; and so we could compute V0 by simulation. More specifically, we could simulate n coin tosses ! = (!1; : : : ; !n ) under the riskneutral probability measure. We could store the value of Vn (! ). We could repeat this several times and take the average value of Vn as an approximation to If EVn. 2. Approximate a manyperiod model by a continuoustime model. Then we can use calculus and partial differential equations. We’ll get to that. 3. Look for Markov structure. Example 4.2 has this. In period 2, the option in Example 4.2 has three possible values v2 (16); v2(4); v2(1), rather than four possible values V2(HH ); V2(HT ); V2(TH ); V2(TT ). If there were 66 periods, then in period 66 there would be 67 possible stock price values (since the final price depends only on the number of upticks of the stock price – i.e., heads – so far) and hence only 67 possible option values, rather than 266 7 1019.
70
4.3 Markov Processes Technical condition always present: We consider only functions on IR and subsets of IR which are Borelmeasurable, i.e., we only consider subsets A of R I that are in B and functions g : IR!IR such that g ,1 is a function B!B. Definition 4.1 () Let ( ; F ; P) be a probability space. Let fF k gnk=0 be a filtration under fXk gnk=0 be a stochastic process on ( ; F ; P). This process is said to be Markov if:
4.3.1
F.
Let
The stochastic process fXk g is adapted to the filtration fF k g, and
(The Markov Property). For each k = 0; 1; : : : ; n , 1, the distribution of Xk+1 conditioned on F k is the same as the distribution of Xk+1 conditioned on Xk .
Different ways to write the Markov property
(a) (Agreement of distributions). For every A 2 B
4 = B(IR), we have
IP (Xk+1 2 AjF k ) = IE [IA(Xk+1)jF k ] = IE [IA(Xk+1 )jXk ] = IP [Xk+1 2 AjXk ]: (b) (Agreement of expectations of all functions). For every (Borelmeasurable) function h : IR!IR for which IE jh(Xk+1 )j < 1, we have
IE [h(Xk+1)jF k ] = IE [h(Xk+1)jXk]: (c) (Agreement of Laplace transforms.) For every u 2 IR for which IEeuXk+1
< 1, we have
IE euXk+1 F k = IE euXk+1 Xk : (If we fix u and define h(x) = eux , then the equations in (b) and (c) are the same. However in (b) we have a condition which holds for every function h, and in (c) we assume this condition only for functions h of the form h(x) = eux . A main result in the theory of Laplace transforms is that if the equation holds for every h of this special form, then it holds for every h, i.e., (c) implies (b).) (d) (Agreement of characteristic functions) For every u 2 IR, we have
h
i
h
i
IE eiuXk+1 jF k = IE eiuXk+1 jXk ;
where i = 1.)
p,1. (Since jeiuxj = j cos x +sin xj 1 we don’t need to assume that IE jeiuxj <
CHAPTER 4. The Markov Property
71
Remark 4.1 In every case of the Markov properties where IE [: : : jXk ] appears, we could just as well write g (Xk ) for some function g . For example, form (a) of the Markov property can be restated as: For every A 2 B, we have
IP (Xk+1 2 AjF k ) = g (Xk); where g is a function that depends on the set A. Conditions (a)(d) are equivalent. The Markov property as stated in (a)(d) involves the process at a “current” time k and one future time k + 1. Conditions (a)(d) are also equivalent to conditions involving the process at time k and multiple future times. We write these apparently stronger but actually equivalent conditions below. Consequences of the Markov property. Let j be a positive integer. (A) For every Ak+1
IR; : : : ; Ak+j IR,
IP [Xk+1 2 Ak+1 ; : : : ; Xk+j 2 Ak+j jF k ] = IP [Xk+1 2 Ak+1; : : : ; Xk+j 2 Ak+j jXk ]: (A’) For every A 2 IRj ,
IP [(Xk+1 ; : : : ; Xk+j ) 2 AjF k ] = IP [(Xk+1 ; : : : ; Xk+j ) 2 AjXk ]: (B) For every function h : IRj !IR for which IE jh(Xk+1; : : : ; Xk+j )j < 1, we have
IE [h(Xk+1; : : : ; Xk+j )jF k ] = IE [h(Xk+1; : : : ; Xk+j )jXk ]: (C) For every u = (uk+1 ; : : : ; uk+j ) 2 IRj for which IE jeuk+1 Xk+1 +:::+uk+j Xk+j j < 1, we have
IE [euk
+1
Xk+1 +:::+uk+j Xk+j jF k ] = IE [euk+1 Xk+1 +:::+uk+j Xk+j jXk ]:
(D) For every u = (uk+1 ; : : : ; uk+j ) 2 IRj we have
IE [ei(uk
+1
Xk+1 +:::+uk+j Xk+j ) jF
i(u X +:::+uk+j Xk+j ) jX ]: k ] = IE [e k+1 k+1 k
Once again, every expression of the form IE (: : : jXk ) can also be written as g (Xk ), where the function g depends on the random variable represented by : : : in this expression. Remark. All these Markov properties have analogues for vectorvalued processes.
72 Proof that (b) Consider
=) (A). (with j = 2 in (A)) Assume (b).
Then (a) also holds (take
h = IA).
IP [Xk+1 2 Ak+1 ; Xk+2 2 Ak+2 jF k ] = IE [IAk+1 (Xk+1 )IAk+2 (Xk+2 )jF k ] (Definition of conditional probability)
= IE [IE [IAk+1 (Xk+1 )IAk+2 (Xk+2 )jF k+1 ]jF k ] (Tower property)
= IE [IAk+1 (Xk+1 ):IE [IAk+2 (Xk+2)jF k+1 ]jF k ] (Taking out what is known)
= IE [IAk+1 (Xk+1 ):IE [IAk+2 (Xk+2)jXk+1 ]jF k ] (Markov property, form (a).)
= IE [IAk+1 (Xk+1 ):g (Xk+1)jF k ] (Remark 4.1)
= IE [IAk+1 (Xk+1 ):g (Xk+1)jXk ] (Markov property, form (b).) Now take conditional expectation on both sides of the above equation, conditioned on (Xk ), and use the tower property on the left, to obtain
IP [Xk+1 2 Ak+1 ; Xk+2 2 Ak+2 jXk ] = IE [IAk+1 (Xk+1):g (Xk+1)jXk ]: Since both
IP [Xk+1 2 Ak+1; Xk+2 2 Ak+2 jF k ]
and
IP [Xk+1 2 Ak+1; Xk+2 2 Ak+2 jXk ]
are equal to the RHS of (3.1)), they are equal to each other, and this is property (A) with j
(3.1)
= 2.
Example 4.3 It is intuitively clear that the stock price process in the binomial model is a Markov process. We will formally prove this later. If we want to estimate the distribution of Sk+1 based on the information in F k , the only relevant piece of information is the value of Sk . For example,
e k+1jF k ] = (~pu + q~d)Sk = (1 + r)Sk IE[S
(3.2)
is a function of Sk . Note however that form (b) of the Markov property is stronger then (3.2); the Markov property requires that for any function h,
e k+1)jF k] IE[h(S is a function of Sk . Equation (3.2) is the case of h(x) = x.
Consider a model with 66 periods and a simple European derivative security whose payoff at time 66 is
V66 = 31 (S64 + S65 + S66 ):
CHAPTER 4. The Markov Property
73
The value of this security at time 50 is
e + r),66V66jF 50] V50 = (1 + r)50IE[(1 e 66jS50]; = (1 + r),16IE[V because the stock price process is Markov. (We are using form (B) of the Markov property here). In other words, the F50measurable random variable V50 can be written as
V50(!1 ; : : : ; !50) = g(S50 (! 1; : : : ; !50)) for some function g, which we can determine with a bit of work.
4.4 Showing that a process is Markov Definition 4.2 (Independence) Let ( ; F ; P) be a probability space, and let G and H be subalgebras of F . We say that G and H are independent if for every A 2 G and B 2 H, we have
IP (A \ B) = IP (A)IP (B): We say that a random variable X is independent of a algebra G if (X ), the algebra generated by X , is independent of G . Example 4.4 Consider the twoperiod binomial model. Recall that F 1 is the algebra of sets determined by the first toss, i.e., F 1 contains the four sets
AH =4 fHH; HT g; AT =4 fTH; TT g; ; : Let H be the algebra of sets determined by the second toss, i.e., H contains the four sets
fHH; T H g; fHT; T T g; ; : Then F 1 and H are independent. For example, if we take A = fHH; HT g from F 1 and B = fHH; T H g from H, then IP(A \ B) = IP(HH) = p2 and IP(A)IP(B) = (p2 + pq)(p2 + pq) = p2 (p + q)2 = p2: Note that F 1 and S2 are not independent (unless p = 1 or p = 0). For example, one of the sets in (S2 ) is f!; S2 (!) = u2S0 g = fHH g. If we take A = fHH; HT g from F 1 and B = fHH g from (S2 ), then IP(A \ B) = IP(HH) = p2 , but
IP(A)IP(B) = (p2 + pq)p2 = p3(p + q) = p3 :
The following lemma will be very useful in showing that a process is Markov: Lemma 4.15 (Independence Lemma) Let X and Y be random variables on a probability space ( ; F ; P). Let G be a subalgebra of F . Assume
74
X is independent of G ; Y is G measurable. Let f (x; y ) be a function of two variables, and define
g (y ) =4 IEf (X; y): Then
IE [f (X; Y )jG] = g (Y ):
Remark. In this lemma and the following discussion, capital letters denote random variables and lower case letters denote nonrandom variables. Example 4.5 (Showing the stock price process is Markov) Consider an nperiod binomial model. Fix a
4 Sk+1 and G = 4 F . Then X = u if ! = H and X = d if ! = T . Since X = k k+1 k+1 Sk 4 depends only on the (k + 1)st toss, X is independent of G . Define Y = Sk , so that Y is G measurable. Let h 4 be any function and set f(x; y) = h(xy). Then time k and define X
4 IEf(X; y) = IEh(Xy) = ph(uy) + qh(dy): g(y) =
The Independence Lemma asserts that
IE[h(Sk+1 )jF k ] = = = =
S
IE[h jF k ] IE[f(X; Y )jG ] g(Y ) ph(uSk ) + qh(dSk ): k+1 Sk :Sk
This shows the stock price is Markov. Indeed, if we condition both sides of the above equation on (Sk ) and use the tower property on the left and the fact that the right hand side is (Sk )measurable, we obtain
IE[h(Sk+1 )jSk ] = ph(uSk ) + qh(dSk ): Thus IE[h(Sk+1 )jF k ] and IE[h(Sk+1 )jXk ] are equal and form (b) of the Markov property is proved.
Not only have we shown that the stock price process is Markov, but we have also obtained a formula for
IE[h(Sk+1 )jF k ] as a function of Sk . This is a special case of Remark 4.1.
4.5 Application to Exotic Options Consider an nperiod binomial model. Define the running maximum of the stock price to be
Mk =4 1max S: j k j
Consider a simple European derivative security with payoff at time n of vn (Sn ; Mn ). Examples:
CHAPTER 4. The Markov Property
75
vn(Sn; Mn) = (Mn , K )+ (Lookback option); vn(Sn; Mn) = IMnB (Sn , K )+ (Knockin Barrier option). Lemma 5.16 The twodimensional process f(Sk ; Mk )gnk=0 is Markov. (Here we are working under the riskneutral measure IP, although that does not matter). Proof: Fix k. We have
Mk+1 = Mk _ Sk+1;
where _ indicates the maximum of two quantities. Let Z
4 Sk+1 = Sk , so
If P (Z = u) = p~; If P (Z = d) = q~;
and Z is independent of F k . Let h(x; y ) be a function of two variables. We have
h(Sk+1 ; Mk+1) = h(Sk+1 ; Mk _ Sk+1 ) = h(ZSk ; Mk _ (ZSk )): Define
g(x; y) =4 If Eh(Zx; y _ (Zx)) = p~h(ux; y _ (ux)) + q~h(dx; y _ (dx)): The Independence Lemma implies
If E[h(Sk+1 ; Mk+1 )jF k ] = g(Sk; Mk ) = p~h(uSk ; Mk _ (uSk )) + q~h(dSk; Mk ); the second equality being a consequence of the fact that Mk ^ dSk = Mk . Since the RHS is a function of (Sk ; Mk ), we have proved the Markov property (form (b)) for this twodimensional process. Continuing with the exotic option of the previous Lemma... Let Vk denote the value of the derivative security at time k. Since (1 + r),k Vk is a martingale under If P , we have
Vk = 1 +1 r If E [Vk+1jF k ]; k = 0; 1; : : : ; n , 1: At the final time, we have
Vn = vn (Sn; Mn):
Stepping back one step, we can compute
Vn,1 = 1 +1 r If E [vn (Sn; Mn)jF n,1 ] = 1 +1 r [~pvn (uSn,1 ; uSn,1 _ Mn,1 ) + q~vn (dSn,1 ; Mn,1)] :
76 This leads us to define
vn,1 (x; y ) =4 1 +1 r [~pvn (ux; ux _ y ) + q~vn(dx; y)] so that
Vn,1 = vn,1 (Sn,1; Mn,1):
The general algorithm is
vk (x; y) = 1 +1 r p~vk+1 (ux; ux _ y ) + q~vk+1 (dx; y ) ; and the value of the option at time k is vk (Sk ; Mk ). Since this is a simple European option, the hedging portfolio is given by the usual formula, which in this case is
k = vk+1 (uSk ; (uSk )(u_,Mdk))S, vk+1 (dSk ; Mk ) k
Chapter 5
Stopping Times and American Options 5.1 American Pricing Let us first review the European pricing formula in a Markov model. Consider the Binomial model with n periods. Let Vn = g (Sn ) be the payoff of a derivative security. Define by backward recursion:
vn (x) = g (x) vk (x) = 1 +1 r [~pvk+1 (ux) + q~vk+1(dx)]: Then vk (Sk ) is the value of the option at time k, and the hedging portfolio is given by k = vk+1 (uS(uk ),,d)vSk+1 (dSk ) ; k = 0; 1; 2; : : : ; n , 1: k Now consider an American option. Again a function g is specified. In any period k, the holder of the derivative security can “exercise” and receive payment g (Sk ). Thus, the hedging portfolio should create a wealth process which satisfies
Xk g (Sk); 8k;
almost surely.
This is because the value of the derivative security at time k is at least g (Sk ), and the wealth process value at that time must equal the value of the derivative security. American algorithm.
vn (x) = g (x) 1 vk (x) = max 1 + r (~pvk+1 (ux) + q~vk+1 (dx)); g (x) Then vk (Sk ) is the value of the option at time k. 77
78 S2 (HH) = 16 v2 (16) = 0
S (H) = 8 1 S2 (HT) = 4 S =4 0
v (4) = 1 2
S2 (TH) = 4 S1 (T) = 2
S2 (TT) = 1
v (1) = 4 2
Figure 5.1: Stock price and final value of an American put option with strike price 5. Example 5.1 See Fig. 5.1. Then
S0 = 4; u = 2; d = 12 ; r = 14 ; p~ = q~ = 12 ; n = 2. Set v2 (x) = g(x) = (5 , x)+ .
4
v (8) = max 5 :0 + :1 ; (5 , 8) 2 1
= = v1 (2) = = = v0 (4) = = =
1 2
1 2
+
max 5 ; 0 0:40 max 45 12 :1 + 12 :4 ; (5 , 2)+ maxf2; 3g 3:00 4 + 1 1 max 5 2 :(0:4) + 2 :(3:0) ; (5 , 4) maxf1:36; 1g 1:36
Let us now construct the hedging portfolio for this option. Begin with initial wealth X0 0 as follows:
0:40 = = = = 3:00 = = = =
v1 (S1 (H)) S1 (H)0 + (1 + r)(X0 , 0S0 ) 80 + 54 (1:36 , 40) 30 + 1:70 =) 0 = ,0:43 v1 (S1 (T )) S1 (T)0 + (1 + r)(X0 , 0S0 ) 20 + 54 (1:36 , 40) ,30 + 1:70 =) 0 = ,0:43
= 1:36. Compute
CHAPTER 5. Stopping Times and American Options
79
Using 0 = ,0:43 results in
X1 (H) = v1(S1 (H)) = 0:40; X1 (T ) = v1 (S1 (T )) = 3:00 Now let us compute 1 (Recall that S1 (T) = 2):
1 = = = = 4 = = = =
v2 (4) S2 (TH)1(T) + (1 + r)(X1 (T ) , 1(T)S1 (T)) 41(T) + 45 (3 , 21(T)) 1:51(T) + 3:75 =) 1(T ) = ,1:83 v2 (1) S2 (TT )1(T) + (1 + r)(X1 (T) , 1 (T )S1 (T)) 1(T ) + 54 (3 , 21(T)) ,1:51(T) + 3:75 =) 1(T) = ,0:16
We get different answers for 1(T )! If we had X1 (T) = 2, the value of the European put, we would have
1 = 1:51(T ) + 2:5 =) 1 (T) = ,1; 4 = ,1:51(T) + 2:5 =) 1(T ) = ,1;
5.2 Value of Portfolio Hedging an American Option
Xk+1 = k Sk+1 + (1 + r)(Xk , Ck , k Sk ) = (1 + r)Xk + k (Sk+1 , (1 + r)Sk ) , (1 + r)Ck Here, Ck is the amount “consumed” at time k.
The discounted value of the portfolio is a supermartingale. The value satisfies Xk
g(Sk); k = 0; 1; : : : ; n.
The value process is the smallest process with these properties.
When do you consume? If
If E((1 + r),(k+1) vk+1 (Sk+1 )jF k ] < (1 + r),k vk (Sk ); or, equivalently,
If E( 1 +1 r vk+1(Sk+1)jF k ] < vk (Sk )
80 and the holder of the American option does not exercise, then the seller of the option can consume to close the gap. By doing this, he can ensure that Xk = vk (Sk ) for all k, where vk is the value defined by the American algorithm in Section 5.1. In the previous example, v1 (S1(T )) = 3; v2(S2(TH )) = 1 and v2(S2 (TT )) = 4. Therefore,
h
If E [ 1 +1 r v2(S2 )jF 1 ](T ) = 45 12 :1 + 21 :4 5 4 =
i
5 2 = 2; v1 (S1(T )) = 3;
so there is a gap of size 1. If the owner of the option does not exercise it at time one in the state ! 1 = T , then the seller can consume 1 at time 1. Thereafter, he uses the usual hedging portfolio
k = vk+1 (uS(uk ),,d)vSk+1 (dSk ) k
In the example, we have v1 (S1(T )) = g (S1(T )). It is optimal for the owner of the American option to exercise whenever its value vk (Sk ) agrees with its intrinsic value g (Sk ). Definition 5.1 (Stopping Time) Let ( ; F ; P) be a probability space and let fF k gnk=0 be a filtration. A stopping time is a random variable : !f0; 1; 2; : : : ; ng [ f1g with the property that:
f! 2 ; (!) = kg 2 F k ; 8k = 0; 1; : : : ; n; 1: Example 5.2 Consider the binomial model with n = 2; S0 = 4; u = 2; d = 12 ; r = 14 , so p~ = v0 ; v1; v2 be the value functions defined for the American put with strike price 5. Define
q~ =
1 2
. Let
(!) = minfk; vk(Sk ) = (5 , Sk )+ g: The stopping time corresponds to “stopping the first time the value of the option agrees with its intrinsic value”. It is an optimal exercise time. We note that
(!) =
1
2
if ! if !
2 AT 2 AH
We verify that is indeed a stopping time:
f!; (!) = 0g = 2 F 0 f!; (!) = 1g = AT 2 F 1 f!; (!) = 2g = AH 2 F 2
Example 5.3 (A random time which is not a stopping time) In the same binomial model as in the previous example, define
(!) = minfk; Sk (!) = m2 (!)g;
CHAPTER 5. Stopping Times and American Options
81
4
where m2 = min0j 2 Sj . In other words, stops when the stock price reaches its minimum value. This random variable is given by 0 if ! 2 AH ; (!) = 1 if ! = TH; 2 if ! = TT
8 < :
We verify that is not a stopping time:
f!; (!) = 0g = AH 62 F 0 f!; (!) = 1g = fTH g 62 F 1 f!; (!) = 2g = fTT g 2 F 2
5.3 Information up to a Stopping Time Definition 5.2 Let be a stopping time. We say that a set A is determined by time provided that
A \ f!; (!) = kg 2 F k ; 8k: The collection of sets determined by is a algebra, which we denote by F . Example 5.4 In the binomial model considered earlier, let
= minfk; vk (Sk ) = (5 , Sk )+ g; i.e.,
(!) =
1
2
if ! if !
2 AT 2 AH
The set fHT g is determined by time , but the set fTH g is not. Indeed,
fHT g \ f!; (!) = 0g = 2 F 0 fHT g \ f!; (!) = 1g = 2 F 1 fHT g \ f!; (!) = 2g = fHT g 2 F 2 but The atoms of F are
fTH g \ f!; (!) = 1g = fTH g 62 F 1 : fHT g; fHH g; AT = fTH; TT g:
Notation 5.1 (Value of Stochastic Process at a Stopping Time) If ( ; F ; P) is a probability space, fF k gnk=0 is a filtration under F , fXk gnk=0 is a stochastic process adapted to this filtration, and is a stopping time with respect to the same filtration, then X is an F measurable random variable whose value at ! is given by
X (!) =4 X (!) (!):
82
n Theorem 3.17 (Optional Sampling) Suppose that fYk ; F k g1 k=0 (or fYk ; F k gk=0 ) is a submartingale. Let and be bounded stopping times, i.e., there is a nonrandom number n such that
n; n; If
almost surely.
almost surely, then
Y IE (Y jF ): Taking expectations, we obtain IEY IEY , and in particular, Y0 = IEY0 IEY . If fYk ; F k g1 k=0 is a supermartingale, then implies Y IE (Y jF ). If fYk ; F k g1 k=0 is a martingale, then implies Y = IE (YjF ). Example 5.5 In the example 5.4 considered earlier, we define (!) = 2 for all ! 2 . Under the riskneutral probability measure, the discounted stock price process ( 54 ),k Sk is a martingale. We compute
" # IEe 45 S F : 2
2
The atoms of F are fHH g; fHT g; and AT . Therefore,
" # eIE 4 S F (HH) = 4 S (HH); 5 5 " # IEe 45 S F (HT) = 45 S (HT ); 2
2
2
2
2
2
2
and for !
2 AT ,
2
" # IEe 45 S F (!) =
4
4 1 5 S2 (TH) + 2 5 = 12 2:56 + 12 0:64 = 1:60
2
2
2
1 2
In every case we have gotten (see Fig. 5.2)
" # ! S ! (!): IEe 45 S F (!) = 45 2
(
2
)
(
)
2
S2 (TT )
CHAPTER 5. Stopping Times and American Options
83
(16/25) S (HH) = 10.24 2
(4/5) S (H) = 6.40 1 (16/25) S2 (HT) = 2.56 S =4 0
(16/25) S2 (TH) = 2.56 (4/5) S (T) = 1.60 1
(16/25) S2 (TT) = 0.64
Figure 5.2: Illustrating the optional sampling theorem.
84
Chapter 6
Properties of American Derivative Securities 6.1 The properties Definition 6.1 An American derivative security is a sequence of nonnegative random variables fGk gnk=0 such that each Gk is F k measurable. The owner of an American derivative security can exercise at any time k, and if he does, he receives the payment Gk . (a) The value Vk of the security at time k is
kf , Vk = max (1 + r) IE [(1 + r) G jF k ];
where the maximum is over all stopping times satisfying
k almost surely.
(b) The discounted value process f(1 + r),k Vk gnk=0 is the smallest supermartingale which satisfies
Vk Gk ; 8k;
(c) Any stopping time which satisfies
almost surely.
V0 = If E [(1 + r), G ]
is an optimal exercise time. In particular
=4 minfk; Vk = Gk g is an optimal exercise time. (d) The hedging portfolio is given by
) , Vk+1 (!1 ; : : : ; !k ; T ) ; k = 0; 1; : : : ; n , 1: k (!1 ; : : : ; !k ) = SVk+1 ((!!1;; :: :: :: ;; !!k ;; H H ) , S (! ; : : : ; ! ; T ) k+1 1
k
k+1 1
85
k
86 (e) Suppose for some k and ! , we have Vk (!) = Gk (!). Then the owner of the derivative security should exercise it. If he does not, then the seller of the security can immediately consume
Vk (!) , 1 +1 r If E[Vk+1jF k ](!) and still maintain the hedge.
6.2 Proofs of the Properties Let fGk gnk=0 be a sequence of nonnegative random variables such that each Gk is F k measurable. Define Tk to be the set of all stopping times satisfying k n almost surely. Define also
Vk =4 (1 + r)k max If E [(1 + r), G jF k ] : 2T k
Lemma 2.18
Vk Gk for every k.
Proof: Take
2 Tk to be the constant k.
Lemma 2.19 The process f(1 + r),k Vk gnk=0 is a supermartingale. Proof: Let attain the maximum in the definition of Vk+1 , i.e.,
h
i
(1 + r),(k+1) Vk+1 = If E (1 + r), G jF k+1 : Because is also in Tk , we have
h i If E[(1 + r),(k+1)Vk+1jF k ] = If E If E [(1 + r), G jF k+1 ]jF k = If E [(1 + r), G jF k ] max If E [(1 + r), G jF k ] 2Tk = (1 + r),k Vk :
Lemma 2.20 If fYk gnk=0 is another process satisfying
Yk Gk ; k = 0; 1; : : : ; n;
a.s.,
and f(1 + r),k Yk gnk=0 is a supermartingale, then
Yk Vk ; k = 0; 1; : : : ; n;
a.s.
CHAPTER 6. Properties of American Derivative Securities
87
Proof: The optional sampling theorem for the supermartingale f(1 + r),k Yk gnk=0 implies
If E[(1 + r), Y jF k ] (1 + r),k Yk ; 8 2 Tk :
Therefore,
Vk = (1 + r)k max If E [(1 + r), G jF k ] 2T
=
k
(1 + r)k max If E [(1 + r), Y jF k ] 2Tk (1 + r),k (1 + r)k Yk Yk :
Lemma 2.21 Define
Ck = Vk , 1 +1 r If E [Vk+1 jF k ] n o = (1 + r)k (1 + r),k Vk , If E[(1 + r),(k+1) Vk+1jF k ] : Since f(1 + r),k Vk gnk=0 is a supermartingale, Ck must be nonnegative almost surely. Define
Set X0
H ) , Vk+1 (!1; : : : ; !k ; T ) : k (!1 ; : : : ; !k ) = SVk+1 ((!!1;; :: :: :: ;; !!k ;; H ) , S (! ; : : : ; ! ; T ) k+1 1
k
k+1 1
= V0 and define recursively Xk+1 = k Sk+1 + (1 + r)(Xk , Ck , k Sk ):
Then
k
Xk = Vk 8k:
We proceed by induction on k. The induction hypothesis is that Xk = Vk k 2 f0; 1; : : : ; n , 1g, i.e., for each fixed (!1; : : : ; !k ) we have Xk (!1 ; : : : ; !k ) = Vk(!1 ; : : : ; !k ): Proof:
We need to show that
Xk+1 (!1; : : : ; !k ; H ) = Vk+1(!1; : : : ; !k ; H ); Xk+1 (!1 ; : : : ; !k ; T ) = Vk+1(!1; : : : ; !k ; T ): We prove the first equality; the proof of the second is similar. Note first that
Vk (!1 ; : : : ; !k ) , Ck (!1; : : : ; !k ) = 1 If E[V jF ](! ; : : : ; !k) 1 + r k+1 k 1
= 1 +1 r (~pVk+1 (!1; : : : ; !k ; H ) + q~Vk+1 (!1 ; : : : ; !k ; T )) :
for some
88 Since (!1 ; : : : ; !k ) will be fixed for the rest of the proof, we will suppress these symbols. For example, the last equation can be written simply as
Vk , Ck = 1 +1 r (~pVk+1 (H ) + q~Vk+1(T )) : We compute
Xk+1 (H ) = k Sk+1(H ) + (1 + r)(Xk , Ck , k Sk ) ) , Vk+1 (T ) (S (H ) , (1 + r)S ) = SVk+1 ((H k+1 k k+1 H ) , Sk+1 (T ) +(1 + r)(Vk , Ck ) V ) , Vk+1 (T ) (uS , (1 + r)S ) = k+1 ((H k k u , d)Sk +~pVk+1 (H ) + q~Vk+1 (T ) = (Vk+1 (H ) , Vk+1 (T ))~q + p~Vk+1(H ) + q~Vk+1 (T ) = Vk+1 (H ): 6.3 Compound European Derivative Securities In order to derive the optimal stopping time for an American derivative security, it will be useful to study compound European derivative securities, which are also interesting in their own right. A compound European derivative security consists of n + 1 different simple European derivative securities (with the same underlying stock) expiring at times 0; 1; : : : ; n; the security that expires at time j has payoff Cj . Thus a compound European derivative security is specified by the process fCj gnj=0 , where each Cj is F j measurable, i.e., the process fCj gnj=0 is adapted to the filtration fF k gnk=0. Hedging a short position (one payment). Here is how we can hedge a short position in the j ’th European derivative security. The value of European derivative security j at time k is given by
Vk(j) = (1 + r)k If E [(1 + r),j Cj jF k ]; k = 0; : : : ; j;
and the hedging portfolio for that security is given by
j ) (! ; : : : ; ! ; H ) , V (j ) (! ; : : : ; ! ; T ) Vk(+1 k k ( j ) k+1 1 k (!1 ; : : : ; !k ) = (j ) 1 ; k = 0; : : : ; j , 1: ( j ) Sk+1 (!1 ; : : : ; !k ; H ) , Sk+1 (!1; : : : ; !k ; T ) (j ) (j ) (j ) Thus, starting with wealth V0 , and using the portfolio (0 ; : : : ; j ,1 ), we can ensure that at time j we have wealth Cj . Hedging a short position (all payments). Superpose the hedges for the individual payments. In P (j) other words, start with wealth V0 = nj=0 V0 . At each time k 2 f0; 1; : : : ; n , 1g, first make the payment Ck and then use the portfolio
k = k (k+1) + k (k+2) + : : : + k (n)
CHAPTER 6. Properties of American Derivative Securities
89
corresponding to all future payments. At the final time n, after making the final payment Cn , we will have exactly zero wealth. Suppose you own a compound European derivative securityfCj gnj=0 . Compute
V0 =
n X j =0
2n 3 X V0(j ) = If E 4 (1 + r),j Cj 5 j =0
,1 . You can borrow V0 and consume it immediately. This leaves and the hedging portfolio is fk gnk=0 you with wealth X0 = ,V0 . In each period k, receive the payment Ck and then use the portfolio ,k . At the final time n, after receiving the last payment Cn, your wealth will reach zero, i.e., you will no longer have a debt.
6.4 Optimal Exercise of American Derivative Security In this section we derive the optimal exercise time for the owner of an American derivative security. Let fGk gnk=0 be an American derivative security. Let be the stopping time the owner plans to use. (We assume that each Gk is nonnegative, so we may assume without loss of generality that the owner stops at expiration – time n– if not before). Using the stopping time , in period j the owner will receive the payment
Cj = If =jg Gj :
In other words, once he chooses a stopping time, the owner has effectively converted the American derivative security into a compound European derivative security, whose value is
V0( )
2n 3 X = If E 4 (1 + r),j Cj 5 2j=0 3 n X = If E 4 (1 + r),j If =j gGj 5 j =0
= If E [(1 + r), G ]: The owner of the American derivative security can borrow this amount of money immediately, if he chooses, and invest in the market so as to exaclty pay off his debt as the payments fCj gnj=0 are
( )
received. Thus, his optimal behavior is to use a stopping time which maximizes V0 . Lemma 4.22
V0( ) is maximized by the stopping time = minfk; Vk = Gk g:
Proof: Recall the definition
V0 =4 max If E [(1 + r), G ] = max V ( ) 2T 2T 0 0
0
90
h
i
Let 0 be a stopping time which maximizes V0 , i.e., V0 = If E (1 + r), G 0 : Because f(1 + r),k Vk gnk=0 is a supermartingale, we have from the optional sampling theorem and the inequality Vk Gk , the following:
( )
h
V0 If E (1 + r), 0 V 0 jF 0 h i = If E (1 + r), 0 V 0 h i IfE (1 + r), 0 G 0 = V0: Therefore,
h
i
0
i
h
i
V0 = If E (1 + r), 0 V 0 = If E (1 + r), 0 G 0 ;
and
V 0 = G 0 ;
a.s.
We have just shown that if 0 attains the maximum in the formula
V0 = max If E [(1 + r), G ] ; 2T
(4.1)
0
then
V 0 = G 0 ;
But we have defined and so we must have
a.s.
= minfk; Vk = Gk g;
0 n almost surely. The optional sampling theorem implies , V (1 + r), G = (1 + r ) h i IfE (1 + r), 0 V 0 jF h i = If E (1 + r), 0 G 0 jF :
Taking expectations on both sides, we obtain
h
i
h
i
If E (1 + r), G If E (1 + r), 0 G 0 = V0 : It follows that also attains the maximum in (4.1), and is therefore an optimal exercise time for the American derivative security.
Chapter 7
Jensen’s Inequality 7.1 Jensen’s Inequality for Conditional Expectations Lemma 1.23 If ' : IR!IR is convex and IE j'(X )j < 1, then
IE ['(X )jG] '(IE [X jG]): For instance, if G
= f; g; '(x) = x2 :
IEX 2 (IEX )2: Proof: Since ' is convex we can express it as follows (See Fig. 7.1):
'(x) = max h(x): h' h is linear
Now let h(x) = ax + b lie below '. Then,
IE ['(X )jG] IE [aX + bjG] = aIE [X jG] + b = h(IE [X jG]) This implies
IE ['(X )jG]
max h(IE [X jG]) h'
h is linear
= '(IE [X jG]):
91
92
ϕ
Figure 7.1: Expressing a convex function as a max over linear functions. Theorem 1.24 If fYk gnk=0 is a martingale and is convex then f'(Yk )gnk=0 is a submartingale. Proof:
IE ['(Yk+1)jF k ] '(IE [Yk+1 jF k ]) = '(Yk ):
7.2 Optimal Exercise of an American Call This follows from Jensen’s inequality. Corollary 2.25 Given a convex function g : [0; 1)!IR where g (0) = 0. For instance, g (x) = (x , K )+ is the payoff function for an American call. Assume that r 0. Consider the American derivative security with payoff g (Sk ) in period k. The value of this security is the same as the value of the simple European derivative security with final payoff g (Sn), i.e., , f If E [(1 + r),n g (Sn)] = max IE [(1 + r) g (S )] ; where the LHS is the European value and the RHS is the American value. In particular optimal exercise time. Proof: Because g is convex, for all 2 [0; 1] we have (see Fig. 7.2):
g(x) = g (x + (1 , ):0) g(x) + (1 , ):g(0) = g (x):
= n is an
CHAPTER 7. Jensen’s Inequality
93
(x,g(x))
( λ x, λ g(x))
x ( λ x, g( λ x)) Figure 7.2: Proof of Cor. 2.25
and
Therefore,
g 1 +1 r Sk+1 1 +1 r g (Sk+1)
h
i
If E (1 + r),(k+1) g(Sk+1)jF k = (1 + r),k If E 1 +1 r g (Sk+1)jF k (1 + r),k IfE g 1 +1 r Sk+1 jF k (1 + r),k g IfE 1 +1 r Sk+1jF k = (1 + r),k g (Sk ); So f(1 + r),k g (Sk )gnk=0 is a submartingale. Let be a stopping time satisfying 0 n. The
optional sampling theorem implies
(1 + r), g (S ) If E [(1 + r),n g (Sn)jF ] :
Taking expectations, we obtain
If E [(1 + r), g(S )] If E If E [(1 + r),n g (Sn)jF ] = If E [(1 + r),ng (Sn)] :
Therefore, the value of the American derivative security is
, ,n f f max IE [(1 + r) g (S )] IE [(1 + r) g (Sn )] ;
and this last expression is the value of the European derivative security. Of course, the LHS cannot be strictly less than the RHS above, since stopping at time n is always allowed, and we conclude that
, ,n f f max IE [(1 + r) g (S )] = IE [(1 + r) g (Sn )] :
94 S2 (HH) = 16
S (H) = 8 1 S2 (HT) = 4 S =4 0
S2 (TH) = 4 S1 (T) = 2
S2 (TT) = 1
Figure 7.3: A three period binomial model.
7.3 Stopped Martingales Let fYk gnk=0 be a stochastic process and let stopped process
be a stopping time.
We denote by
fYk^ gnk=0 the
Yk^ (!)(!); k = 0; 1; : : : ; n:
Example 7.1 (Stopped Process) Figure 7.3 shows our familiar 3period binomial example. Define
(!) =
1
2
if if
!1 = T; !1 = H:
8 S (HH) = 16 > < =4 S ^ ! (!) = > SS (HT) (T) = : S (T) = 22
Then
2
2
(
2
)
1 1
if if if if
! = HH; ! = HT; ! = TH; ! = TT:
Theorem 3.26 A stopped martingale (or submartingale, or supermartingale) is still a martingale (or submartingale, or supermartingale respectively). Proof: Let fYk gnk=0 be a martingale, and be a stopping time. Choose some k 2 f0; 1; : : : ; ng. The set f kg is in F k , so the set f k + 1g = f kgc is also in F k . We compute
h
i
h
IE Y(k+1)^ jF k = IE If kgY + If k+1gYk+1 jF k = If kgY + If k+1gIE [Yk+1 jF k ] = If kgY + If k+1gYk = Yk^ :
i
CHAPTER 7. Jensen’s Inequality
95
96
Chapter 8
Random Walks 8.1 First Passage Time Toss a coin infinitely many times. Then the sample space is the set of all infinite sequences ! = (!1; !2; : : : ) of H and T . Assume the tosses are independent, and on each toss, the probability of H is 12 , as is the probability of T . Define
Yj (!) =
(
1
,1
if if
! j = H; ! j = T;
M0 = 0; Mk =
k X
j =1
Yj ; k = 1; 2; : : :
The process fMk g1 k=0 is a symmetric random walk (see Fig. 8.1) Its analogue in continuous time is Brownian motion. Define
= minfk 0; Mk = 1g: If Mk never gets to 1 (e.g., ! = (TTTT : : : )), then = 1. The random variable is called the
first passage time to 1. It is the first time the number of heads exceeds by one the number of tails.
8.2 is almost surely finite 1 It is shown in a Homework Problem that fMk g1 k=0 and fNk gk=0 where
(
, Nk = exp Mk , k log e +2 e
=
eMk
2 k e + e, 97
!)
98
Mk k
Figure 8.1: The random walk process Mk
eθ+ e−θ 2
2 eθ+ e−θ 1
1
θ
θ Figure 8.2: Illustrating two functions of are martingales. (Take Mk = ,Sk in part (i) of the Homework Problem and take (v).) Since N0 = 1 and a stopped martingale is a martingale, we have
"
k^ 2 M k ^ 1 = IENk^ = IE e e + e,
= , in part
#
(2.1)
for every fixed 2 IR (See Fig. 8.2 for an illustration of the various functions involved). We want to let k!1 in (2.1), but we have to worry a bit that for some sequences ! 2 , (! ) = 1. We consider fixed
As k!1,
Furthermore, Mk^
> 0, so
2 < 1: e + e,
(
2 k^ ! e +2e, e + e, 0
if if
< 1; =1
1, because we stop this martingale when it reaches 1, so 0 eMk^ e
CHAPTER 8. Random Walks and
99
0 eMk^
In addition,
lim k!1
eMk^
2 k^ e : e + e,
2 k ^ = e + e,
"
Recall Equation (2.1):
IE eMk^
( e
0
2 e +e,
if if
< 1; = 1:
#
2 k^ = 1 e + e,
Letting k!1, and using the Bounded Convergence Theorem, we obtain
IE For all
2 (0; 1], we have
e
2 I e + e, f<1g = 1:
(2.2)
0 e e +2 e, If<1g e;
so we can let #0 in (2.2), using the Bounded Convergence Theorem again, to conclude
h
i
IE If < 1g = 1; i.e.,
IP f < 1g = 1:
We know there are paths of the symmetric random walk fMk g1 k=0 which never reach level 1. We have just shown that these paths collectively have no probability. (In our infinite sample space , each path individually has zero probability). We therefore do not need the indicator If < 1g in (2.2), and we rewrite that equation as
IE
2 = e, : e + e,
8.3 The moment generating function for Let 2 (0; 1) be given. We want to find
> 0 so that
2 = e + e, :
Solution:
e + e, , 2 = 0 (e, )2 , 2e, + = 0
(2.3)
100
p
2 e, = 1 1 , :
< 1. Now 0 < < 1, so
We want > 0, so we must have e,
0 < (1 , )2 < (1 , ) < 1 , 2 ;
p
1 , < 1 , 2 ; p 1 , 1 , 2 < ; p 1 , 1 , 2 < 1
We take the negative square root:
Recall Equation (2.3):
IE With 2 (0; 1) and
p
2 e, = 1 , 1 , :
> 0 related by
2 = e, ; > 0: e + e,
p
2 e, = 1 , 1 , ;
= this becomes
IE
2 ; e + e,
p
2 = 1 , 1 , ; 0 < < 1:
We have computed the moment generating function for the first passage time to 1.
8.4 Expectation of Recall that
IE so
p
2 = 1 , 1 , ; 0 < < 1;
d IE = IE ( ,1) d p 2! d 1 , 1, = d p 2 = 1 ,2 p 1 , 2 : 1,
(3.1)
CHAPTER 8. Random Walks
101
Using the Monotone Convergence Theorem, we can let "1 in the equation
IE ( ,1) = to obtain
p
1 ,p 1 , 2 ; 2 1 , 2
IE = 1:
Thus in summary:
=4 minfk; Mk = 1g; IP f < 1g = 1; IE = 1:
8.5 The Strong Markov Property The random walk process fMk g1 k=0 is a Markov process, i.e.,
IE [
Mk+1 ; Mk+2; : : : j F k ] jMk ] :
random variable depending only on
= IE [
same random variable
In discrete time, this Markov property implies the Strong Markov property:
IE [
M +1 ; M +2; : : : j F ] j M ] :
random variable depending only on
= IE [
same random variable
for any almost surely finite stopping time .
8.6 General First Passage Times Define
m =4 minfk 0; Mk = mg; m = 1; 2; : : :
Then 2 , 1 is the number of periods between the first arrival at level 1 and the first arrival at level 2. The distribution of 2 , 1 is the same as the distribution of 1 (see Fig. 8.3), i.e.,
IE2 ,1
p
2 = 1 , 1 , ; 2 (0; 1):
102
Mk k τ1
τ2
τ 2 − τ1 Figure 8.3: General first passage times. For 2 (0; 1),
IE [ jF ] = IE , jF = IE [ , jF ] 2
1
2
1
2
1
1
1
1
1
(taking out what is known)
= IE [ , jM ] 2
1
=
1
1
(strong Markov property)
1 IE [2,1 ]
(M = 1; not random p 2! ) = 1 , 1 , : 1
1
Take expectations of both sides to get
p
2 IE = IE : 1 , 1 , 2
1
= In general,
IEm
!
p
1 , 1 , 2
!2
p 2 !m 1 , 1, = ; 2 (0; 1):
8.7 Example: Perpetual American Put Consider the binomial model, with u = 2; d = 12 ; r neutral probabilities are p~ = 12 , q~ = 12 , and thus
=
1 , and payoff function (5 , S )+ . The risk k 4
Sk = S0 uMk ;
CHAPTER 8. Random Walks where
103
Mk is a symmetric random walk under the riskneutral measure, denoted by If P.
S0 = 4. Here are some possible exercise rules: Rule 0: Stop immediately. 0
Suppose
= 0; V ( ) = 1. 0
Rule 1: Stop as soon as stock price falls to 2, i.e., at time
,1 =4 minfk; Mk = ,1g: Rule 2: Stop as soon as stock price falls to 1, i.e., at time
,2 =4 minfk; Mk = ,2g: Because the random walk is symmetric under If P , ,m has the same distribution under If P as the stopping time m in the previous section. This observation leads to the following computations of value. Value of Rule 1:
V (, ) = If E (1 + r),, (5 , S, )+ h i = (5 , 2)+IE ( 54 ), 1
1
1
1
q
1 , 1 , ( 45 )2
= 3: = 23 :
4 5
Value of Rule 2:
h
V (, ) = (5 , 1)+If E ( 45 ), = 4:( 12 )2 = 1: 2
2
i
This suggests that the optimal rule is Rule 1, i.e., stop (exercise the put) as soon as the stock price falls to 2, and the value of the put is 32 if S0 = 4.
Suppose instead we start with S0 = 8, and stop the first time the price falls to 2. This requires 2 down steps, so the value of this rule with this initial stock price is
i h (5 , 2)+ If E ( 54 ), = 3:( 12 )2 = 34 : 2
In general, if S0 = 2j for some j 1, and we stop when the stock price falls to 2, then j , 1 down steps will be required and the value of the option is
h
i
(5 , 2)+ If E ( 54 ), j, = 3:( 12 )j,1 : We define
(
1)
v(2j ) =4 3:( 12 )j,1 ; j = 1; 2; 3; : : :
104 If S0 = 2j for some j 1, then the initial price is at or below 2. In this case, we exercise immediately, and the value of the put is
v(2j ) =4 5 , 2j ; j = 1; 0; ,1; ,2; : : : Proposed exercise rule: Exercise the put whenever the stock price is at or below 2. The value of this rule is given by v (2j ) as we just defined it. Since the put is perpetual, the initial time is no different from any other time. This leads us to make the following: Conjecture 1 The value of the perpetual put at time k is v (Sk ). How do we recognize the value of an American derivative security when we see it? There are three parts to the proof of the conjecture. We must show:
v (Sk ) (5 , Sk )+ 8k; n o1 (b) ( 45 )k v (Sk ) is a supermartingale, k=0 (c) fv (Sk )g1 k=0 is the smallest process with properties (a) and (b). (a)
Note: To simplify matters, we shall only consider initial stock prices of the form S0 always of the form 2j , with a possibly different j . Proof: (a). Just check that
v(2j ) =4 3:( 21 )j,1 (5 , 2j )+ v(2j ) =4 5 , 2j (5 , 2j )+
for for
j 1; j 1:
This is straightforward. Proof: (b). We must show that
h
i
v(Sk ) If E 45 v (Sk+1 )jF k = 45 : 12 v (2Sk ) + 45 : 12 v ( 12 Sk ): By assumption, Sk = 2j for some j . We must show that v(2j ) 52 v (2j +1) + 25 v (2j ,1 ): If j 2, then v (2j ) = 3:( 12 )j ,1 and 2 v (2j +1 ) + 2 v (2j ,1) 5 5 = 25 :3:( 21 )j + 25 :3:( 12 )j ,2 1 = 3: 52 : 4 + 25 ( 12 )j ,2 = 3: 12 :( 12 )j ,2 = v (2j ):
= 2j , so Sk is
CHAPTER 8. Random Walks If j
105
= 1, then v (2j ) = v (2) = 3 and 2 v (2j +1 ) + 2 v (2j ,1) 5 5 = 52 v (4) + 25 v (1) = 25 :3: 21 + 25 :4
= 3=5 + 8=5 = 2 15 < v (2) = 3
There is a gap of size 45 .
If j
0, then v(2j ) = 5 , 2j and 2 v (2j +1 ) + 2 v (2j ,1) 5 5 = 25 (5 , 2j +1 ) + 52 (5 , 2j ,1 ) = 4 , 25 (4 + 1)2j ,1 = 4 , 2j < v (2j ) = 5 , 2j :
There is a gap of size 1. This concludes the proof of (b). Proof: (c). Suppose fYk gnk=0 is some other process satisfying: (a’) (b’)
Yk (5 , Sk )+ 8k; f( 54 )k Yk g1k=0 is a supermartingale.
We must show that
Yk v (Sk ) 8k:
(7.1)
Actually, since the put is perpetual, every time k is like every other time, so it will suffice to show
Y0 v(S0);
(7.2)
provided we let S0 in (7.2) be any number of the form 2j . With appropriate (but messy) conditioning on F k , the proof we give of (7.2) can be modified to prove (7.1).
For j
1,
so if S0
v(2j ) = 5 , 2j = (5 , 2j )+;
= 2j for some j 1, then (a’) implies
Y0 (5 , 2j )+ = v (S0): Suppose now that S0
= 2j for some j 2, i.e., S0 4. Let
= minfk; Sk = 2g = minfk; Mk = j , 1g:
106 Then
v(S0) = v (2j ) = 3:( 12 )j,1 h i = IE ( 45 ) (5 , S )+ : Because f( 45 )k Yk g1 k=0 is a supermartingale
h
i
h
i
Y0 IE ( 45 ) Y IE ( 45 ) (5 , S )+ = v (S0): Comment on the proof of (c): If the candidate value process is the actual value of a particular exercise rule, then (c) will be automatically satisfied. In this case, we constructed v so that v (Sk ) is the value of the put at time k if the stock price at time k is Sk and if we exercise the put the first time (k, or later) that the stock price is 2 or less. In such a situation, we need only verify properties (a) and (b).
8.8 Difference Equation If we imagine stock prices which can fall at any point in (0; 1), not just at points of the form 2j for integers j , then we can imagine the function v (x), defined for all x > 0, which gives the value of the perpetual American put when the stock price is x. This function should satisfy the conditions:
v (x) (K , x)+ ; 8x, 1 [~ (b) v (x) 1+ r pv (ux) + q~v (dx)] ; 8x; (c) At each x, either (a) or (b) holds with equality.
(a)
In the example we worked out, we have For
j 1 : v (2j ) = 3:( 21 )j,1 = 26j ; For
This suggests the formula
v(x) =
j 1 : v (2j ) = 5 , 2j :
(
x 3; 5 , x; 0 < x 3: 6 x;
We then have (see Fig. 8.4): (a)
v (x) (5 , x)+ ; 8x;
(b)
v (x) 54 12 v(2x) + 12 v( x2 )
h
i
for every x except for 2 < x < 4.
CHAPTER 8. Random Walks
107
v(x) 5 (3,2)
5
x
Figure 8.4: Graph of v (x). Check of condition (c):
If 0 < x 3, then (a) holds with equality. If x 6, then (b) holds with equality: x 4 1 v (2x) + 1 v ( ) 5 2 2 2
=
4 1 6 + 1 12 5 2 2x 2 x
6 = x:
If 3 < x < 4 or 4 < x < 6, then both (a) and (b) are strict.
This is an artifact of the discreteness of the binomial model. This artifact will disappear in the continuous model, in which an analogue of (a) or (b) holds with equality at every point.
8.9 Distribution of First Passage Times Let fMk g1 k=0 be a symetric random walk under a probability measure IP , with M0
= 0. Defining
= minfk 0; Mk = 1g; we recall that
p
2 IE = 1 , 1 , ; 0 < < 1:
We will use this moment generating function to obtain the distribution of . We first obtain the Taylor series expasion of IE as follows:
108
p
f (x) = 1 , 1 , x; 1 f 0(x) = 21 (1 , x), 2 ; 3 f 00(x) = 14 (1 , x), 2 ;
f (0) = 0 f 0 (0) = 21 f 00(0) = 14
f 000(x) = 38 (1 , x), ; f 000(0) = 38 ::: ( j ) f (x) = 1 3 : :2:j (2j , 3) (1 , x), f (j)(0) = 1 3 : : : (2j , 3) 5 2
j,1)
(2
2
;
2j
= 1 3 : :2: j (2j , 3) : 2 42j ,:1:(j: ,(21)!j , 2) 2j,1 (2j , 2)! = 21 (j , 1)!
The Taylor series expansion of f (x) is given by
p
f (x) = 1 , 1 , x 1 1 X (j ) j = j ! f (0)x =
j =0 1 X j =1
1 2j ,1 (2j , 2)! xj 2 j !(j , 1)!
!
1 2j ,1 1 X 2j , 2 xj : 1 = x2 + 2 (j , 1) j j =2 So we have
IE
p
2 = 1 , 1 , = 1 f (2 )
! 1 2j ,1 1 X 2 j , 2 = + : 2 j =2 2 (j , 1) j But also,
IE =
1 X j =1
2j ,1 IP f = 2j , 1g:
CHAPTER 8. Random Walks
109
Figure 8.5: Reflection principle.
Figure 8.6: Example with j
= 2.
Therefore,
IP f = 1g =
1; 2
1 2j,1 1 2j , 2! IP f = 2j , 1g = 2 (j , 1) j ; j = 2; 3; : : : 8.10 The Reflection Principle To count how many paths reach level 1 by time 2j , 1, count all those for which M2j ,1 double count all those for which M2j ,1 3. (See Figures 8.5, 8.6.)
= 1 and
110 In other words,
IP f 2j , 1g = IP fM2j,1 = 1g + 2IP fM2j,1 3g = IP fM2j ,1 = 1g + IP fM2j ,1 3g + IP fM2j ,1 ,3g = 1 , IP fM2j ,1 = ,1g: For j
2, IP f = 2j , 1g = IP f 2j , 1g , IP f 2j , 3g = [1 , IP fM2j ,1 = ,1g] , [1 , IP fM2j ,3 = ,1g] = IP fM2j ,3 = ,1g , IP fM2j ,1 = ,1g 2j,3 (2j , 3)! 1 2j,1 (2j , 1)! = 12 , (j , 1)!(j , 2)! 2 j !(j , 1)! 1 2j,1 (2j , 3)! = 2 j !(j , 1)! [4j (j , 1) , (2j , 1)(2j , 2)] 2j,1 (2j , 3)! = 12 j !(j , 1)! [2j (2j , 2) , (2j , 1)(2j , 2)] 2j,1 (2j , 2)! = 12 j !(j , 1)! 1 2j,1 1 2j , 2! : = 2 (j , 1) j
Chapter 9
Pricing in terms of Market Probabilities: The RadonNikodym Theorem. 9.1 RadonNikodym Theorem
P be two probability measures on a space ( ; F ). Theorem 1.27 (RadonNikodym) Let IP and If Assume that for every A 2 F satisfying IP (A) = 0, we also have If P (A) = 0. Then we say that If P is absolutely continuous with respect to IP. Under this assumption, there is a nonegative random variable Z such that If P (A) =
Z
A
ZdIP; 8A 2 F ;
(1.1)
and Z is called the RadonNikodym derivative of If P with respect to IP. Remark 9.1 Equation (1.1) implies the apparently stronger condition
If EX = IE [XZ ] for every random variable X for which IE jXZ j < 1. P is absolutely continuous with respect to PI , and IP is absolutely continuous with Remark 9.2 If If f respect to IP , we say that IP and If P are equivalent. IP and If P are equivalent if and only if IP (A) = 0 exactly when If P (A) = 0; 8A 2 F : If IP and If P are equivalent and Z is the RadonNikodym derivative of If P w.r.t. PI , then Z1 is the P , i.e., RadonNikodym derivative of IP w.r.t. If If EX = IE [XZ ] 8X; (1.2) IEY = If E [Y: Z1 ] 8Y: (Let X and Y be related by the equation Y = XZ to see that (1.2) and (1.3) are the same.) 111
(1.3)
112 Example 9.1 (RadonNikodym Theorem) Let = fHH; HT; T H; T T g, the set of coin toss sequences of length 2. Let P correspond to probability 13 for H and 23 for T , and let IP correspond to probability 12 for
H and 12
e
e for T . Then Z(!) = IP ! , so (
)
IP (! )
9: Z(HH) = 94 ; Z(HT) = 98 ; Z(TH) = 89 ; Z(TT) = 16
9.2 RadonNikodym Martingales Let be the set of all sequences of n coin tosses. Let P I be the market probability measure and let If P be the riskneutral probability measure. Assume
IP (!) > 0; If P (!) > 0; 8! 2 ;
so that P I and If P are equivalent. The RadonNikodym derivative of If P with respect to IP is
f Z (!) = IIPP ((!!)) :
Define the P I martingale
Zk =4 IE [Z jF k ]; k = 0; 1; : : : ; n:
We can check that Zk is indeed a martingale:
IE [Zk+1jF k ] = IE [IE [Z jF k+1 ]jF k ] = IE [Z jF k ] = Zk :
Lemma 2.28 If X is F k measurable, then If EX Proof:
= IE [XZk ].
If EX = IE [XZ ] = IE [IE [XZ jF k ]] = IE [X:IE [Z jF k ]] = IE [XZk ]:
Note that Lemma 2.28 implies that if X is F k measurable, then for any A 2 F k ,
If E [IA X ] = IE [Zk IA X ];
or equivalently,
Z A
XdIf P=
Z A
XZk dIP:
CHAPTER 9. Pricing in terms of Market Probabilities
113 Z2 (HH) = 9/4
1/3 Z (H) = 3/2 1
Z =1 0
1/3
2/3
2/3
1/3
Z2 (HT) = 9/8 Z2 (TH) = 9/8
Z1 (T) = 3/4 2/3 Z2 (TT) = 9/16
Figure 9.1: Showing the Zk values in the 2period binomial model example. The probabilities shown are for IP, not If P. Lemma 2.29 If X is F k measurable and 0 j
k, then
If E [X jF j ] = Z1 IE [XZk jF j ]: j
Proof: Note first that Z1
j
IE [XZk jF j ] is F j measurable. So for any A 2 F j , we have
Z Z 1 f IE [ XZ jF ] d I P = IE [XZk jF j ]dIP k j A A Zj Z = =
ZA A
XZk dIP XdIf P
(Lemma 2.28)
(Partial averaging) (Lemma 2.28)
Example 9.2 (RadonNikodym Theorem, continued) We show in Fig. 9.1 the values of the martingale Zk . We always have Z0 = 1, since
Z0 = IEZ =
Z
e = 1: ZdIP = IP( )
9.3 The State Price Density Process In order to express the value of a derivative security in terms of the market probabilities, it will be useful to introduce the following state price density process:
k = (1 + r),k Zk ; k = 0; : : : ; n:
114 We then have the following pricing formulas: For a Simple European derivative security with payoff Ck at time k,
h
i
V0 = If E (1 + r),k Ck h i = IE (1 + r),k Zk Ck = IE [k Ck ]: More generally for 0 j
k,
h
(Lemma 2.28)
i
Vj = (1 + r)j If E (1 + r),k Ck jF j r)j IE h(1 + r),k Z C jF i = (1 + k k j Zj = 1 IE [k Ck jF j ]
(Lemma 2.29)
j
Remark 9.3
fj Vj gkj=0 is a martingale under IP, as we can check below: IE [j+1Vj+1jF j ] = IE [IE [kCk jF j +1]jF j ] = IE [k Ck jF j ] = j Vj :
Now for an American derivative security fGk gnk=0 :
V0 = sup If E [(1 + r), G ] 2T = sup IE [(1 + r), Z G ] 2T = sup IE [ G ]: 0
0
2T0
More generally for 0 j
n,
Vj = (1 + r)j sup If E [(1 + r), G jF j ] 2Tj
= (1 + r)j sup Z1 IE [(1 + r), Z G jF j ] 2Tj j
= 1 sup IE [ G jF j ]: j 2Tj
Remark 9.4 Note that
fj Vj gnj=0 is a supermartingale under IP, (b) j Vj j Gj 8j;
(a)
CHAPTER 9. Pricing in terms of Market Probabilities
115 ζ (ΗΗ) = 1.44 2 S2 (HH) = 16
1/3 ζ (Η) = 1.20 1 S (H) = 8 1 1/3
2/3
S =4 0 2/3 ζ0 = 1.00
1/3 S1 (T) = 2 ζ (Τ) = 0.6 1
2/3
ζ (ΗΤ) = 0.72 2 S2 (HT) = 4 S2 (TH) = 4 ζ (ΤΗ) = 0.72 2
ζ (ΤΤ) = 0.36 2 S2 (TT) = 1
Figure 9.2: Showing the state price values k . The probabilities shown are for IP, not If P. (c)
fj Vj gnj=0 is the smallest process having properties (a) and (b).
We interpret k by observing that k (! )IP (! ) is the value at time zero of a contract which pays $1 at time k if ! occurs. Example 9.3 (RadonNikodymTheorem, continued) We illustrate the use of the valuation formulas for European and American derivative securities in terms of market probabilities. Recall that p = 31 , q = 23 . The state price values k are shown in Fig. 9.2. For a European Call with strike price 5, expiration time 2, we have
V2 (HH) = 11; 2(HH)V2 (HH) = 1:44 11 = 15:84: V2 (HT ) = V2(TH) = V2 (TT ) = 0: V0 = 13 13 15:84 = 1:76: 2 (HH) 1:44 1 (HH) V2(HH) = 1:20 11 = 1:20 11 = 13:20 V1 (H) = 13 13:20 = 4:40 Compare with the riskneutral pricing formulas:
V1(H) = 25 V1 (HH) + 25 V1(HT) = 25 11 = 4:40; V1 (T ) = 52 V1 (TH) + 25 V1 (TT ) = 0; V0 = 25 V1 (H) + 52 V1 (T) = 25 4:40 = 1:76: Now consider an American put with strike price 5 and expiration time 2. Fig. 9.3 shows the values of k (5 , Sk )+ . We compute the value of the put under various stopping times : (0) Stop immediately: value is 1. (1) If (HH) = (HT) = 2;
(TH) = (TT ) = 1, the value is 1 2 0:72 + 2 1:80 = 1:36: 3 3 3
116 (5  S2(HH))+= 0 (5  S (H))+ = 0 1 ζ1(H) (5  S (H))+ = 0 1 1/3 (5S 0)+=1 ζ0 (5S 0)+=1
ζ2(HH) (5  S2(HH))+= 0
1/3
(5  S2(HT))+= 1
ζ2(HT) (5  S2(HT))+= 0.72
2/3
2/3
(5  S (TH))+= 1 2 ζ (TH) (5  S (TH))+= 0.72 2 2
1/3
(5  S1(T))+ = 3 ζ1(T) (5  S1(T))+ = 1.80
2/3
+ (5  S2(TT)) = 4 + ζ2(TT) (5  S2(TT)) = 1.44
Figure 9.3: Showing the values k (5 , Sk )+ for an American put. The probabilities shown are for IP, not If P. (2) If we stop at time 2, the value is
1 2 0:72 + 2 1 0:72 + 2 2 1:44 = 0:96 3 3 3 3 3 3 We see that (1) is optimal stopping rule.
9.4 Stochastic Volatility Binomial Model Let be the set of sequences of n tosses, and let 0 < dk are F k measurable. Also let
< 1+ rk < uk , where for each k, dk ; uk ; rk
p~k = 1 +u r,k ,d dk ; q~k = uk u, (1, +d rk) : k
k
Let If P be the riskneutral probability measure:
and for 2 k n,
k
k
If P f!1 = H g = p~0 ; If P f!1 = T g = q~0 ;
If P [!k+1 = H jF k ] = p~k ; If P [!k+1 = T jF k ] = q~k : Let IP be the market probability measure, and assume IP f! g > 0 8! 2 .
equivalent. Define
If P (!) 8! 2 ; Z (!) = IP (! )
Then IP and
If P are
CHAPTER 9. Pricing in terms of Market Probabilities
117
Zk = IE [Z jF k ]; k = 0; 1; : : : ; n: We define the money market price process as follows:
M0 = 1; Mk = (1 + rk,1)Mk,1 ; k = 1; : : : ; n:
Note that Mk is Fk,1 measurable.
We then define the state price process to be
k = M1 Zk ; k = 0; : : : ; n:
k ,1 . The selffinancing value process (wealth process) As before the portfolio process is fk gnk=0 consists of X0, the nonrandom initial wealth, and
Xk+1 = k Sk+1 + (1 + rk )(Xk , k Sk ); k = 0; : : : ; n , 1: Then the following processes are martingales under If P:
1 n M Sk k
k=0
and
1 n ; M Xk k
k=0
and the following processes are martingales under IP:
fk Sk gnk=0
and
fk Xk gnk=0:
We thus have the following pricing formulas: Simple European derivative security with payoff Ck at time k:
C f Vj = Mj IE Mk F j k 1 = IE [k Ck jF j ] j
American derivative security
fGk gnk=0:
G f Vj = Mj sup IE M F j 2Tj
= 1 sup IE [ G jF j ] : j 2Tj
The usual hedging portfolio formulas still work.
118
9.5 Another Applicaton of the RadonNikodym Theorem Let ( ; F ; Q) be a probability space. Let G be a subalgebra of F , and let X be a nonnegative R random variable with X dQ = 1. We construct the conditional expectation (under Q) of X given G . On G , define two probability measures
IP (A) = Q(A) 8A 2 G ; If P (A) =
Z
XdQ 8A 2 G :
A
Whenever Y is a G measurable random variable, we have
Z
1
Y dIP =
Z
Y dQ;
if Y = A for some A 2 G , this is just the definition of IP , and the rest follows from the “standard machine”. If A 2 G and IP (A) = 0, then Q(A) = 0, so If P (A) = 0. In other words, the measure If P f is absolutely continuous with respect to the measure IP . The RadonNikodym theorem implies that there exists a G measurable random variable Z such that
If P (A) =4 i.e.,
Z A
Z
X dQ =
Z dIP 8A 2 G ;
A
Z A
Z dIP 8A 2 G :
This shows that Z has the “partial averaging” property, and since Z is G measurable, it is the conditional expectation (under the probability measure Q) of X given G . The existence of conditional expectations is a consequence of the RadonNikodym theorem.
Chapter 10
Capital Asset Pricing 10.1 An Optimization Problem Consider an agent who has initial wealth X0 and wants to invest in the stock and money markets so as to maximize
IE log Xn:
Remark 10.1 Regardless of the portfolio used by the agent, fk Xk g1 k=0 is a martingale under IP, so
IEnXn = X0
(BC )
Here, (BC) stands for “Budget Constraint”. Remark 10.2 If is any random variable satisfying (BC), i.e.,
IEn = X0 ; then there is a portfolio which starts with initial wealth X0 and produces Xn = at time n. To see this, just regard as a simple European derivative security paying off at time n. Then X0 is its value at time 0, and starting from this value, there is a hedging portfolio which produces Xn = . Remarks 10.1 and 10.2 show that the optimal obtained by solving the following
Xn
for the capital asset pricing problem can be
Constrained Optimization Problem: Find a random variable which solves:
IE log IEn = X0:
Maximize Subject to Equivalently, we wish to Maximize
X ! 2
(log (! )) IP (! )
119
120
X
Subject to
n(!)(!)IP (!) , X0 = 0:
!2
n There are 2 sequences ! in . Call them ! 1 ; !2 ; : : : ; !2n . Adopt the notation x1 = (!1 ); x2 = (!2); : : : ; x2n = (!2n ): We can thus restate the problem as:
2n X
Maximize
Subject to
2n X
k=1
k=1
(log xk )IP (! k )
n (!k )xk IP (! k ) , Xo = 0:
In order to solve this problem we use: Theorem 1.30 (Lagrange Multiplier) If (x1; : : : ; xm) solve the problem Maxmize Subject to then there is a number such that
f (x1 ; : : : ; xm)
g (x1; : : : ; xm) = 0;
@ f (x; : : : ; x ) = @ g (x; : : : ; x ); k = 1; : : : ; m; m m @xk 1 @xk 1
(1.1)
g (x1; : : : ; xm) = 0:
(1.2)
and
For our problem, (1.1) and (1.2) become
1 n xk IP (! k ) = n (!k )IP (!k ); k = 1; : : : ; 2 ; 2n X
k=1
n (!k )xk IP (!k ) = X0 :
Equation (1.1’) implies
(1:20 )
xk = 1(! ) : n k
Plugging this into (1.2’) we get
2 1X 1 IP (! k ) = X0 =) = X0: n
k=1
(1:10 )
CHAPTER 10. Capital Asset Pricing Therefore,
121
xk = X(!0 ) ; k = 1; : : : ; 2n: n k
Thus we have shown that if solves the problem Maximize Subject to
IE log IE ( n ) = X0;
(1.3)
then
= X 0 :
(1.4)
n
Theorem 1.31 If is given by (1.4), then solves the problem (1.3). Proof: Fix Z
> 0 and define
f (x) = log x , xZ:
We maximize f over x > 0:
f 0 (x) = x1 , Z = 0 () x = Z1 ; The function f is maximized at x
f 00(x) = , x12 < 0; 8x 2 IR: = Z1 , i.e.,
log x , xZ f (x ) = log Z1 , 1; 8x > 0; 8Z > 0:
Let be any random variable satisfying
IE (n ) = X0 and let
= X 0 : n
From (1.5) we have
log , Xn log X 0 , 1: 0
n
Taking expectations, we have
IE log , X1 IE (n ) IE log , 1; 0
and so
IE log IE log :
(1.5)
122 In summary, capital asset pricing works as follows: Consider an agent who has initial wealth and wants to invest in the stock and money market so as to maximize
IE log Xn: The optimal Xn is Xn
= Xn , i.e., 0
n Xn = X0:
Since fk Xk gnk=0 is a martingale under IP, we have
k Xk = IE [nXn jF k ] = X0; k = 0; : : : ; n; so
Xk = X 0 ; k
and the optimal portfolio is given by
) , k+1 (!1 ; : : : ; !k ;T ) : k (!1 ; : : : ; !k ) = S k+1(!(!1; ;: :: :: :; ;!!k; ;H k+1 1 k H ) , Sk+1 (!1; : : : ; !k ; T )
X0
X0
X0
Chapter 11
General Random Variables 11.1 Law of a Random Variable Thus far we have considered only random variables whose domain and range are discrete. We now consider a general random variable X : !IR defined on the probability space ( ; F ; P). Recall that:
F is a algebra of subsets of . IP is a probability measure on F , i.e., IP (A) is defined for every A 2 F . A function X : !IR is a random variable if and only if for every Borel subsets of R I ), the set
B 2 B(IR) (the algebra of
fX 2 Bg =4 X ,1(B) =4 f!; X (!) 2 Bg 2 F ; i.e., X 11.1)
: !IR is a random variable if and only if X ,1 is a function from B(IR) to F (See Fig.
Thus any random variable X induces a measure X on the measurable space by
(IR; B(IR)) defined
X (B) = IP X ,1(B) 8B 2 B(IR); where the probabiliy on the right is defined since X ,1(B ) 2 F . X is often called the Law of X –
in Williams’ book this is denoted by LX .
11.2 Density of a Random Variable The density of X (if it exists) is a function fX
X (B) =
Z
B
: IR![0; 1) such that
fX (x) dx 8B 2 B (IR): 123
124
X
R
{X ε B}
B
Ω
Figure 11.1: Illustrating a realvalued random variable X . We then write
dX (x) = fX (x)dx;
where the integral is with respect to the Lebesgue measure on IR. fX is the RadonNikodym derivative of X with respect to the Lebesgue measure. Thus X has a density if and only if X is absolutely continuous with respect to Lebesgue measure, which means that whenever B 2 B(IR) has Lebesgue measure zero, then
IP fX 2 Bg = 0:
11.3 Expectation Theorem 3.32 (Expectation of a function of X ) Let h : IR!IR be given. Then
Z IEh(X ) =4 h(X (!)) dIP (! ) = =
1
Z
ZIR IR
h(x) dX (x) h(x)fX (x) dx:
IR, then these equations are IE 1B (X ) =4 P fX 2 B g = Z X (B )
Proof: (Sketch). If h(x) = B (x) for some B
=
B
fX (x) dx;
which are true by definition. Now use the “standard machine” to get the equations for general h.
CHAPTER 11. General Random Variables
125
(X,Y)
C y { (X,Y)ε C}
Ω x
Figure 11.2: Two realvalued random variables X; Y .
11.4 Two random variables Let X; Y be two random variables !IR defined on the space ( ; F ; P). Then measure on B(IR2 ) (see Fig. 11.2) called the joint law of (X; Y ), defined by
X; Y
induce a
X;Y (C ) =4 IP f(X; Y ) 2 C g 8C 2 B(IR2 ): The joint density of (X; Y ) is a function
fX;Y : IR2 ![0; 1) that satisfies
X;Y (C ) = fX;Y
ZZ C
fX;Y (x; y) dxdy 8C 2 B(IR2 ):
is the RadonNikodym derivative of X;Y with respect to the Lebesgue measure (area) on IR2 .
We compute the expectation of a function of X; Y in a manner analogous to the univariate case:
Z IEk(X; Y ) =4 k(X (!); Y (!)) dIP (!) =
ZZ
IR ZZ
k(x; y ) dX;Y (x; y )
2
=
IR
2
k(x; y )fX;Y (x; y ) dxdy
126
11.5 Marginal Density
IR be given. Then
Suppose (X; Y ) has joint density fX;Y . Let B
Y (B) = IP fY 2 Bg = IP f(X; Y ) 2 IR B g = Z X;YZ (IR B ) = fX;Y (x; y ) dxdy
ZB
= where
fY
B
IR
fY (y ) dy;
Z
4 (y ) =
IR
fX;Y (x; y) dx:
Therefore, fY (y ) is the (marginal) density for Y .
11.6 Conditional Expectation 4 (X; Y ) has joint density fX;Y . Let h : IR!IR be given. Recall that IE [h(X )jY ] = IE [h(X )j(Y )] depends on ! through Y , i.e., there is a function g (y ) (g depending on h) such that Suppose
IE [h(X )jY ](!) = g (Y (!)): How do we determine g ?
We can characterize g using partial averaging: Recall that A 2 (Y )()A = B 2 B(IR). Then the following are equivalent characterizations of g :
Z
A
g (Y ) dIP =
Z
Z
A
fY 2 Bg for some
h(X ) dIP 8A 2 (Y );
Z
1 (Y )g(Y ) dIP = 1B (Y )h(X ) dIP 8B 2 B(IR);
B
Z
1B (y)g(y)Y (dy) =
IR
Z B
g(y)fY (y) dy =
(6.1)
ZZ IR2
Z Z B IR
(6.2)
1B (y)h(x) dX;Y (x; y) 8B 2 B(IR);
(6.3)
h(x)fX;Y (x; y) dxdy 8B 2 B (IR):
(6.4)
CHAPTER 11. General Random Variables
127
11.7 Conditional Density A function fX jY (xjy ) : IR2 ![0; 1) is called a conditional density for X given Y provided that for any function h : IR!IR:
g (y ) = (Here g is the function satisfying
Z
IR
h(x)fX jY (xjy ) dx:
(7.1)
IE [h(X )jY ] = g (Y );
and g depends on h, but fX jY does not.)
Theorem 7.33 If (X; Y ) has a joint density fX;Y , then
(x; y ) : fX jY (xjy ) = fX;Y f (y )
(7.2)
Y
Proof: Just verify that g defined by (7.1) satisfies (6.4): For B
Z Z
B  IR
h(x)fX jY (xjy) dx fY (y ) dy =
{z
g(y)
}
Z Z
B IR
2 B(IR);
h(x)fX;Y (x; y ) dxdy:
Notation 11.1 Let g be the function satisfying
IE [h(X )jY ] = g (Y ): The function g is often written as
g(y ) = IE [h(X )jY = y]; and (7.1) becomes
IE [h(X )jY = y ] =
Z IR
h(x)fX jY (xjy ) dx:
In conclusion, to determine IE [h(X )jY ] (a function of ! ), first compute
g (y ) =
Z
IR
h(x)fX jY (xjy ) dx;
and then replace the dummy variable y by the random variable Y :
IE [h(X )jY ](!) = g (Y (!)): Example 11.1 (Jointly normal random variables) Given parameters: (X; Y ) have the joint density
1 > 0; 2 > 0; ,1 < < 1.
1 1 x2 , 2 x y + y2 p fX;Y (x; y) = exp , 2(1 , 2 ) 12 1 2 22 212 1 , 2
:
Let
128 The exponent is
x
1
2 , 2(1 , 2 ) 2 , 2 x y + y 2 1 " 1 2 2 2 2 # 1 x y y 2 = , 2(1 , 2 ) , + 2 (1 , ) 1 2 2 2 2 1 1 y = , 2(1 ,1 2 ) 12 x , y , 2 2 2 2 : 1 2
We can compute the Marginal density of Y as follows
fY (y) =
Z1 , e
1 p
2(1
1 1 ,2)12 x , 2 y
2
, 12 y
2
dx:e
2 2
21 2 1 , 2 ,1 Z1 u ,1 y 1 e, 2 du:e 2 = 2 2 ,1 using the substitution u = p 1 x , y , du = p1,dx 1, 2
2
2 2
2
, = p1 e 2 2
1 2
1
2
1
2 1 y 2 2 2 :
Thus Y is normal with mean 0 and variance 22 . Conditional density. From the expressions
fX;Y (x; y) =
212
1 p
,
, 2(1,1 2 ) 12 x, 21 y
e 1 , 2
1
, e 2
1 2
y2 22 ;
1 y fY (y) = p 1 e, 2 ; 2 2 2
2 2
we have
(x; y) fX jY (xjy) = fX;Y fY (y) 2 1 1 , x, y 1 1 p e 2(1, ) = p : 2 2 1 1 , 2 2 In the xvariable, fX jY (xjy) is a normal density with mean y and variance (1 , )1 . Therefore, Z1 1 IE[X jY = y] = xfX jY (xjy) dx = 2 y; ,1 2
2 1
1
2
"
#
2 1 IE X , y Y = y 2 Z 1 1 2 = x , y fX jY (xjy) dx 2 ,1 2 2 = (1 , )1 :
1 2
CHAPTER 11. General Random Variables
129
From the above two formulas we have the formulas 1 IE[X jY ] = Y;
(7.3)
2
IE
"
1 X , 2 Y
# Y = (1 , ) : 2
2
2 1
(7.4)
Taking expectations in (7.3) and (7.4) yields 1 IEX = IEY = 0;
(7.5)
2
IE
"
1 X , 2 Y
# 2
= (1 , 2 ) 21:
(7.6)
1 Based on Y , the best estimator of X is 2 Y . This estimator is unbiased (has expected error zero) and the 2 2 expected square error is (1 , ) 1. No other estimator based on Y can have a smaller expected square error (Homework problem 2.1).
11.8 Multivariate Normal Distribution Please see Oksendal Appendix A.
X
x
Let denote the column vector of random variables (X1; X2; : : : ; Xn )T , and the corresponding column vector of values (x1 ; x2; : : : ; xn )T . has a multivariate normal distribution if and only if the random variables have the joint density
X
p A exp n, 1 (X , )T:A:(X , )o : fX(x) = (2det 2 )n=2
Here,
4 4 = (1 ; : : : ; n )T = IE X = (IEX1; : : : ; IEXn)T ; and A is an n n nonsingular matrix. A,1 is the covariance matrix
h
i
A,1 = IE (X , ):(X , )T ; i.e. the (i; j )th element of A,1 is IE (Xi , i )(Xj , j ). The random variables in if and only if A,1 is diagonal, i.e.,
A,1 = diag(21 ; 22; : : : ; 2n ); where 2j
= IE (Xj , j )2 is the variance of Xj .
X are independent
130
11.9 Bivariate normal distribution Take n = 2 in the above definitions, and let
=4 IE (X1 , 1)(X2 , 2) : 1 2
Thus,
A,1 =
2 A=4
"
#
21 1 2 ; 1 2 22
1 (1,2 ) , 1 2 (1 ,2)
p
2 1
det A =
3 , (1 , ) 5 ; 1 2
1 2
22 (1,2 )
p1
12 1 , 2
;
and we have the formula from Example 11.1, adjusted to account for the possibly nonzero expectations:
(
"
1p 1 (x1 , 1 )2 , 2(x1 , 1 )(x2 , 2 ) + (x2 , 2 )2 fX ;X (x1; x2) = exp , 2(1 , 2) 1 2 21 22 212 1 , 2 1
2
11.10 MGF of jointly normal random variables
u
X
Let = (u1; u2; : : : ; un )T denote a column vector with components in IR, and let have a multivariate normal distribution with covariance matrix A,1 and mean vector . Then the moment generating function is given by
T IEeu :X =
Z1
Z1
T eu :XfX1; X2; : : : ; Xn (x1; x2; : : : ; xn) dx1 : : :dxn ,1n ,1 o = exp 12 uT A,1 u + uT :
:::
If any n random variables X1 ; X2; : : : ; Xn have this moment generating function, then they are jointly normal, and we can read out the means and covariances. The random variables are jointly normal and independent if and only if for any real column vector = (u1 ; : : : ; un )T
u 8n 9 8n 9 < = < = X X IEeuT :X =4 IE exp : uj Xj ; = exp : [ 12 2j u2j + uj j ]; : j =1 j =1
#)
:
Chapter 12
SemiContinuous Models 12.1 Discretetime Brownian Motion Let fYj gnj=1 be a collection of independent, standard normal random variables defined on ( ; F ; P), where P I is the market measure. As before we denote the column vector (Y1 ; : : : ; Yn )T by . We therefore have for any real colum vector = (u1 ; : : : ; un )T ,
u 8n 9 8n 9 < =
Y
Define the discretetime Brownian motion (See Fig. 12.1):
B0 = 0; Bk =
k X
j =1
Yj ; k = 1; : : : ; n:
If we know Y1 ; Y2; : : : ; Yk , then we know B1 ; B2; : : : ; Bk . Conversely, if we know B1 ; B2; : : : ; Bk , then we know Y1 = B1 ; Y2 = B2 , B1 ; : : : ; Yk = Bk , Bk,1 . Define the filtration
F 0 = f; g; F k = (Y1; Y2; : : : ; Yk ) = (B1; B2; : : : ; Bk ); k = 1; : : : ; n:
Theorem 1.34
fBk gnk=0 is a martingale (under IP).
Proof:
IE [Bk+1 jF k ] = IE [Yk+1 + Bk jF k ] = IEYk+1 + Bk = Bk :
131
132
B k Y2 Y1 0
Y4
Y3
1
2
3
4
k
Figure 12.1: Discretetime Brownian motion. Theorem 1.35
fBk gnk=0 is a Markov process.
Proof: Note that
IE [h(Bk+1 )jF k ] = IE [h(Yk+1 + Bk )jF k ]:
Use the Independence Lemma. Define
Z1 1 1 g(b) = IEh(Yk+1 + b) = p h(y + b)e, 2 y dy: 2 ,1 2
Then which is a function of Bk alone.
IE [h(Yk+1 + Bk )jF k ] = g (Bk );
12.2 The Stock Price Process Given parameters:
2 IR, the mean rate of return. > 0, the volatility. S0 > 0, the initial stock price. The stock price process is then given by
n
o
Sk = S0 exp Bk + ( , 12 2 )k ; k = 0; : : : ; n: Note that
n
o
Sk+1 = Sk exp Yk+1 + ( , 21 2 ) ;
CHAPTER 12. SemiContinuous Models
133
1 IE [Sk+1jF k ] = Sk IE [eYk+1 jF k ]:e, 2 1 1 = Sk e 2 e, 2 = e Sk : 2
2
2
Sk+1 IE [ S jF ] k +1 k = log = log IE S F k ; Sk k
Thus
Sk+1 var log = var Y
and
2 k+1 + ( , 21 )
Sk
= 2:
12.3 Remainder of the Market The other processes in the market are defined as follows. Money market process:
Mk = erk ; k = 0; 1; : : : ; n:
Portfolio process:
0; 1; : : : ; n,1; Each k is F k measurable. Wealth process:
X0 given, nonrandom. Xk+1 = k Sk+1 + er (Xk , k Sk ) = k (Sk+1 , er Sk ) + er Xk
Each Xk is F k measurable.
Discounted wealth process:
Xk+1 Sk+1 Sk Xk Mk+1 = k Mk+1 , Mk + Mk : 12.4 RiskNeutral Measure Definition n o 12.1 Let IfP be a probability measure on ( ; F ), equivalent to the market measure IP. If
Sk Mk
n is a martingale under If P , we say that If P is a riskneutral measure. k=0
134
P is a riskneutral measure, then every discounted wealth process Theorem 4.36 If If
a martingale under If P , regardless of the portfolio process used to generate it.
n Xk on Mk
k=0
is
Proof:
X S S X k +1 k +1 k k f f IE Mk+1 F k = IE k Mk+1 , Mk + Mk F k S S Xk k +1 k f = k IE Mk+1 F k , Mk + M k Xk : = M k
12.5 RiskNeutral Pricing Let Vn be the payoff at time n, and say it is F n measurable. Note that Vn may be pathdependent. Hedging a short position:
Sell the simple European derivative security Vn . Receive X0 at time 0.
Construct a portfolio process 0 ; : : : ; n,1 which starts with X0 and ends with Xn
P , then If there is a riskneutral measure If
= Vn .
Xn = If Vn : X0 = If EM EM n n Remark 12.1 Hedging in this “semicontinuous” model is usually not possible because there are not enough trading dates. This difficulty will disappear when we go to the fully continuous model.
12.6 Arbitrage Definition 12.2 An arbitrage is a portfolio which starts with X0
= 0 and ends with Xn satisfying
IP (Xn 0) = 1; IP (Xn > 0) > 0: (IP here is the market measure). Theorem 6.37 (Fundamental Theorem of Asset Pricing: Easy part) If there is a riskneutral measure, then there is no arbitrage.
CHAPTER 12. SemiContinuous Models
135
Proof: Let If P be a riskneutralnmeasure, on let X0 = 0, and let Xnfbe the final wealth corresponding X k to any portfolio process. Since is a martingale under IP ,
Mk
k=0
Xn = If E MX = 0: If EM n 0 0
Suppose IP (Xn
(6.1)
0) = 1. We have
IP (Xn 0) = 1 =) IP (Xn < 0) = 0 =) If P (Xn < 0) = 0 =) If P (Xn 0) = 1: (6.1) and (6.2) imply If P (Xn
(6.2)
= 0) = 1. We have
If P (Xn = 0) = 1 =) If P (Xn > 0) = 0 =) IP (Xn > 0) = 0: This is not an arbitrage.
12.7 Stalking the RiskNeutral Measure Recall that
Y1; Y2; : : : ; Yn are independent, standard normal random variables on some probability space ( ; F ; P). n o Sk = S0 exp Bk + ( , 12 2)k . n o Sk+1 = S0 exp (Bk + Yk+1 ) + ( , 12 2)(k + 1) n o = Sk exp Yk+1 + ( , 12 2) : Sk+1 = Sk : exp nYk+1 + ( , r , 1 2)o ; 2 Mk+1 Mk
Therefore,
Sk+1 F k = Sk :IE [exp fYk+1 gjF k ] : expf , r , 1 2g IE M 2 Mk k+1 Sk : expf 1 2 g: expf , r , 1 2g = M 2 2 k S = e,r : Mkk : If = r, the market measure is risk neutral. If 6= r, we must seek further.
136
Sk+1 = Sk : exp nYk+1 + ( , r , 1 2 )o 2 Mk+1 Mk n Sk : exp (Yk+1 + ,r ) , 1 2 o = M 2 k n o Sk : exp Y~k+1 , 1 2 ; = M 2 k where
Y~k+1 = Yk+1 + , r :
r The quantity , is denoted and is called the market price of risk. We want a probability measure If P under which Y~1 ; : : : ; Y~n are independent, standard normal random variables. Then we would have
S h i Sk :If f F k = M ~k+1 gjF k : expf, 21 2g IE Mkk+1 E exp f Y +1 k S k = Mk : expf 21 2 g: expf, 12 2g Sk : = M k
CameronMartinGirsanov’s Idea: Define the random variable
2n 3 X Z = exp 4 (,Yj , 21 2 )5 : j =1
Properties of Z :
Z 0.
Define
8n 9
Then If P (A) 0 for all A 2 F and
Z A
Z dIP 8A 2 F :
If P ( ) = IEZ = 1:
In other words, If P is a probability measure.
CHAPTER 12. SemiContinuous Models
137
We show that If P is a riskneutral measure. For this, it suffices to show that
Y~ 1 = Y1 + ; : : : ; Y~n = Yn + are independent, standard normal under If P. Verification:
Y1; Y2; : : : ; Yn : Independent, standard normal under IP, and 2n 3 2n 3 X X IE exp 4 uj Yj 5 = exp 4 12 u2j 5 : j =1
j =1
Y~ = Y1 + ; : : : ; Y~n = Yn + : Z > 0 almost surely. i h Z = exp Pnj=1(,Yj , 12 2) ; Z
If P (A) =
A
Z dIP 8A 2 F ;
If EX = IE (XZ ) for every random variable X . Compute the moment generating function of (Y~1; : : : ; Y~n ) under IfP :
3 2n 3 2n n X X X If E exp 4 uj Y~j 5 = IE exp 4 uj (Yj + ) + (,Yj , 12 2)5 j =1 j =1 2j=1 3 2n 3 n X X = IE exp 4 (uj , )Yj 5 : exp 4 (uj , 12 2 )5 2 n j=1 3 2 nj=1 3 X X = exp 4 12 (uj , )25 : exp 4 (uj , 21 2 )5 j =1 2j=1 3 n X = exp 4 ( 12 u2j , uj + 12 2 ) + (uj , 12 2) 5 2j=1 3 n X = exp 4 12 u2j 5 : j =1
138
12.8 Pricing a European Call Stock price at time n is
n o 9 8 n = < X = S0 exp : Yj + ( , 12 2 )n; 8 j=1 9 n < X = = S0 exp : (Yj + , r ) , ( , r)n + ( , 12 2)n; 8 j=1 9 n < X = = S0 exp : Y~j + (r , 12 2 )n; : j =1
Sn = S0 exp Bn + ( , 12 2)n
Payoff at time n is (Sn , K )+ . Price at time zero is
2
0
8
Z1
,rn
n
9
1+3
n = , K )+ = IfE 4e,rn @S exp < X ~j + (r , 21 2)n , K A 5 If E (Sn M Y 0 ; : n
=
j =1
o
+
S0 exp b + (r , 12 2 )n , K : p 1 e, n db 2n ,1 P n ~ since j =1 Yj is normal with mean 0, variance n, under If P. e
This is the BlackScholes price. It does not depend on .
b
2 2 2
Chapter 13
Brownian Motion 13.1 Symmetric Random Walk Toss a fair coin infinitely many times. Define
(
Xj (!) = 1
,1
if if
! j = H; ! j = T:
Set
M0 = 0 Mk =
k X j =1
Xj ;
k 1:
13.2 The Law of Large Numbers We will use the method of moment generating functions to derive the Law of Large Numbers:
Theorem 2.38 (Law of Large Numbers:)
1 M !0 k k
almost surely, as 139
k!1:
140 Proof:
u
'k (u) = IE exp k Mk 8k 9
(Def. of
(Independence of the Xj ’s)
k
1
u 1 ,u 2e k + 2e k ;
= which implies,
u
u
log 'k (u) = k log 21 e k + 12 e, k Let x = k1 . Then
lim log 'k (u) = xlim !0
k!1
M k :)
log 12 eux + 12 e,ux
u eux , 2 = xlim !0 12 eux +
= 0:
x
u e,ux 2 1 e,ux 2
(L’Hˆopital’s Rule)
Therefore,
0 = 1;
lim ' (u) = e k!1 k
which is the m.g.f. for the constant 0.
13.3 Central Limit Theorem We use the method of moment generating functions to prove the Central Limit Theorem. Theorem 3.39 (Central Limit Theorem)
p1 Mk ! k
Proof:
Standard normal, as
k!1:
'k (u) = IE exp pu Mk 1 pu 1k , pu k = 2e k + 2e k ;
CHAPTER 13. Brownian Motion
141
so that,
u
u
log 'k (u) = k log 12 e pk + 21 e, pk : Let x = p1k . Then
lim log 'k (u) = xlim !0
k!1
log 21 eux + 12 e,ux
x2 u eux , u e,ux 2 2 = xlim !0 2x 12 eux + 12 e,ux
(L’Hˆopital’s Rule)
u eux , u e,ux 1 2 2 = xlim : lim !0 12 eux + 21 e,ux x!0 2x u eux , u e,ux 2
2 = xlim !0
= xlim !0 1 = 2 u2 :
2x
u2 eux , u2 e,ux 2 2
2
Therefore,
1
lim ' (u) = e 2 u ; k!1 k 2
which is the m.g.f. for a standard normal random variable.
13.4 Brownian Motion as a Limit of Random Walks Let n be a positive integer. If t 0 is of the form nk , then set
B (n) (t) = p1n Mtn = p1n Mk : If t 0 is not of the form nk , then define B (n) (t) by linear interpolation (See Fig. 13.1). Here are some properties of B (100)(t):
(L’Hˆopital’s Rule)
142
k/n
(k+1)/n
Figure 13.1: Linear Interpolation to define B (n) (t).
Properties of
B (100)(1) :
1 B(100)(1) = 10
1 IEB (100)(1) = 10
100 X
j =1 100 X
Xj IEXj = 0:
j =1 100 1 X
var(B (100)(1)) = 100 Properties of
B (100)(2) :
1 B(100)(2) = 10
IEB (100)(2) = 0: var(B (100)(2)) = 2:
j =1
200 X
j =1
var(Xj ) = 1
Xj
Also note that:
B(100)(1) and B(100)(2) , B(100)(1) are independent. B(100)(t) is a continuous function of t. To get Brownian motion, let n!1 in B (n) (t); t 0. 13.5 Brownian Motion (Please refer to Oksendal, Chapter 2.)
(Approximately normal)
(Approximately normal)
CHAPTER 13. Brownian Motion
143
B(t) = B(t,ω) ω
t
(Ω, F,P) Figure 13.2: Continuoustime Brownian Motion. A random variable properties: 1. 2. 3.
B(t) (see Fig.
13.2) is called a Brownian Motion if it satisfies the following
B (0) = 0, B (t) is a continuous function of t; B has independent, normally distributed increments: If 0 = t0 < t1 < t2 < : : : < tn and
Y1 = B(t1 ) , B (t0 ); Y2 = B(t2 ) , B(t1 ); : : : Yn = B(tn ) , B(tn,1 ); then
Y1; Y2; : : : ; Yn are independent, IEYj = 0 8j; var(Yj ) = tj , tj,1 8j: 13.6 Covariance of Brownian Motion
0 s t be given. Then B (s) and B (t) , B (s) are independent, so B (s) and B (t) = (B (t) , B (s)) + B (s) are jointly normal. Moreover, IEB (s) = 0; var(B(s)) = s; IEB(t) = 0; var(B(t)) = t; IEB(s)B (t) = IEB(s)[(B(t) , B(s)) + B(s)] 2 = IEB  (s)(B{z(t) , B(s))} + IEB  {z(s)}
Let
= s:
0
s
144 Thus for any s 0, t 0 (not necessarily s t), we have
IEB(s)B(t) = s ^ t: 13.7 FiniteDimensional Distributions of Brownian Motion Let
0 < t1 < t2 < : : : < tn
be given. Then
(B (t1 ); B (t2); : : : ; B (tn ))
is jointly normal with covariance matrix
2 3 IEB 2 (t1 ) IEB(t1)B(t2 ) : : : IEB(t1 )B (tn ) 6 (t2 )B(t1) IEB2(t2) : : : IEB(t2 )B(tn )77 C = 664:IEB : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :75 IEB(tn)B (t1 ) IEB (tn )B(t2 ) : : : IEB 2 (tn ) 2 3 t1 t1 : : : t1 6 7 = 664t: 1: : : t:2: : ::::::: : : t:2:775 t1 t2 : : : tn
13.8 Filtration generated by a Brownian Motion
fF (t)gt0 Required properties:
For each t, B (t) is F (t)measurable,
< t2 < < tn , the Brownian motion increments B (t1) , B (t); B(t2 ) , B(t1 ); : : :; B(tn ) , B(tn,1 ) are independent of F (t). For each t and for t < t1
Here is one way to construct F (t). First fix t. Let s 2 [0; t] and C
2 B(IR) be given. Put the set fB(s) 2 C g = f! : B(s; !) 2 C g in F (t). Do this for all possible numbers s 2 [0; t] and C 2 B(IR). Then put in every other set
required by the algebra properties.
This F (t) contains exactly the information learned by observing the Brownian motion upto time t. fF (t)gt0 is called the filtration generated by the Brownian motion.
CHAPTER 13. Brownian Motion
145
13.9 Martingale Property Theorem 9.40 Brownian motion is a martingale. Proof: Let 0 s t be given. Then
IE [B(t)jF (s)] = IE [(B(t) , B(s)) + B(s)jF (s)] = IE [B (t) , B (s)] + B (s) = B (s): Theorem 9.41 Let
2 IR be given. Then
n
o
Z (t) = exp ,B(t) , 12 2 t is a martingale. Proof: Let 0 s t be given. Then
IE [Z (t)jF (s)] = IE expf,(B(t) , B(s) + B(s)) ,
1 2 ((t , s) + s)g F (s) 2
= IE Z (s) expf,(B (t) , B (s)) , 12 2 (t , s)g F (s) h i = Z (s)IE expf,(B (t) , B (s)) , 12 2 (t , s)g n o = Z (s) exp 12 (,)2 var(B (t) , B (s)) , 12 2 (t , s) = Z (s):
13.10 The Limit of a Binomial Model Consider the n’th Binomial model with the following parameters:
un = 1 + pn : “Up” factor. ( > 0). dn = 1 , pn : “Down” factor. r = 0. p p~n = u1n,,ddnn = 2==pnn = 12 . q~n = 12 .
146 Let ]k (H ) denote the number of H in the first k tosses, and let ]k (T ) denote the number of T in the first k tosses. Then
]k (H ) + ]k (T ) = k; ]k (H ) , ]k (T ) = Mk ;
which implies,
]k (H ) = 12 (k + Mk ) ]k (T ) = 12 (k , Mk ): (n) = 1. Let t = k for some k, and let n 21 (nt+Mnt) 12 (nt,Mnt)
In the n’th model, take n steps per unit time. Set S0
S (n)(t) = 1 + pn
Under If P , the price process S (n) is a martingale.
1 , pn
:
Theorem 10.42 As n!1, the distribution of S (n) (t) converges to the distribution of
expfB (t) , 12 2 tg;
where B is a Brownian motion. Note that the correction martingale.
, 12 2t is necessary in order to have a
Proof: Recall that from the Taylor series we have
log(1 + x) = x , 12 x2 + O(x3); so
log S (n)(t) = 21 (nt + Mnt ) log(1 + p ) + 12 (nt , Mnt ) log(1 , p )
n n = nt 12 log(1 + p ) + 12 log(1 , p ) n n + Mnt 12 log(1 + p ) , 12 log(1 , p ) n n ! 2 2 1 1 , 3 = 2 1 1 = nt 2 pn , 4 n , 2 pn , 4 n + O(n ) ! 2 2 1 1 , 3 = 2 + Mnt 12 pn , 4 n + 12 pn + 4 n + O(n ) 1 = , 12 2 t + O(n, 2 ) 1 1 1 + pn Mnt + n Mnt O(n, 2 )
 {z }  {z } !0 !Bt
As n!1, the distribution of log S (n) (t) approaches the distribution of B (t) , 21 2t.
CHAPTER 13. Brownian Motion
147
B(t) = B(t,ω) x ω
t
(Ω, F, Px) Figure 13.3: Continuoustime Brownian Motion, starting at x 6= 0.
13.11 Starting at Points Other Than 0 (The remaining sections in this chapter were taught Dec 7.) For a Brownian motion B (t) that starts at 0, we have:
IP (B(0) = 0) = 1:
For a Brownian motion B (t) that starts at x, denote the corresponding probability measure by IP x (See Fig. 13.3), and for such a Brownian motion we have:
IP x (B(0) = x) = 1: Note that:
If x 6= 0, then IP x puts all its probability on a completely different set from IP. The distribution of B(t) under IP x is the same as the distribution of x + B(t) under IP. 13.12 Markov Property for Brownian Motion We prove that Theorem 12.43 Brownian motion has the Markov property. Proof:
Let s 0;
t 0 be given (See Fig. 13.4).
2 6 IE h(B (s + t)) F (s) = IE 64h( B  (s + t{z) , B(s)} +
Independent of
F (s)
B  {z(s)}
F (s)measurable
3 7 ) F (s)75
148
B(s) s
s+t
restart Figure 13.4: Markov Property of Brownian Motion. Use the Independence Lemma. Define
g(x) = IE [h( B (s + t) , B(s) + x )]
2 6 = IE 64h( x +
B (t) {z}
same distribution as B (s + t)
= IE xh(B (t)): Then
, B (s)
3 7 )75
IE h (B(s + t) ) F (s) = g (B(s))
= E B (s)h(B (t)):
In fact Brownian motion has the strong Markov property. Example 13.1 (Strong Markov Property) See Fig. 13.5. Fix x > 0 and define
= min ft 0; B(t) = xg : Then we have:
IE h( B( + t) ) F () = g(B()) = IE x h(B(t)):
CHAPTER 13. Brownian Motion
149
x τ
τ+t
restart Figure 13.5: Strong Markov Property of Brownian Motion.
13.13 Transition Density Let p(t; x; y ) be the probability that the Brownian motion changes value from x to y in time t, and let be defined as in the previous section.
p(t; x; y) = p 1 e, 2t g(x) = IE xh(B(t)) =
Z1
,1
y,x)2 2t
(
h(y )p(t; x; y) dy:
Z1 IE h(B (s + t)) F (s) = g (B(s)) = h(y )p(t; B (s); y ) dy: ,1 Z1 IE h(B ( + t)) F ( ) = h(y )p(t; x; y ) dy:
,1
13.14 First Passage Time Fix x > 0. Define Fix
> 0. Then
is a martingale, and
= min ft 0; B (t) = xg :
n
o
exp B (t ^ ) , 12 2 (t ^ )
n
o
IE exp B (t ^ ) , 12 2 (t ^ ) = 1:
150 We have
8 n 12 o
(14.1)
Let t!1 in (14.1), using the Bounded Convergence Theorem, to get
h
i
IE expfx , 12 2 g1f<1g = 1:
1
Let #0 to get IE f<1g
= 1, so
IP f < 1g = 1; IE expf, 12 2 g = e,x :
(14.2)
Let = 21 2 . We have the m.g.f.:
p
IEe, = e,x
2;
> 0:
(14.3)
Differentiation of (14.3) w.r.t. yields
,IE e, = , px2 e,x
p
2 :
Letting #0, we obtain
IE = 1:
(14.4)
Conclusion. Brownian motion reaches level x with probability 1. The expected time to reach level
x is infinite.
We use the Reflection Principle below (see Fig. 13.6).
IP f t; B (t) < xg = IP fB (t) > xg IP f tg = IP f t; B (t) < xg + IP f t; B (t) > xg = IP fB (t) > xg + IP fB (t) > xg = 2IP fB (t) > xg
Z y = p2 e, t dy 2t x 1
2
2
CHAPTER 13. Brownian Motion
151
shadow path
x τ
t Brownian motion
Figure 13.6: Reflection Principle in Brownian Motion. Using the substitution z
= pyt ; dz = pdyt we get
IP f tg = p2
Z1
2 px
e, z dz: 2
2
t
Density:
@ IP f tg = p x e, x t ; f (t) = @t 2t3 2
2
which follows from the fact that if
F (t) = then
Zb a(t)
g (z ) dz;
@F = , @a g (a(t)): @t @t
Laplace transform formula:
IEe, =
Z1 0
p
e,tf (t)dt = e,x
2 :
152
Chapter 14
The Itˆo Integral The following chapters deal with Stochastic Differential Equations in Finance. References: 1. B. Oksendal, Stochastic Differential Equations, SpringerVerlag,1995 2. J. Hull, Options, Futures and other Derivative Securities, Prentice Hall, 1993.
14.1 Brownian Motion (See Fig. 13.3.) ( ; F ; P) is given, always in the background, even when not explicitly mentioned. Brownian motion, B (t; ! ) : [0; 1) !IR, has the following properties: 1. 2. 3.
B (0) = 0; Technically, IP f! ; B (0; !) = 0g = 1, B (t) is a continuous function of t, If 0 = t0 t1 : : : tn , then the increments B (t1) , B(t0 ); : : : ; B(tn ) , B(tn,1 ) are independent,normal, and
IE [B (tk+1 ) , B(tk )] = 0; IE [B (tk+1 ) , B(tk )]2 = tk+1 , tk : 14.2 First Variation Quadratic variation is a measure of volatility. First we will consider first variation, FV (f ), of a function f (t). 153
154
f(t)
t2 t
T
t 1
Figure 14.1: Example function f (t). For the function pictured in Fig. 14.1, the first variation over the interval
[0; T ] is given by:
FV[0;T ](f ) = [f (t1) , f (0)] , [f (t2) , f (t1 )] + [f (T ) , f (t2 )]
Zt
Zt2 ZT 0 0 = f (t) dt + (,f (t)) dt + f 0(t) dt: t1 t2 0 T Z = jf 0(t)j dt: 1
0
Thus, first variation measures the total amount of up and down motion of the path. The general definition of first variation is as follows: Definition 14.1 (First Variation) Let = ft0 ; t1 ; : : : ; tn g be a partition of [0; T ], i.e.,
0 = t0 t1 : : : tn = T: The mesh of the partition is defined to be
jjjj = k=0max (t , t ): ;:::;n,1 k+1 k We then define
FV[0;T ](f ) = jjlim jj!0
nX ,1 k=0
jf (tk+1) , f (tk )j:
Suppose f is differentiable. Then the Mean Value Theorem implies that in each subinterval [tk ; tk+1 ], there is a point tk such that
f (tk+1 ) , f (tk ) = f 0 (tk )(tk+1 , tk ):
CHAPTER 14. The Itˆo Integral Then
nX ,1 k=0
155
nX ,1
jf (tk+1) , f (tk )j =
k=0
jf 0(tk )j(tk+1 , tk );
and
FV[0;T ](f ) = jjlim jj!0
ZT
nX ,1 k=0
jf 0(tk )j(tk+1 , tk )
= jf 0 (t)j dt: 0
14.3 Quadratic Variation Definition 14.2 (Quadratic Variation) The quadratic variation of a function f on an interval [0; T ] is
hf i(T ) = jjlim jj!0
nX ,1 k=0
jf (tk+1 ) , f (tk )j2:
Remark 14.1 (Quadratic Variation of Differentiable Functions) If f is differentiable, then hf i(T ) = 0, because
nX ,1 k=0
jf (tk+1) , f (tk )j2 =
nX ,1 k=0
jf 0(tk )j2(tk+1 , tk )2
jjjj:
nX ,1 k=0
jf 0(tk )j2(tk+1 , tk )
and
nX ,1
hf i(T ) jjlim jjjj: jjlim jf 0(tk )j2(tk+1 , tk ) jj!0 jj!0 k=0 ZT = lim jjjj jf 0(t)j2 dt jjjj!0 0 = 0:
Theorem 3.44
hBi(T ) = T; or more precisely,
IP f! 2 ; hB(:; !)i(T ) = T g = 1:
In particular, the paths of Brownian motion are not differentiable.
156 Proof: (Outline) Let = ft0 ; t1; : : : ; tn g be a partition of [0; T ]. To simplify notation, set Dk B(tk+1 ) , B (tk ). Define the sample quadratic variation
nX ,1
Q = Then
Q , T = We want to show that
k=0
Dk2 :
nX ,1 k=0
[Dk2 , (tk+1 , tk )]:
lim (Q , T ) = 0: jjjj!0
Consider an individual summand
Dk2 , (tk+1 , tk ) = [B(tk+1 ) , B (tk )]2 , (tk+1 , tk ): This has expectation 0, so
IE (Q , T ) = IE For j
6= k, the terms
nX ,1 k=0
Dj2 , (tj+1 , tj )
[Dk2 , (tk+1 , tk )] = 0:
and
Dk2 , (tk+1 , tk )
are independent, so
var(Q , T ) = = =
nX ,1 k=0 nX ,1 k=0 nX ,1 k=0
var[Dk2 , (tk+1 , tk )]
IE [Dk4 , 2(tk+1 , tk )Dk2 + (tk+1 , tk )2] [3(tk+1 , tk )2 , 2(tk+1 , tk )2 + (tk+1 , tk )2 ]
(if X is normal with mean 0 and variance 2, then IE (X 4) = 3 4)
=2
nX ,1 k=0
(tk+1 , tk )2
2jjjj
nX ,1 k=0
(tk+1 , tk )
= 2jjjj T: Thus we have
IE (Q , T ) = 0; var(Q , T ) 2jjjj:T:
=
CHAPTER 14. The Itˆo Integral
157
As jjjj!0, var(Q , T )!0, so
lim (Q , T ) = 0: jjjj!0 Remark 14.2 (Differential Representation) We know that
IE [(B(tk+1 ) , B(tk ))2 , (tk+1 , tk )] = 0: We showed above that
var[(B (tk+1 ) , B (tk ))2 , (tk+1 , tk )] = 2(tk+1 , tk )2 : When (tk+1 , tk ) is small, (tk+1 , tk )2 is very small, and we have the approximate equation
(B (tk+1 ) , B (tk ))2 ' tk+1 , tk ; which we can write informally as
dB(t) dB(t) = dt:
14.4 Quadratic Variation as Absolute Volatility On any time interval [T1; T2], we can sample the Brownian motion at times
T1 = t0 t1 : : : tn = T2 and compute the squared sample absolute volatility
1
nX ,1
T2 , T1 k=0(B(tk+1 ) , B(tk )) : 2
This is approximately equal to
1
T2 , T1
[hB i(T2) , hB i(T1 )] = T2 , T1 = 1:
T2 , T1
As we increase the number of sample points, this approximation becomes exact. In other words, Brownian motion has absolute volatility 1. Furthermore, consider the equation
ZT
hBi(T ) = T = 1 dt; 0
8T 0:
This says that quadratic variation for Brownian motion accumulates at rate 1 at all times along almost every path.
158
14.5 Construction of the Itˆo Integral The integrator is Brownian motion following properties: 1. 2. 3.
B(t); t 0, with associated filtration F (t); t 0, and the
s t=) every set in F (s) is also in F (t), B (t) is F (t)measurable, 8t, For t t1 : : : tn , the increments B (t1 ) , B (t); B (t2 ) , B (t1 ); : : : ; B (tn ) , B (tn,1 ) are independent of F (t).
The integrand is (t); t 0, where 1. 2.
(t) is F (t)measurable 8t (i.e., is adapted) is squareintegrable:
ZT
IE 2(t) dt < 1; 0
8T:
We want to define the Itˆo Integral:
I (t) =
Zt 0
(u) dB (u);
Remark 14.3 (Integral w.r.t. a differentiable function) If we can define
Zt 0
(u) df (u) =
Zt 0
t 0: f (t) is a differentiable function, then
(u)f 0(u) du:
This won’t work when the integrator is Brownian motion, because the paths of Brownian motion are not differentiable.
14.6 Itˆo integral of an elementary integrand Let = ft0 ; t1; : : : ; tn g be a partition of [0; T ], i.e.,
0 = t0 t1 : : : tn = T: Assume that (t) is constant on each subinterval [tk ; tk+1 ] (see Fig. 14.2). We call such a elementary process. The functions B (t) and (tk ) can be interpreted as follows:
Think of B (t) as the price per unit share of an asset at time t.
an
CHAPTER 14. The Itˆo Integral
159
δ( t ) = δ( t 1 )
δ( t )= δ( t 3 )
δ( t ) = δ( t ) 0
0=t0
t2
t1
t3
t4 = T
δ( t ) = δ( t 2 ) Figure 14.2: An elementary function .
Think of t0 ; t1; : : : ; tn as the trading dates for the asset.
Think of (tk ) as the number of shares of the asset acquired at trading date tk and held until trading date tk+1 .
Then the Itˆo integral I (t) can be interpreted as the gain from trading at time t; this gain is given by:
8 > (t0 )[B(t) , B 0 t t1 >  ({zt0)} ]; < =B (0)=0 I (t) = > ( t )[ B ( t ) , B(t )] + (t )[B(t) , B(t )]; t tt > :(t00)[B(t11) , B(t00)] + (t11)[B(t2) , B(1t1 )] + (t2)[B(t) , B(t2)]; t12 t t23: In general, if tk t tk+1 , kX ,1 I (t) = (tj )[B (tj +1 ) , B(tj )] + (tk )[B(t) , B(tk )]: j =0
14.7 Properties of the Itˆo integral of an elementary process Adaptedness For each t; I (t) is F (t)measurable. Linearity If
Zt
I (t) = (u) dB(u); 0
then
I (t) J (t) =
Zt 0
Zt
J (t) = (u) dB (u) 0
( (u) (u)) dB (u)
160 s t
t t l+1
l
.....
t
k
t
k+1
Figure 14.3: Showing s and t in different partitions. and
cI (t) = Martingale
Zt 0
c(u)dB (u):
I (t) is a martingale.
We prove the martingale property for the elementary process case. Theorem 7.45 (Martingale Property)
I (t) =
kX ,1 j =0
(tj )[B (tj +1 ) , B(tj )] + (tk )[B (t) , B(tk )];
tk t tk+1
is a martingale. Proof: Let 0 s t be given. We treat the more difficult case that s and t are in different subintervals, i.e., there are partition points t` and tk such that s 2 [t` ; t`+1 ] and t 2 [tk ; tk+1 ] (See Fig. 14.3). Write
I (t) =
`X ,1 j =0
(tj )[B(tj+1 ) , B (tj )] + (t` )[B (t`+1 ) , B(t` )]
+
kX ,1 j =`+1
(tj )[B(tj +1 ) , B (tj )] + (tk )[B(t) , B(tk )]
We compute conditional expectations:
2`,1 3 `X ,1 X 4 5 IE (tj )(B (tj+1 ) , B(tj )) F (s) = (tj )(B(tj +1 ) , B(tj )): j =0 j=0 IE (t`)(B(t`+1 ) , B(t` )) F (s) = (t` ) (IE [B(t`+1 )jF (s)] , B(t` )) = (t` )[B (s) , B (t` )]
CHAPTER 14. The Itˆo Integral
161
These first two terms add up to I (s). We show that the third and fourth terms are zero.
2 k,1 3 kX ,1 X 4 5 IE (tj )(B(tj+1 ) , B(tj )) F (s) = IE IE (tj )(B(tj+1 ) , B(tj )) F (tj ) F (s) j =`+1 j =`+1 2 3 kX ,1 = IE 64(tj ) (IE [B (tj +1)jF{z(tj )] , B(tj ))} F (s)75 j =`+1 =0 2 3 IE (tk )(B(t) , B(tk )) F (s) = IE 64(tk ) (IE [B(t)jF ({ztk )] , B (tk ))} F (s)75 =0
Theorem 7.46 (Itˆo Isometry)
IEI 2(t) = IE
Zt
Proof: To simplify notation, assume t = tk , so
I (t) =
k X
j =0
0
2(u) du:
(tj )[B  (tj+1){z, B(tj )}] Dj
Each Dj has expectation 0, and different Dj are independent.
0k 12 X I 2 (t) = @ (tj )Dj A
j =0 k X X = 2(tj )Dj2 + 2 (ti ) (tj )DiDj : j =0 i
Since the cross terms have expectation zero,
IEI 2(t) = = =
k X j =0 k X j =0 k X j =0
= IE
IE [ 2(tj )Dj2]
IE 2(tj )IE (B(tj+1 ) , B (tj ))2 F (tj ) IE 2(tj )(tj +1 , tj )
k tZj X
+1
2 (u) du
j =0 tj Zt = IE 2 (u) du 0
162
path of δ
path of δ 4
t2
t1
0=t0
t3
t4 = T
Figure 14.4: Approximating a general process by an elementary process 4 , over [0; T ].
14.8 Itˆo integral of a general integrand Fix T
> 0. Let be a process (not necessarily an elementary process) such that
(t) is F (t)measurable, 8t 2 [0; T ], IE R0T 2(t) dt < 1: Theorem 8.47 There is a sequence of elementary processes fn g1 n=1 such that
nlim !1 IE
ZT 0
jn(t) , (t)j2 dt = 0:
Proof: Fig. 14.4 shows the main idea.
In the last section we have defined
In (T ) = for every n. We now define
ZT 0
ZT 0
n (t) dB (t)
(t) dB(t) = nlim !1
ZT 0
n(t) dB(t):
CHAPTER 14. The Itˆo Integral
163
The only difficulty with this approach is that we need to make sure the above limit exists. Suppose n and m are large positive integers. Then
var(In (T ) , Im (T )) = IE (Itˆo Isometry:) = IE
= IE
ZT 0 T
Z 0
ZT
((a + b)2 2a2 + 2b2 :) 2IE
0
!2
[n (t) , m (t)] dB (t)
[n (t) , m (t)]2 dt [ jn (t) , (t)j + j (t) , m (t)j ]2 dt
ZT 0
jn(t) , (t)j2 dt + 2IE
ZT 0
jm(t) , (t)j2 dt;
which is small. This guarantees that the sequence fIn (T )g1 n=1 has a limit.
14.9 Properties of the (general) Itˆo integral
I (t) =
Zt 0
(u) dB (u):
Here is any adapted, squareintegrable process. Adaptedness. For each t, I (t) is F (t)measurable. Linearity. If
Zt
Zt
I (t) = (u) dB(u); 0
then
I (t) J (t) =
and
Zt
cI (t) =
0
J (t) = (u) dB (u) 0
( (u) (u)) dB (u)
Zt 0
c(u)dB (u):
I (t) is a martingale. Continuity. I (t) is a continuous function of the upper limit of integration t. R Itˆo Isometry. IEI 2(t) = IE 0t 2 (u) du.
Martingale.
Example 14.1 () Consider the Itˆo integral
ZT 0
B(u) dB(u):
We approximate the integrand as shown in Fig. 14.5
164
2T/4
T/4
T
3T/4
Figure 14.5: Approximating the integrand B (u) with 4 , over [0; T ].
8 > B(0) = 0 > : : : > : n, T B
By definition,
ZT 0
B(u) dB(u) = nlim !1
(
ZT 0
,
(n 1)T
if
n
u < T:
X kT (k + 1)T kT B n B ,B n : n
n,1 k=0
Bk =4 B kT n ;
To simplify notation, we denote
so
1)
T
0 u < T=n; T=n u < 2T=n;
if if
B(u) dB(u) = nlim !1
X
n,1 k=0
Bk (Bk+1 , Bk ):
We compute 1 2
X
X
n,1
n,1
k=0
k=0
(Bk+1 , Bk )2 = 12
Bk2+1 ,
= 12 Bn2 + 12 = 12 Bn2 + = 12 Bn2 ,
X
n,1
j =0 n,1
X
k=0 n,1
X
k=0
X
n,1 k=0
Bk Bk+1 + 12
Bj2 ,
Bk2 ,
X
n,1
k=0 n,1
X
k=0
X
n,1 k=0
Bk2
Bk Bk+1 + 21
Bk Bk+1
Bk (Bk+1 , Bk ):
X
n,1 k=0
Bk2
CHAPTER 14. The Itˆo Integral Therefore,
X
n,1 k=0
165
Bk (Bk+1 , Bk ) = 12 Bn2 , 12
X
n,1
(Bk+1 , Bk )2 ;
k=0
or equivalently
n , (k + 1)T k X kT (k + 1)T kT X B n B , B n = B (T) , B n n T :
n,1
1
1 2
k=0
2
1 2
2
k=0
Let n!1 and use the definition of quadratic variation to get
ZT 0
B(u) dB(u) = 12 B 2 (T) , 12 T:
Remark 14.4 (Reason for the 12 T term) If f is differentiable with f (0) = 0, then
ZT 0
f (u) df (u) =
ZT 0
f (u)f 0 (u) du
T
= 12 f 2 (u)
0
= 12 f 2 (T ): In contrast, for Brownian motion, we have
ZT 0
B (u)dB (u) = 12 B 2(T ) , 21 T:
The extra term 12 T comes from the nonzero quadratic variation of Brownian motion. It has to be there, because Z
IE
but
T
0
B (u) dB (u) = 0
(Itˆo integral is a martingale)
IE 12 B 2 (T ) = 21 T:
14.10 Quadratic variation of an Itˆo integral Theorem 10.48 (Quadratic variation of Itˆo integral) Let
I (t) = Then
Zt 0
hI i(t) =
(u) dB (u):
Zt 0
2(u) du:
166 This holds even if is not an elementary process. The quadratic variation formula says that at each time u, the instantaneous absolute volatility of I is 2 (u). This is the absolute volatility of the Brownian motion scaled by the size of the position (i.e. (t)) in the Brownian motion. Informally, we can write the quadratic variation formula in differential form as follows:
dI (t) dI (t) = 2(t) dt: Compare this with
dB(t) dB(t) = dt:
Proof: (For an elementary process ). Let = ft0 ; t1; : : : ; tn g be the partition for , i.e., (t) (tk ) for tk t tk+1 . To simplify notation, assume t = tn . We have
hI i(t) =
nX ,1 k=0
[hI i(tk+1 ) , hI i(tk )] :
Let us compute hI i(tk+1) , hI i(tk ). Let = fs0; s1 ; : : : ; sm g be a partition
tk = s0 s1 : : : sm = tk+1 :
Then
I (sj+1 ) , I (sj ) =
sZj+1 sj
(tk ) dB(u)
= (tk ) [B (sj +1 ) , B (sj )] ; so
hI i(tk+1) , hI i(tk) =
mX ,1 j =0
[I (sj +1 ) , I (sj )]2 mX ,1
= 2 (tk )
[B (sj +1 ) , B (sj )]2
j =0 2 (t
jjjj!0 ,,,,,!
k )(tk+1 , tk ):
It follows that
hI i(t) = =
nX ,1
2(tk )(tk+1 , tk )
k=0 nX ,1 tZk+1 k=0 tk
jj!! 0 ,jj,,,,,
2 (u) du
Zt 0
2(u) du:
=
Chapter 15
Itˆo’s Formula 15.1 Itˆo’s formula for one Brownian motion We want a rule to “differentiate” expressions of the form f (B (t)), where f (x) is a differentiable function. If B (t) were also differentiable, then the ordinary chain rule would give
d f (B(t)) = f 0 (B(t))B0 (t); dt which could be written in differential notation as
df (B(t)) = f 0 (B(t))B0 (t) dt = f 0 (B (t))dB (t) However, B (t) is not differentiable, and in particular has nonzero quadratic variation, so the correct formula has an extra term, namely,
df (B(t)) = f 0(B(t)) dB(t) + 12 f 00 (B(t)) {z} dt : dB (t) dB (t)
This is Itˆo’s formula in differential form. Integrating this, we obtain Itˆo’s formula in integral form:
f (B(t)) , f (B{z(0))} = f (0)
Zt 0
f 0 (B (u)) dB(u) + 21
Zt 0
f 00 (B (u)) du:
Remark 15.1 (Differential vs. Integral Forms) The mathematically meaningful form of Itˆo’s formula is Itˆo’s formula in integral form:
f (B(t)) , f (B (0)) =
Zt 0
f 0 (B(u)) dB (u) + 1
Zt
2 0
167
f 00 (B (u)) du:
168 This is because we have solid definitions for both integrals appearing on the righthand side. The first,
Zt 0
f 0(B(u)) dB (u)
is an Itˆo integral, defined in the previous chapter. The second,
Zt 0
f 00 (B (u)) du;
is a Riemann integral, the type used in freshman calculus. For paper and pencil computations, the more convenient form of Itˆo’s rule is Itˆo’s formula in differential form:
df (B (t)) = f 0 (B(t)) dB(t) + 21 f 00 (B(t)) dt: There is an intuitive meaning but no solid definition for the terms df (B (t)); dB (t) and dt appearing in this formula. This formula becomes mathematically respectable only after we integrate it.
15.2 Derivation of Itˆo’s formula Consider f (x) = 12 x2 , so that
f 0 (x) = x; f 00 (x) = 1:
Let xk ; xk+1 be numbers. Taylor’s formula implies
f (xk+1 ) , f (xk ) = (xk+1 , xk )f 0(xk ) + 12 (xk+1 , xk )2f 00 (xk ): In this case, Taylor’s formula to second order is exact because f is a quadratic function.
In the general case, the above equation is only approximate, and the error is of the order of (xk+1 , xk )3. The total error will have limit zero in the last step of the following argument. Fix T
> 0 and let = ft0 ; t1; : : : ; tng be a partition of [0; T ]. Using Taylor’s formula, we write: f (B(T )) , f (B(0)) = 12 B 2 (T ) , 21 B 2 (0) = = =
nX ,1 k=0 nX ,1 k=0 nX ,1 k=0
[f (B (tk+1 )) , f (B (tk ))] [B (tk+1 ) , B (tk )] f 0 (B (tk )) + 12
B(tk ) [B(tk+1 ) , B(tk )] + 12
nX ,1 k=0
nX ,1 k=0
[B (tk+1 ) , B (tk )]2 f 00(B (tk ))
[B (tk+1 ) , B (tk )]2 :
CHAPTER 15. Itˆo’s Formula
169
We let jjjj!0 to obtain
f (B(T )) , f (B(0)) = =
ZT
i(T }) B(u) dB (u) + 12 hB{z
0
T ZT f 0 (B(u)) dB(u) + 12 f 00(B{z(u))} du: 0 0 1
ZT
This is Itˆo’s formula in integral form for the special case
f (x) = 12 x2 : 15.3 Geometric Brownian motion
n
o
Definition 15.1 (Geometric Brownian Motion) Geometric Brownian motion is where and
S (t) = S (0) exp B(t) + , 12 2 t ;
> 0 are constant.
n
Define
o
f (t; x) = S (0) exp x + , 21 2 t ; so
S (t) = f (t; B (t)): Then
ft = , 12 2 f; fx = f; fxx = 2f: According to Itˆo’s formula,
dS (t) = df (t; B(t)) = ft dt + fx dB + 21 fxx dBdB  {z } dt
= ( , dt + f dB + 21 2 f dt = S (t)dt + S (t) dB (t) 1 2 )f 2
Thus, Geometric Brownian motion in differential form is
dS (t) = S (t)dt + S (t) dB (t); and Geometric Brownian motion in integral form is
S (t) = S (0) +
Zt 0
S (u) du +
Zt 0
S (u) dB(u):
170
15.4 Quadratic variation of geometric Brownian motion In the integral form of Geometric Brownian motion,
S (t) = S (0) +
Zt 0
S (u) du +
the Riemann integral
F (t) =
Zt 0
Zt 0
S (u) dB(u);
S (u) du
is differentiable with F 0 (t) = S (t). This term has zero quadratic variation. The Itˆo integral
G(t) =
Zt
is not differentiable. It has quadratic variation
hGi(t) =
0
S (u) dB (u)
Zt 0
2 S 2(u) du:
Thus the quadratic variation of S is given by the quadratic variation of G. In differential notation, we write
dS (t) dS (t) = (S (t)dt + S (t)dB(t))2 = 2S 2(t) dt
15.5 Volatility of Geometric Brownian motion Fix 0 T1 T2 . Let = volatility of S on [T1; T2] is
1
ft0; : : : ; tng be a partition of [T1; T2]. The squared absolute sample nX ,1
T2 , T1 k=0 [S (tk+1) , S (tk
)]2 '
1
ZT
2
2 2 T2 , T1 S (u) du T 2 2 ' S (T1) 1
As T2 # T1 , the above approximation becomes exact. In other words, the instantaneous relative volatility of S is 2 . This is usually called simply the volatility of S .
15.6 First derivation of the BlackScholes formula Wealth of an investor. An investor begins with nonrandom initial wealth X0 and at each time t, holds (t) shares of stock. Stock is modelled by a geometric Brownian motion:
dS (t) = S (t)dt + S (t)dB(t):
CHAPTER 15. Itˆo’s Formula
171
(t) can be random, but must be adapted. lending at interest rate r.
The investor finances his investing by borrowing or
Let X (t) denote the wealth of the investor at time t. Then
dX (t) = (t)dS (t) + r [X (t) , (t)S (t)] dt = (t) [S (t)dt + S (t)dB (t)] + r [X (t) , (t)S (t)] dt = rX (t)dt + (t)S (t) ( {z , r)} dt + (t)S (t)dB(t): Risk premium
Value of an option. Consider an European option which pays g (S (T )) at time T . Let v (t; x) denote the value of this option at time t if the stock price is S (t) = x. In other words, the value of the option at each time t 2 [0; T ] is
v (t; S (t)):
The differential of this value is
dv (t; S (t)) = vt dt + vx dS + 12 vxxdS dS = vt dt + vx [S dt + S dB ] + 12 vxx 2 S 2 dt h i = vt + Svx + 21 2 S 2vxx dt + SvxdB A hedging portfolio starts with some initial wealth X0 and invests so that the wealth X (t) at each time tracks v (t; S (t)). We saw above that dX (t) = [rX + ( , r)S ] dt + S dB: To ensure that X (t) = v (t; S (t)) for all t, we equate coefficients in their differentials. Equating the dB coefficients, we obtain the hedging rule: (t) = vx (t; S (t)): Equating the dt coefficients, we obtain: vt + Svx + 12 2 S 2vxx = rX + ( , r)S: But we have set = vx , and we are seeking to cause X to agree with v . Making these substitutions, we obtain
vt + Svx + 21 2S 2vxx = rv + vx ( , r)S; (where v = v (t; S (t)) and S = S (t)) which simplifies to vt + rSvx + 12 2 S 2vxx = rv: In conclusion, we should let v be the solution to the BlackScholes partial differential equation vt(t; x) + rxvx (t; x) + 21 2 x2vxx(t; x) = rv (t; x) satisfying the terminal condition
v(T; x) = g (x):
If an investor starts with X0 = v (0; S (0)) and uses the hedge (t) = vx (t; S (t)), then he will have X (t) = v(t; S (t)) for all t, and in particular, X (T ) = g (S (T )).
172
15.7 Mean and variance of the CoxIngersollRoss process The CoxIngersollRoss model for interest rates is
q
dr(t) = a(b , cr(t))dt + r(t) dB (t); where a; b; c; and r(0) are positive constants. In integral form, this equation is
Zt
r(t) = r(0) + a (b , cr(u)) du +
Z tq
r(u) dB(u):
0 0 2 We apply Itˆo’s formula to compute dr (t). This is df (r(t)), where f (x) = x2 . We obtain
dr2(t) = df (r(t)) = f 0 (r(t)) dr(t) + 21 f 00(r(t)) dr(t) dr(t)
q
q
2
= 2r(t) a(b , cr(t)) dt + r(t) dB (t) + a(b , cr(t)) dt + r(t) dB (t) = 2abr(t) dt , 2acr2(t) dt + 2r (t) dB (t) + 2 r(t) dt = (2ab + 2 )r(t) dt , 2acr2(t) dt + 2r (t) dB (t) 3 2
3 2
The mean of r(t). The integral form of the CIR equation is
Zt
r(t) = r(0) + a (b , cr(u)) du + 0
Z tq 0
r(u) dB(u):
Taking expectations and remembering that the expectation of an Itˆo integral is zero, we obtain
Zt
IEr(t) = r(0) + a (b , cIEr(u)) du: 0
Differentiation yields
which implies that
d dt IEr(t) = a(b , cIEr(t)) = ab , acIEr(t);
d heactIEr(t)i = eact acIEr(t) + d IEr(t) = eactab: dt dt
Integration yields
eact IEr(t) , r(0) = ab We solve for IEr(t):
Zt 0
eacu du = cb (eact , 1):
IEr(t) = bc + e,act r(0) , bc :
If r(0) = cb , then IEr(t) = bc for every t. If r(0) 6= cb , then r(t) exhibits mean reversion:
b: lim IEr ( t ) = t!1 c
CHAPTER 15. Itˆo’s Formula
173
Variance of r(t). The integral form of the equation derived earlier for dr2(t) is
r2 (t) = r2(0) + (2ab + 2)
Zt 0
r(u) du , 2ac
Taking expectations, we obtain
IEr2(t) = r2(0) + (2ab + 2) Differentiation yields
Zt 0
Zt 0
r2(u) du + 2
IEr(u) du , 2ac
Zt 0
Zt 0
r (u) dB(u): 3 2
IEr2(u) du:
d IEr2(t) = (2ab + 2)IEr(t) , 2acIEr2(t); dt
which implies that
d e2actIEr2(t) = e2act 2acIEr2(t) + d IEr2(t) dt dt 2 act 2 = e (2ab + )IEr(t): Using the formula already derived for IEr(t) and integrating the last equation, after considerable algebra we obtain
!
b2 + b2 + r(0) , b 2 + 2b e,act 2ac2 c2 c ac c 2 2 2b b , 2 act ,2act + + r(0) , c ac e ac 2c , r(0) e : var r(t) = IEr2(t) , (IEr(t))2 2 2 2b b b , act = 2ac2 + r(0) , c ac e + ac 2c , r(0) e,2act : IEr2(t) =
15.8 Multidimensional Brownian Motion Definition 15.2 (ddimensional Brownian Motion) A ddimensional Brownian Motion is a process
B (t) = (B1 (t); : : : ; Bd (t))
with the following properties:
Each Bk (t) is a onedimensional Brownian motion; If i 6= j , then the processes Bi(t) and Bj (t) are independent. Associated with a ddimensional Brownian motion, we have a filtration fF (t)g such that
For each t, the random vector B (t) is F (t)measurable; For each t t1
: : : tn , the vector increments B(t1 ) , B(t); : : : ; B(tn) , B(tn,1 ) are independent of F (t).
174
15.9 Crossvariations of Brownian motions Because each component Bi is a onedimensional Brownian motion, we have the informal equation
dBi(t) dBi (t) = dt: However, we have: Theorem 9.49 If i 6= j ,
dBi(t) dBj (t) = 0
Proof: Let = ft0 ; : : : ; tn g be a partition of [0; T ]. For i of Bi and Bj on [0; T ] to be
C =
nX ,1
6= j , define the sample cross variation
[Bi (tk+1 ) , Bi (tk )] [Bj (tk+1 ) , Bj (tk )] :
k=0
The increments appearing on the righthand side of the above equation are all independent of one another and all have mean zero. Therefore,
IEC = 0: We compute var(C). First note that
C2 = +2
nX ,1 k=0 nX ,1 `
2
2
Bi (tk+1 ) , Bi(tk ) Bj (tk+1) , Bj (tk )
[Bi (t`+1 ) , Bi (t` )][Bj (t`+1 ) , Bj (t` )] : [Bi (tk+1 ) , Bi (tk )] [Bj (tk+1 ) , Bj (tk )]
All the increments appearing in the sum of cross terms are independent of one another and have mean zero. Therefore,
var(C ) = IEC2 = IE
nX ,1 k=0
[Bi (tk+1 ) , Bi (tk )]2 [Bj (tk+1 ) , Bj (tk )]2 :
But [Bi (tk+1 ) , Bi (tk )]2 and [Bj (tk+1 ) , Bj (tk )]2 are independent of one another, and each has expectation (tk+1 , tk ). It follows that
var(C ) =
nX ,1
nX ,1
k=0
k=0
(tk+1 , tk )2 jjjj
(tk+1 , tk ) = jjjj:T:
As jjjj!0, we have var(C )!0, so C converges to the constant IEC
= 0.
CHAPTER 15. Itˆo’s Formula
175
15.10 Multidimensional Itˆo formula To keep the notation as simple as possible, we write the Itˆo formula for two processes driven by a twodimensional Brownian motion. The formula generalizes to any number of processes driven by a Brownian motion of any number (not necessarily the same number) of dimensions. Let X and Y be processes of the form
X (t) = X (0) + Y (t) = Y (0) +
Zt Z0
t
0
(u) du + (u) du +
Zt Z0t 0
11(u) dB1(u) + 21 (u) dB1(u) +
Zt Z0t 0
12 (u) dB2(u); 22 (u) dB2(u):
Such processes, consisting of a nonrandom initial condition, plus a Riemann integral, plus one or more Itˆo integrals, are called semimartingales. The integrands (u); (u); and ij (u) can be any adapted processes. The adaptedness of the integrands guarantees that X and Y are also adapted. In differential notation, we write
dX = dt + 11 dB1 + 12 dB2 ; dY = dt + 21 dB1 + 22 dB2: Given these two semimartingales X and Y , the quadratic and cross variations are: dX dX = ( dt + 11 dB1 + 12 dB2)2; 2 dB dB +2 dB dB + 2 dB dB = 11  1{z 1} 11 12  1{z 2} 12  2{z 2} dY dY dX dY
0 dt dt 2 2 2 = (11 + 12 ) dt; = ( dt + 21 dB1 + 22 dB2 )2 2 + 2 )2 dt; = (21 22 = ( dt + 11 dB1 + 12 dB2)( dt + 21 dB1 + 22 dB2 ) = (1121 + 1222 ) dt
Let f (t; x; y ) be a function of three variables, and let X (t) and Y (t) be semimartingales. Then we have the corresponding Itˆo formula:
df (t; x; y) = ft dt + fx dX + fy dY + 21 [fxx dX dX + 2fxy dX dY + fyy dY dY ] : In integral form, with X and Y as decribed earlier and with all the variables filled in, this equation is
f (t; X (t); Y (t)) , f (0; X (0); Y (0)) Zt 2 + 2 )f + ( + )f + 1 ( 2 + 2 )f ] du = [ft + fx + fy + 12 (11 11 21 12 22 xy 2 21 12 xx 22 yy 0Z
+ where f
t
0
[11 fx + 21 fy ] dB1 +
Zt 0
[12 fx + 22 fy ] dB2 ;
= f (u; X (u); Y (u), for i; j 2 f1; 2g, ij = ij (u), and Bi = Bi (u).
176
Chapter 16
Markov processes and the Kolmogorov equations 16.1 Stochastic Differential Equations Consider the stochastic differential equation:
dX (t) = a(t; X (t)) dt + (t; X (t)) dB(t):
(SDE)
Here a(t; x) and (t; x) are given functions, usually assumed to be continuous in (t; x) and Lipschitz continuous in x,i.e., there is a constant L such that
ja(t; x) , a(t; y)j Ljx , yj;
j(t; x) , (t; y)j Ljx , yj
for all t; x; y .
Let (t0 ; x) be given. A solution to (SDE) with the initial condition (t0; x) is a process fX (t)gtt0 satisfying
X (t0) = x;
Zt
Zt
t0
t0
X (t) = X (t0) + a(s; X (s)) ds + (s; X (s)) dB(s);
t t0
The solution process fX (t)gtt0 will be adapted to the filtration fF (t)gt0 generated by the Brownian motion. If you know the path of the Brownian motion up to time t, then you can evaluate X (t). Example 16.1 (Drifted Brownian motion) Let a be a constant and = 1, so
dX(t) = a dt + dB(t): If (t0 ; x) is given and we start with the initial condition
X(t0 ) = x; 177
178 then
X(t) = x + a(t , t0) + (B(t) , B(t0 )); To compute the differential w.r.t. t, treat t0 and B(t0 ) as constants:
t t0:
dX(t) = a dt + dB(t):
Example 16.2 (Geometric Brownian motion) Let r and be constants. Consider
dX(t) = rX(t) dt + X(t) dB(t): Given the initial condition
X(t0 ) = x;
the solution is
X(t) = x exp (B(t) , B(t0 )) + (r , 12 2 )(t , t0) : Again, to compute the differential w.r.t. t, treat t0 and B(t0 ) as constants: dX(t) = (r , 12 2)X(t) dt + X(t) dB(t) + 12 2 X(t) dt = rX(t) dt + X(t) dB(t):
16.2 Markov Property Let 0 t0
< t1 be given and let h(y) be a function. Denote by IE t ;x h(X (t1)) 0
the expectation of h(X (t1)), given that X (t0) condition
= x. Now let 2 IR be given, and start with initial
X (0) = :
We have the Markov property
IE 0;
h(X (t1)) F (t0) = IE t ;X (t )h(X (t1)):
0
0
In other words, if you observe the path of the driving Brownian motion from time 0 to time t0 , and based on this information, you want to estimate h(X (t1)), the only relevant information is the value of X (t0). You imagine starting the (SDE ) at time t0 at value X (t0), and compute the expected value of h(X (t1)).
CHAPTER 16. Markov processes and the Kolmogorov equations
179
16.3 Transition density Denote by
p(t0 ; t1; x; y) the density (in the y variable) of X (t1), conditioned on X (t0) = x. In other words, IE t ;xh(X (t1)) = 0
The Markov property says that for 0 t0
IE 0;
Z
h(y )p(t0; t1; x; y ) dy:
t1 and for every , Z
h(X (t1)) F (t0 ) =
IR
IR
h(y )p(t0; t1; X (t0); y ) dy:
Example 16.3 (Drifted Brownian motion) Consider the SDE
dX(t) = a dt + dB(t): Conditioned on (t1 , t0 ), i.e.,
X(t0 ) = x, the random variable X(t1 ) is normal with mean x + a(t1 , t0 ) and variance
(y , (x + a(t1 , t0)))2 1 p(t0; t1; x; y) = p exp , : 2(t1 , t0 ) 2(t1 , t0) Note that p depends on t0 and t1 only through their difference t1 , t0 . This is always the case when a(t; x) and (t; x) don’t depend on t. Example 16.4 (Geometric Brownian motion) Recall that the solution to the SDE
dX(t) = rX(t) dt + X(t) dB(t); with initial condition X(t0 ) = x, is Geometric Brownian motion:
X(t1 ) = x exp (B(t1 ) , B(t0 )) + (r , 21 2 )(t1 , t0) : The random variable B(t1 ) , B(t0 ) has density
IP fB(t1 ) , B(t0 ) 2 dbg = p 1 exp , 2(t b, t ) db; 2(t1 , t0) 1 0 and we are making the change of variable
2
y = x exp b + (r , 12 2)(t1 , t0 ) or equivalently,
The derivative is
h i b = 1 log xy , (r , 12 2)(t1 , t0) : dy = y; db
or equivalently,
dy : db = y
180 Therefore,
p(t0 ; t1; x; y) dy = IP fX(t1 ) 2 dyg h i2 = p 1 exp , 2(t ,1 t )2 log xy , (r , 12 2 )(t1 , t0 ) dy: y 2(t1 , t0) 1 0 Using the transition density and a fair amount of calculus, one can compute the expected payoff from a European call:
, K) =
IE t;x (X(T )
+
where
Z1
(y , K)+ p(t; T; x; y) dy 0 h x i r = e (T ,t) xN p 1 log K + r(T , t) + 21 2(T , t) 1 T , ht x i , KN p log K + r(T , t) , 12 2 (T , t) T ,t
Z 1 Z1 1 N() = p1 e, 2 x dx = p1 e, 2 x dx: 2 ,1 2 , 2
Therefore,
IE
0;
2
e,r(T ,t) (X(T ) , K)+ F (t) = e,r(T ,t) IE t;X (t) (X(T) , K)+
X(t) log + r(T , t) + (T , t) T , t K 1 X(t)
1 = X(t)N p
, e,r(T ,t) K N
1 2
p
T ,t
2
16.4 The Kolmogorov Backward Equation Consider
log K + r(T , t) , (T , t) 2
1 2
dX (t) = a(t; X (t)) dt + (t; X (t)) dB(t);
and let p(t0; t1 ; x; y ) be the transition density. Then the Kolmogorov Backward Equation is:
2
@ p(t ; t ; x; y) + 1 2(t ; x) @ p(t ; t ; x; y ): , @t@ p(t0; t1; x; y) = a(t0; x) @x 0 1 0 @x2 0 1 2 0
(KBE)
The variables t0 and x in (KBE ) are called the backward variables.
In the case that a and are functions of x alone, p(t0; t1 ; x; y ) depends on t0 and t1 only through their difference = t1 , t0 . We then write p( ; x; y ) rather than p(t0 ; t1; x; y ), and (KBE ) becomes
@ p( ; x; y) = a(x) @ p( ; x; y ) + 1 2(x) @ 2 p( ; x; y ): 2 @ @x @x2
(KBE’)
CHAPTER 16. Markov processes and the Kolmogorov equations
181
Example 16.5 (Drifted Brownian motion)
dX(t) = a dt + dB(t) (y , (x + a))2 1 p(; x; y) = p exp , : 2 2 @ p = p = @ p 1 exp , (y , x , a)2 @ @ 2 (y , x , a)2 @ (y , x , a)2 21 p , @ exp , 2 2 1 a(y , x , a) (y , x , a) 2 = , 2 + + p: 2 2 @ p = p = y , x , a p: x @x @ 2 p = p = @ y , x , a p + y , x , a p xx x @x2 @x 2 = , 1 p + (y , x ,2 a) p: Therefore,
2 apx + 21 pxx = a(y , x , a) , 21 + (y , x2,2 a) p = p :
This is the Kolmogorov backward equation. Example 16.6 (Geometric Brownian motion)
dX(t) = rX(t) dt + X(t) dB(t): 1 h y i2 1 1 2 p(; x; y) = p exp , 22 log x , (r , 2 ) : y 2 It is true but very tedious to verify that p satisfies the KBE p = rxpx + 21 2x2 pxx:
16.5 Connection between stochastic calculus and KBE Consider
dX (t) = a(X (t)) dt + (X (t)) dB(t): Let h(y ) be a function, and define
v (t; x) = IE t;xh(X (T ));
(5.1)
182 where 0 t T . Then
Z
v(t; x) = h(y ) p(T , t; x; y ) dy;
Z
vt(t; x) = , h(y ) p (T , t; x; y ) dy;
Z
vx(t; x) = h(y ) px(T , t; x; y) dy;
Z
vxx(t; x) = h(y ) pxx(T , t; x; y ) dy: Therefore, the Kolmogorov backward equation implies
vt(t; x) +Za(x)vx(t; x) + 12 2 (x)vxx(t; x) = h i h(y ) ,p (T , t; x; y ) + a(x)px(T , t; x; y ) + 12 2(x)pxx(T , t; x; y ) dy = 0 Let (0; ) be an initial condition for the SDE (5.1). We simplify notation by writing IE rather than IE 0;. Theorem 5.50 Starting at X (0) = , the process v (t; X (t)) satisfies the martingale property:
IE v (t; X (t)) F (s) = v (s; X (s));
0 s t T:
Proof: According to the Markov property,
IE h(X (T )) F (t) = IE t;X (t)h(X (T )) = v (t; X (t));
so
IE [v (t; X (t))jF(s)] = IE IE h(X (T )) F (t) F (s) = IE h(X (T )) F (s)
= IE s;X (s)h(X (T )) = v (s; X (s)):
Itˆo’s formula implies
dv(t; X (t)) = vt dt + vxdX + 12 vxx dX dX = vt dt + avx dt + vx dB + 21 2 vxx dt:
(Markov property)
CHAPTER 16. Markov processes and the Kolmogorov equations
183
In integral form, we have
v (t; X (t)) = v(0; X (0)) Z th i + vt (u; X (u)) + a(X (u))vx(u; X (u)) + 12 2 (X (u))vxx(u; X (u)) du +
Z0t 0
(X (u))vx(u; X (u)) dB (u):
R We know that v (t; X (t)) is a martingale, so the integral 0t for all t. This implies that the integrand is zero; hence
h
i
vt + avx + 12 2 vxx du must be zero
vt + avx + 12 2 vxx = 0: Thus by two different arguments, one based on the Kolmogorov backward equation, and the other based on Itˆo’s formula, we have come to the same conclusion. Theorem 5.51 (FeynmanKac) Define
v(t; x) = IE t;xh(X (T )); where
0 t T;
dX (t) = a(X (t)) dt + (X (t)) dB(t):
Then
vt(t; x) + a(x)vx(t; x) + 12 2 (x)vxx(t; x) = 0 and
(FK)
v (T; x) = h(x):
The BlackScholes equation is a special case of this theorem, as we show in the next section. Remark 16.1 (Derivation of KBE) We plunked down the Kolmogorov backward equation without any justification. In fact, one can use Itˆo’s formula to prove the FeynmanKac Theorem, and use the FeynmanKac Theorem to derive the Kolmogorov backward equation.
16.6 BlackScholes Consider the SDE
dS (t) = rS (t) dt + S (t) dB (t):
With initial condition the solution is
S (t) = x;
n
o
S (u) = x exp (B (u) , B(t)) + (r , 12 2 )(u , t) ;
u t:
184 Define
v(t; x) = IE t;xh(S (T )) n o = IEh x exp (B (T ) , B (t)) + (r , 12 2)(T , t) ; where h is a function to be specified later.
Recall the Independence Lemma: If G is a field, X is G measurable, and Y is independent of G , then
IE h(X; Y ) G = (X );
where
(x) = IEh(x; Y ): With geometric Brownian motion, for 0 t T , we have
n
o
S (t) = S (0) exp B(t) + (r , 12 2)t ; o n S (T ) = S (0) exp B(T ) + (r , 12 2 )T o n = S{z} (t) exp (B (T ) , B (t)) + (r , 21 2 )(T , t) F (t)measurable

{z
independent of
F (t)
We thus have
S (T ) = XY; where
X = S (t) n o Y = exp (B(T ) , B(t)) + (r , 12 2 )(T , t) : Now
IEh(xY ) = v(t; x): The independence lemma implies
IE h(S (T )) F (t) = IE [h(XY )jF (t)]
= v (t; X ) = v (t; S (t)):
}
CHAPTER 16. Markov processes and the Kolmogorov equations We have shown that
v(t; S (t)) = IE h(S (T )) F (t) ;
185
0 t T:
Note that the random variable h(S (T )) whose conditional expectation is being computed does not depend on t. Because of this, the tower property implies that v (t; S (t)); 0 t T , is a martingale: For 0 s t T ,
IE v (t; S (t)) F (s) = IE IE h(S (T )) F (t) F (s) = IE h(S (T )) F (s)
= v (s; S (s)):
This is a special case of Theorem 5.51. Because v (t; S (t)) is a martingale, the sum of the formula,
dt terms in dv (t; S (t)) must be 0.
h
By Itˆo’s
i
dv (t; S (t)) = vt(t; S (t)) dt + rS (t)vx(t; S (t)) + 21 2 S 2(t)vxx(t; S (t)) dt + S (t)vx(t; S (t)) dB (t): This leads us to the equation
vt(t; x) + rxvx(t; x) + 12 2x2vxx(t; x) = 0;
0 t < T; x 0:
This is a special case of Theorem 5.51 (FeynmanKac). Along with the above partial differential equation, we have the terminal condition
v(T; x) = h(x); x 0: Furthermore, if S (t) = 0 for some t 2 [0; T ], then also S (T ) = 0. condition
v(t; 0) = h(0);
This gives us the boundary
0 t T:
Finally, we shall eventually see that the value at time t of a contingent claim paying h(S (T )) is
u(t; x) = e,r(T ,t) IE t;xh(S (T )) = e,r(T ,t) v (t; x) at time t if S (t) = x. Therefore,
v(t; x) = er(T ,t)u(t; x); vt(t; x) = ,rer(T ,t)u(t; x) + er(T ,t)ut(t; x); vx(t; x) = er(T ,t)ux(t; x); vxx(t; x) = er(T ,t)uxx(t; x):
186 Plugging these formulas into the partial differential equation for v and cancelling the er(T ,t) appearing in every term, we obtain the BlackScholes partial differential equation:
,ru(t; x) + ut(t; x) + rxux(t; x) + 12 2x2uxx(t; x) = 0;
0 t < T; x 0: (BS)
Compare this with the earlier derivation of the BlackScholes PDE in Section 15.6. In terms of the transition density
(
y 2 1 1 2 1 p(t; T ; x; y ) = p exp , 2(T , t)2 log x , (r , 2 )(T , t) y 2(T , t)
)
for geometric Brownian motion (See Example 16.4), we have the “stochastic representation”
u(t; x) = e,r(T ,t)IE t;xh(S (T )) Z1 = e,r(T ,t) h(y )p(t; T ; x; y) dy:
(SR)
0
In the case of a call, and
h(y ) = (y , K )+
x 2 1 log K + r(T , t) + 2 (T , t) u(t; x) = x N T ,t 1 x , r ( T , t ) 2 1 ,e KN p log K + r(T , t) , 2 (T , t) T ,t Even if h(y ) is some other function (e.g., h(y ) = (K , y )+ , a put), u(t; x) is still given by and p1
satisfies the BlackScholes PDE (BS) derived above.
16.7 BlackScholes with pricedependent volatility
dS (t) = rS (t) dt + (S (t)) dB(t); v(t; x) = e,r(T ,t) IE t;x(S (T ) , K )+ : The FeynmanKac Theorem now implies that
,rv(t; x) + vt(t; x) + rxvx(t; x) + 12 2(x)vxx(t; x) = 0; v also satisfies the terminal condition v(T; x) = (x , K )+;
x 0;
0 t < T; x > 0:
CHAPTER 16. Markov processes and the Kolmogorov equations and the boundary condition
v(t; 0) = 0;
187
0 t T:
An example of such a process is the following from J.C. Cox, Notes on options pricing I: Constant elasticity of variance diffusions, Working Paper, Stanford University, 1975:
dS (t) = rS (t) dt + S (t) dB(t); where 0 < 1. The “volatility” S ,1(t) decreases with increasing stock price. The corresponding BlackScholes equation is
,rv + vt + rxvx + 12 2x2 vxx = 0;
0 t < T x > 0; v (t; 0) = 0; 0 t T v (T; x) = (x , K )+ ; x 0:
188
Chapter 17
Girsanov’s theorem and the riskneutral measure (Please see Oksendal, 4th ed., pp 145–151.) Theorem 0.52 (Girsanov, Onedimensional) Let B (t); 0 t T , be a Brownian motion on a probability space ( ; F ; P). Let F (t); 0 t T , be the accompanying filtration, and let (t); 0 t T , be a process adapted to this filtration. For 0 t T , define
Be (t) =
Zt 0
(u) du + B(t);
Zt Zt 2 1 Z (t) = exp , (u) dB (u) , 2 (u) du ; 0
0
and define a new probability measure by
If P (A) =
Z A
8A 2 F :
Z (T ) dIP;
Under If P , the process Be (t); 0 t T , is a Brownian motion. Caveat: This theorem requires a technical condition on the size of . If
IE exp
( ZT 1 2 0
2 (u) du
)
< 1;
everything is OK. We make the following remarks:
Z (t) is a matingale. In fact, dZ (t) = ,(t)Z (t) dB(t) + 21 2 (t)Z (t) dB(t) dB(t) , 12 2 (t)Z (t) dt = ,(t)Z (t) dB (t): 189
190
If P is a probability measure.
Since Z (0) = 1, we have IEZ (t) = 1 for every t 0. In particular
If P ( ) =
Z
so If P is a probability measure.
If E in terms of IE .
Z (T ) dIP = IEZ (T ) = 1;
E denote expectation under If P . If X is a random variable, then Let If
If EZ = IE [Z (T )X ] : To see this, consider first the case X = 1A , where A 2 F . We have Z Z If EX = If P (A) = Z (T ) dIP = Z (T )1A dIP = IE [Z (T )X ] : A
Now use Williams’ “standard machine”.
If P and IP .
Z
The intuition behind the formula
If P (A) =
is that we want to have
A
Z (T ) dIP
8A 2 F
If P (!) = Z (T; !)IP (!);
but since IP (! ) = 0 and If P (!) = 0, this doesn’t really tell us anything useful about If P . Thus, we consider subsets of , rather than individual elements of .
e (T ). If is constant, then Distribution of B
n
o
Z (T ) = exp ,B (T ) , 12 2 T Be (T ) = T + B(T ): e (T ) is normal with mean T and Under IP , B (T ) is normal with mean 0 and variance T , so B variance T : ( ~ 2) 1 ( b , T ) IP (Be (T ) 2 d~b) = p exp , 2T d~b: 2T e (T ). The change of measure from IP to IfP removes the drift from Be (T ). Removal of Drift from B To see this, we compute
If EBe (T ) = IE [Z (T )(T + B (T ))] h n o i = IE exp ,B (T ) , 21 2 T (T + B (T ))
(
)
Z1 2 = p1 (T + b) expf,b , 21 2 T g exp , 2bT db 2T ,1 ( Z1 2) 1 ( b + T ) =p (T + b) exp , 2T db 2T ,1 ( 2) Z1 1 y exp , y2 dy (Substitute y = T + b) (y = T + b) = p 2T ,1 = 0:
CHAPTER 17. Girsanov’s theorem and the riskneutral measure
191
We can also see that If E Be (T ) = 0 by arguing directly from the density formula
ne
)
(
o ~
2 ~ IP B (t) 2 db = p 1 exp , (b ,2TT ) d~b: 2T Because
Z (T ) = expf,B (T ) , 21 2 T g = expf,(Be (T ) , T ) , 12 2 T g = expf,Be (T ) + 12 2 T g; we have
n
o
n
o
n
o
If P Be (T ) 2 d~b = IP Be (T ) 2 d~b exp ,~b + 21 2 T ( ~ ) 2 1 ( b , T ) 2 1 ~ =p exp , 2T , b + 2 T d~b: 2T ( ~2 ) 1 =p exp , 2bT d~b: 2T Under If P , Be (T ) is normal with mean zero and variance mean T and variance T .
T . Under IP , Be (T ) is normal with
Means change, variances don’t. When we use the Girsanov Theorem to change the probability measure, means change but variances do not. Martingales may be destroyed or created. Volatilities, quadratic variations and cross variations are unaffected. Check:
dBe dBe = ((t) dt + dB(t))2 = dB:dB = dt: 17.1 Conditional expectations under If P Lemma 1.53 Let 0 t T . If X is F (t)measurable, then
If EX = IE [X:Z (t)]:
Proof:
If EX = IE [X:Z (T )] = IE [ IE [X:Z (T )jF (t)] ] = IE [X IE [Z (T )jF (t)] ] = IE [X:Z (t)] because Z (t); 0 t T , is a martingale under IP .
192 Lemma 1.54 (Baye’s Rule) If X is F (t)measurable and 0 s t T , then
If E [X jF (s)] = Z (1s) IE [XZ (t)jF (s)]:
Proof: It is clear that Z 1(s) IE [XZ (t)jF (s)] is property. For A 2 F (s), we have
F (s)measurable.
(1.1)
We check the partial averaging
1 1 IE [XZ (t)jF (s)] dIf f P = IE 1A Z (s) IE [XZ (t)jF(s)] A Z (s) = IE [1A IE [XZ (t)jF (s)]] (Lemma 1.53) = IE [IE [1A XZ (t)jF (s)]] (Taking in what is known) = IE [1A XZ (t)] = If ZE [1AX ] (Lemma 1.53 again) = X dIf P:
Z
A
Although we have proved Lemmas 1.53 and 1.54, we have not proved Girsanov’s Theorem. We will not prove it completely, but here is the beginning of the proof. Lemma 1.55 Using the notation of Girsanov’s Theorem, we have the martingale property
If E [Be (t)jF (s)] = Be (s);
0 s t T:
e (t)Z (t) is a martingale under IP . Recall Proof: We first check that B dBe (t) = (t) dt + dB(t); dZ (t) = ,(t)Z (t) dB(t):
Therefore,
e ) = Be dZ + Z dBe + dBe dZ d(BZ e dB + Z dt + Z dB , Z dt = ,BZ e + Z ) dB: = (,BZ
Next we use Bayes’ Rule. For 0 s t T ,
If E[Be (t)jF (s)] = Z 1(s) IE [Be (t)Z (t)jF (s)] = Z 1(s) Be (s)Z (s) = Be (s):
CHAPTER 17. Girsanov’s theorem and the riskneutral measure
193
Definition 17.1 (Equivalent measures) Two measures on the same probability space which have the same measurezero sets are said to be equivalent. The probability measures defined by
IP
R
and
If P of the Girsanov Theorem are equivalent.
Z f IP (A) = Z (T ) dIP;
If IP (A) = 0, then A Z (T ) P to obtain of If
R
If P is
A 2 F:
dIP = 0: Because Z (T ) > 0 for every !, we can invert the definition IP (A) =
If If P (A) = 0, then A Z (1T )
Recall that
Z
1
f
dIP ; A Z (T )
A 2 F:
dIP = 0:
17.2 Riskneutral measure As usual we are given the Brownian motion: B (t); 0 t T , with filtration F (t); 0 defined on a probability space ( ; F ; P). We can then define the following.
t T,
Stock price:
dS (t) = (t)S (t) dt + (t)S (t) dB(t): The processes (t) and (t) are adapted to the filtration. The stock price model is completely general, subject only to the condition that the paths of the process are continuous.
r(t); 0 t T . The process r(t) is adapted. Wealth of an agent, starting with X (0) = x. We can write the wealth process differential in Interest rate: several ways:
dX (t) =
(  t){zdS (t})
Capital gains from Stock
+ r (t)[X (t) ,{z(t)S (t)] dt} Interest earnings
= r(t)X (t) dt + (t)[dS (t) , rS (t) dt] = r(t)X (t) dt + (t) ((t) {z , r(t))} S (t) dt + (t)(t)S (t) dB(t)
2 3 66 (t) , r(t) 77 6 = r(t)X (t) dt + (t)(t)S (t) 6 dt + dB (t)77 4  {z(t) } 5 Risk premium
Market price of risk= (t)
194 Discounted processes:
Rt Rt d e, r(u) du S (t) = e, r(u) du [,r(t)S (t) dt + dS (t)]
0
0
Rt Rt d e, r(u) du X (t) = e, r(u) du [,r(t)X (t) dt + dX (t)] 0
0
Rt = (t)d e, r(u) du S (t) : 0
Notation:
R t r(u) du
(t) = e
0
1 = e, R t r(u) du ; (t) 1 r(t) d (t) = , (t) dt:
;
0
d (t) = r(t) (t) dt; The discounted formulas are
S (t)
d (t) = 1(t) [,r(t)S (t) dt + dS (t)] = 1(t) [((t) , r(t))S (t) dt + (t)S (t) dB (t)]
= 1(t) (t)S (t) [(t) dt + dB (t)] ; X (t) S (t) d (t) = (t) d (t) t) (t)S (t) [(t) dt + dB(t)]: = ( (t) Changing the measure. Define
Be (t) = Then
Zt 0
(u) du + B(t):
S (t)
d (t) = 1(t) (t)S (t) dBe (t); t) (t)S (t) dBe (t): d X ((tt)) = ( (t) Under If P , S ((tt)) and X ((tt)) are martingales. Definition 17.2 (Riskneutral measure) A riskneutral measure (sometimes called a martingale measure) is any probability measure, equivalent to the market measure IP , which makes all discounted asset prices martingales.
CHAPTER 17. Girsanov’s theorem and the riskneutral measure For the market model considered here,
If P (A) = where
Z A
Zt
Z (t) = exp ,
0
Z (T ) dIP;
195
A 2 F;
(u) dB(u) , 12
is the unique riskneutral measure. Note that because (t) = 0.
Zt
2(u) du ;
0 (t),r(t) ; we must assume that (t) 6= (t)
Riskneutral valuation. Consider a contingent claim paying an F (T )measurable random variable V at time T . Example 17.1
V = (S(T) , K)+ ; European call + V = (K , S(T)) ; European put !+ Z T 1 S(u) du , K ; Asian call V= T 0 V = 0max S(t); Look back tT If there is a hedging portfolio, i.e., a process (t); 0 t T , whose corresponding wealth process satisfies X (T ) = V , then
X (0) = If E (VT ) : (t) f This is because X (t) is a martingale under IP , so
X (T ) V X (0) f f X (0) = (0) = IE (T ) = IE (T ) :
196
Chapter 18
Martingale Representation Theorem 18.1 Martingale Representation Theorem See Oksendal, 4th ed., Theorem 4.11, p.50. Theorem 1.56 Let B (t); 0 t T; be a Brownian motion on ( ; F ; P). Let F (t); 0 t T , be the filtration generated by this Brownian motion. Let X (t); 0 t T , be a martingale (under IP ) relative to this filtration. Then there is an adapted process (t); 0 t T , such that
X (t) = X (0) +
Zt 0
(u) dB(u);
0 t T:
In particular, the paths of X are continuous. Remark 18.1 We already know that if X (t) is a process satisfying
dX (t) = (t) dB (t); then X (t) is a martingale. Now we see that if X (t) is a martingale adapted to the filtration generated by the Brownian motion B (t), i.e, the Brownian motion is the only source of randomness in X , then dX (t) = (t) dB(t) for some (t). 18.2 A hedging application Homework Problem 4.5. In the context of Girsanov’s Theorem, suppse that F (t); 0 t T; is the filtration generated by the Brownian motion B (under IP ). Suppose that Y is a If P martingale. Then there is an adapted process (t); 0 t T , such that
Y (t) = Y (0) +
Zt 0
(u) dBe (u); 197
0 t T:
198
dS (t) = (t)S (t) dt + (t)S (t) dB(t); Z t (t) = exp r(u) du ; 0 (t) = (t),(t)r(t) ; Zt Be (t) = (u) du + B(t); 0
Zt
Z (t) = exp , If P (A) = Then
Z
A
0
(u) dB (u) , 12
8A 2 F :
Z (T ) dIP;
Zt 0
2 (u) du ;
S (t) S (t) d (t) = (t) (t) dBe (t):
Let (t); 0 t T; be a portfolio process. The corresponding wealth process X (t) satisfies
d X ((tt)) = (t)(t) S ((tt)) dBe (t);
i.e.,
X (t) = X (0) + Z t (u)(u) S (u) dBe (u); (t) (u) 0
0 t T:
Let V be an F (T )measurable random variable, representing the payoff of a contingent claim at time T . We want to choose X (0) and (t); 0 t T , so that Define the If P martingale
X (T ) = V:
Y (t) = If E (VT ) F (t) ;
0 t T:
According to Homework Problem 4.5, there is an adapted process (t); 0 t T , such that
Y (t) = Y (0) +
h
i
Zt 0
(u) dBe (u);
Set X (0) = Y (0) = If E (VT ) and choose (u) so that
(u)(u) S (u) = (u): (u)
0 t T:
CHAPTER 18. Martingale Representation Theorem
199
With this choice of (u); 0 u T , we have
V X (t) = Y (t) = If E (T ) F (t) ; (t)
In particular,
0 t T:
V V X (T ) = If E (T ) (T ) F (T ) = (T ) ;
so
X (T ) = V: The Martingale Representation Theorem guarantees the existence of a hedging portfolio, although it does not tell us how to compute it. It also justifies the riskneutral pricing formula
V f X (t) = (t)IE (T ) F (t) Z (T ) ( t ) = Z (t) IE (T ) V F (t) = (1t) IE (T )V F (t) ;
0 t T;
where
(t) = Z ((tt))
Zt
= exp ,
0
(u) dB(u) ,
Zt 0
(r(u) + 12 2 (u)) du
18.3 ddimensional Girsanov Theorem Theorem 3.57 (ddimensional Girsanov) B(t) dimensional Brownian motion on ( ; F ; P);
= (B1 (t); : : : ; Bd (t)); 0 t T , a d
F (t); 0 t T; the accompanying filtration, perhaps larger than the one generated by B; (t) = (1(t); : : : ; d(t)); 0 t T , ddimensional adapted process. For 0 t T; define Z eBj (t) = t j (u) du + Bj (t); j = 1; : : : ; d; 0 Z Zt t 2 1 Z (t) = exp , (u): dB (u) , 2 jj(u)jj du ; 0 0 Z If P (A) = Z (T ) dIP: A
200 Then, under If P , the process
Be (t) = (Be1 (t); : : : ; Bed (t));
0 t T;
is a ddimensional Brownian motion.
18.4 ddimensional Martingale Representation Theorem Theorem 4.58 on ( ; F ; P);
B(t) = (B1 (t); : : : ; Bd(t)); 0 t T; a ddimensional Brownian motion
F (t); 0 t T; the filtration generated by the Brownian motion B. If X (t); 0 t T , is a martingale (under IP ) relative to F (t); 0 t T , then there is a
ddimensional adpated process (t) = (1(t); : : : ; d(t)), such that X (t) = X (0) +
Zt 0
(u): dB (u);
0 t T:
Corollary 4.59 If we have a ddimensional adapted process (t) = (1 (t); : : : ; d (t)); then we can e Z and IfP as in Girsanov’s Theorem. If Y (t); 0 t T , is a martingale under IfP relative define B; to F (t); 0 t T , then there is a ddimensional adpated process (t) = ( 1(t); : : : ; d(t)) such that Z
Y (t) = Y (0) +
t
0
(u): dBe (u);
0 t T:
18.5 Multidimensional market model
B (t) = (B1(t); : : : ; Bd (t)); 0 t T , be a ddimensional Brownian motion on some ( ; F ; P), and let F (t); 0 t T , be the filtration generated by B . Then we can define the Let
following:
Stocks
dSi (t) = i (t)Si(t) dt + Si (t) Accumulation factor
(t) = exp
Z t 0
r(u) du :
Here, i (t); ij (t) and r(t) are adpated processes.
d X j =1
ij (t) dBj (t);
i = 1; : : : ; m
CHAPTER 18. Martingale Representation Theorem
201
Discounted stock prices
d X d S i((tt)) = (i (t){z, r(t))} S i((tt)) dt + S i((tt)) ij (t) dBj (t) j =1 Risk Premium
d X S ( t ) ? i = (t) ij (t) [j (t) +{zdBj (t)]} j =1 dBej (t)
(5.1)
For 5.1 to be satisfied, we need to choose 1 (t); : : : ; d (t), so that
d X j =1
ij (t)j (t) = i (t) , r(t);
i = 1; : : : ; m:
(MPR)
Market price of risk. The market price of risk is an adapted process (t) = (1 (t); : : : ; d (t)) satisfying the system of equations (MPR) above. There are three cases to consider: Case I: (Unique Solution). For Lebesguealmost every t and IP almost every ! , (MPR) has a unique solution (t). Using (t) in the ddimensional Girsanov Theorem, we define a unique riskneutral probability measure If P . Under If P , every discounted stock price is a martingale. Consequently, the discounted wealth process corresponding to any portfolio process is a If Pmartingale, and this implies that the market admits no arbitrage. Finally, the Martingale Representation Theorem can be used to show that every contingent claim can be hedged; the market is said to be complete. Case II: (No solution.) If (MPR) has no solution, then there is no riskneutral probability measure and the market admits arbitrage. Case III: (Multiple solutions). If (MPR) has multiple solutions, then there are multiple riskneutral probability measures. The market admits no arbitrage, but there are contingent claims which cannot be hedged; the market is said to be incomplete. Theorem 5.60 (Fundamental Theorem of Asset Pricing) Part I. (Harrison and Pliska, Martingales and Stochastic integrals in the theory of continuous trading, Stochastic Proc. and Applications 11 (1981), pp 215260.): If a market has a riskneutral probability measure, then it admits no arbitrage. Part II. (Harrison and Pliska, A stochastic calculus model of continuous trading: complete markets, Stochastic Proc. and Applications 15 (1983), pp 313316): The riskneutral measure is unique if and only if every contingent claim can be hedged.
202
Chapter 19
A twodimensional market model Let B (t)
= (B1 (t); B2(t)); 0 t T; be a twodimensional Brownian motion on ( ; F ; P). Let
F (t); 0 t T; be the filtration generated by B.
In what follows, all processes can depend on t and ! , but are adapted to simplify notation, we omit the arguments whenever there is no ambiguity.
F (t); 0 t T .
To
Stocks:
dS1 = S1 [1 dt + 1 dB1] ; q 2 dS2 = S2 2 dt + 2 dB1 + 1 , 2 dB2 : We assume 1
> 0; 2 > 0; ,1 1: Note that dS1 dS2 = S1221 dB1 dB1 = 21 S12 dt; dS2 dS2 = S22222 dB1 dB1 + S22(1 , 2)22 dB2 dB2 = 22 S22 dt; dS1 dS2 = S11S2 2 dB1 dB1 = 12 S1S2 dt:
In other words,
dS1 has instantaneous variance 2 , 1 S1 dS2 has instantaneous variance 2 , 2 S2 dS1 and dS2 have instantaneous covariance 1 2 . S1 S2
Accumulation factor:
(t) = exp
Z t 0
r du :
The market price of risk equations are
1 1 = 1 , r 21 + 1 , 2 2 2 = 2 , r
q
203
(MPR)
204 The solution to these equations is
1 = 1, r ; 1 ( , 2(1 , r) ; 2 = 1 2 , r ) p 1 2 1 , 2
provided ,1 < < 1.
Suppose ,1 < < 1. Then (MPR) has a unique solution (1 ; 2); we define
Zt
Z (t) = exp , If P (A) =
Z
0
1 dB1 ,
Z (T ) dIP;
A
Zt 0
2 dB2 , 12
8A 2 F :
Zt 0
(12 + 22 ) du ;
If P is the unique riskneutral measure. Define Be1 (t) = Be2 (t) = Then
Zt Z0 0
t
1 du + B1 (t); 2 du + B2 (t):
h
i
dS1 = S1 r dt + 1 dBe1 ;
q
dS2 = S2 r dt + 2 dBe1 + 1 , 22 dBe2 : We have changed the mean rates of return of the stock prices, but not the variances and covariances.
19.1 Hedging when ,1 < < 1
XdX = 1 1 dS1 + 2 dS2 + r(X , 1S1 , 2S2) dt d = (dX , rX dt)
= 1 1(dS1 , rS1 dt) + 1 2 (dS2 , rS2 dt) q 1 1 e e 2 e = 1S1 1 dB1 + 2 S2 2 dB1 + 1 , 2 dB2 :
Let V be F (T )measurable. Define the If P martingale
Y (t) = If E (VT ) F (t) ;
0 t T:
CHAPTER 19. A twodimensional market model
205
The Martingale Representation Corollary implies
Y (t) = Y (0) + We have
Zt 0
1 dBe1 +
Zt 0
2 dBe2:
X 1 1 d = 1 S1 1 + 2S2 2 dBe1 q + 1 2S2 1 , 2 2 dBe2 ; dY = 1 dBe1 + 2 dBe2 :
We solve the equations
1 S + 1 S = 1 1 1 q 2 2 2 1 1 S 1 , 2 = 2 2
2
2
for the hedging portfolio (1; 2). With this choice of (1 ; 2) and setting
X (0) = Y (0) = If E (VT ) ; we have X (t) = Y (t);
0 t T; and in particular,
X (T ) = V: Every F (T )measurable random variable can be hedged; the market is complete.
19.2 Hedging when = 1 The case = ,1 is analogous. Assume that = 1. Then
dS1 = S1 [1 dt + 1 dB1 ] dS2 = S2 [2 dt + 2 dB1 ] The stocks are perfectly correlated. The market price of risk equations are
11 = 1 , r 21 = 2 , r The process 2 is free. There are two cases:
(MPR)
206 1 ,r Case I: 6= 2 ,2 r : There is no solution to (MPR), and consequently, there is no riskneutral 1 measure. This market admits arbitrage. Indeed
X 1 d = 1(dS1 , rS1 dt) + 1 2 (dS2 , rS2 dt)
= 1 1S1 [(1 , r) dt + 1 dB1] + 1 2S2 [(2 , r) dt + 2 dB1]
1 ,r Suppose 1
> ,r : Set 2
2
1 = 1S ; 2 = , 1S : 1 1
Then
2 2
1 , r X 1 1 , r 2 d = dt + dB1 , dt + dB1 2 1 ,1 r 2 , r 1 = , dt  1 {z 2 } Positive
1 ,r Case II: 1
= ,r : The market price of risk equations 2
2
11 = 1 , r 21 = 2 , r 1 = 1, r = 2, r ;
have the solution
1
2
2 is free; there are infinitely many riskneutral measures. Let If P be one of them. Hedging:
X 1 d = 1S1 [(1 , r) dt + 1 dB1] + 1 2S2 [(2 , r) dt + 2 dB1] = 1 1S1 1[1 dt + dB1 ] + 1 2 S2 2 [1 dt + dB1 ] 1 1 = S + S dBe :
1 1 1
e2 does not appear. Notice that B
2 2 2
1
Let V be an F (T )measurable random variable. If V depends on B2 , then it can probably not be hedged. For example, if
V = h(S1 (T ); S2(T ));
and 1 or 2 depend on B2 , then there is trouble.
CHAPTER 19. A twodimensional market model
207
More precisely, we define the If P martingale
Y (t) = If E (VT ) F (t) ;
We can write
Y (t) = Y (0) +
Zt 0
1 dBe1 +
so
dY = 1 dBe1 + 2 dBe2 :
To get d X to match dY , we must have
2 = 0:
0 t T:
Zt 0
2 dBe2 ;
208
Chapter 20
Pricing Exotic Options 20.1 Reflection principle for Brownian motion Without drift. Define
M (T ) = 0max B (t): tT Then we have:
IP fM (T ) > m; B(T ) < bg
= IP fB (T ) > 2m , bg ( 2) Z1 1 =p exp , 2xT dx; m > 0; b < m 2T 2m,b
So the joint density is
(
) !
Z1 2 2 IP fM (T ) 2 dm; B (T ) 2 dbg = , @m@ @b p 1 exp , x dx dm db 2T 2T 2m,b ( 2 )! @ 1 (2 m , b ) = , @m p exp , 2T dm db; 2T ( 2) 2(2 m , b ) (2 m , b ) = p exp , 2T dm db; m > 0; b < m: T 2T With drift. Let
Be (t) = t + B(t); 209
210
shadow path
2mb
m b Brownian motion Figure 20.1: Reflection Principle for Brownian motion without drift
m m=b (B(T), M(T)) lies in here
b
Figure 20.2: Possible values of B (T ); M (T ).
CHAPTER 20. Pricing Exotic Options where B (t);
211
0 t T , is a Brownian motion (without drift) on ( ; F ; P). Define
Z (T ) = expf,B (T ) , 12 2 T g = expf,(B (T ) + T ) + 21 2 T g = expf,Be (t) + 12 2 T g; Z If P (A) = Z (T ) dIP; 8A 2 F : A
f(T ) = max0tT SetM
Be (T ):
Under If P ; Be is a Brownian motion (without drift), so
~b) ( (2m ~b)2 ) 2(2 m ~ , ~ , f f e ~ IP fM (T ) 2 dm; ~ B (T ) 2 dbg = p exp , 2T dm~ d~b; m~ > 0; ~b < m: ~
T 2T
Let h(m; ~ ~b) be a function of two variables. Then
f
e
f(T ); Be (T )) = If IEh(M E h(M (ZT()T; B) (T )) h
f(T ); Be (T )) expfBe (T ) , 21 2T g = If E h(M
=
m~Z=1 ~bZ=m~
m~ =0 ~b=,1
i
f(T ) 2 dm; h(m; ~ ~b) expf~b , 12 2 T g If P fM ~ Be (T ) 2 d~bg:
But also,
f(T ); Be (T )) = IEh(M
m~Z=1 ~bZ=m~ m~ =0 ~b=,1
f(T ) 2 dm; h(m; ~ ~b) IP fM ~ Be (T ) 2 d~bg:
Since h is arbitrary, we conclude that (MPR)
f(T ) 2 dm; IP fM ~ Be (T ) 2 d~bg f(T ) 2 dm; = expf~b , 12 2 T g If P fM ~ Be (T ) 2 d~bg ( ) ~ , ~b) exp , (2m ~ , ~b)2 : expf~b , 1 2 T gdm = 2(2pm ~ d~b; m ~ > 0; ~b < m: ~ 2 2T T 2T
212
20.2 Up and out European call. Let 0 < K
< L be given. The payoff at time T is (S (T ) , K )+ 1fS (T )
where
S (T ) = 0max S (t): tT
To simplify notation, assume that IP is already the riskneutral measure, so the value at time zero of the option is
h
i
v (0; S (0)) = e,rT IE (S (T ) , K )+1fS(T )
8 2 39 > > > < 66 77> = r = S0 exp > 64B (t) + , 2 t75> >  {z } > : ; = S0 expf Be (t)g;
where
r = ,2 ;
Be (t) = t + B (t): Consequently,
f(t)g; S (t) = S0 expf M where,
f(t) = max Be (u): M 0ut We compute,
h
v (0; S (0)) = e,rT IE (S (T ) , K )+1fS(T )
i +
e S (0) expfB (T )g , K 1fS(0)expfMe (T )g < Lg # " = e,rT IE S (0) expf Be (T )g , K 1 1 K ; Me (T )< 1 log L Be(T )> log  {zS (0)}  {zS (0)} = e,rT IE
b
~
m~
CHAPTER 20. Pricing Exotic Options
213
~ M(T) y
x=y
~ m (B(T), M(T)) lies in here
~ b
x ~ B(T)
e (T ); Mf(T ). Figure 20.3: Possible values of B We consider only the case
S (0) K < L; so 0 ~b < m: ~ The other case, K < S (0) L leads to ~b < 0 m ~ and the analysis is similar. R R m ~ m ~ We compute ~b x : : :dy dx:
Z m~ Z m~
(
)
2(2py , x) exp , (2y , x)2 + x , 1 2 T dy dx ( S (0) exp f x g , K ) 2 ~b x 2T T 2T( ) y=m~ Z m~ 2 1 (2 y , x ) 1 2 , rT + x , 2 T dx = ,e (S (0) expfxg , K ) p exp , ~b 2T 2T y=x " ( 2 ) Z m~ = e,rT ~ (S (0) expfxg , K ) p 1 exp , x + x , 12 2 T 2T b 2T ( )# 2 , exp , (2m~2,T x) + x , 12 2 T dx ) Z m~ ( 2 1 x , rT 2 1 = p e S (0) ~ exp x , 2T + x , 2 T dx 2T b ( 2 ) Z m~ 1 x , rT 2 1 , p e K ~ exp , 2T + x , 2 T dx 2T b ) Z m~ ( 2 1 (2 m ~ , x ) , rT 2 1 , p e S (0) ~ exp x , 2T + x , 2 T dx 2T b ( ) Z 2 m ~ 1 (2 m ~ , x ) , rT 2 1 + p e K ~ exp , 2T + x , 2 T dx: 2T b
v (0; S (0)) = e,rT
The standard method for all these integrals is to complete the square in the exponent and then recognize a cumulative normal distribution. We carry out the details for the first integral and just
214 give the result for the other three. The exponent in the first integrand is
2 x , 2xT + x , 12 2 T
= , 21T (x , T , T )2 + 12 2T + T rT T 2 1 = , 2T x , , 2 + rT:
In the first integral we make the change of variable
p
p
y = (x , rT= , T=2)= T; dy = dx= T; to obtain
(
)
e,prT S (0) Z m~ exp x , x2 + x , 1 2 T dx 2 2T 2T ~b ( rT T 2) Z m~ 1 1 = p S (0) ~ exp , 2T x , , 2 dx 2T b p p pmT , r T , T Z 2 = p 1 S (0): expf, y2 g dy 2T p p pbT , r T , T " p ! p ~b rpT pT !# T T m ~ r : = S (0) N p , , 2 , N p , , 2 T T ~
2
~
2
Putting all four integrals together, we have
"
p
p !
p
p !#
~ v (0; S (0)) = S (0) N pm~ , r T , 2 T , N pb , r T , 2 T " T p p ! T ~ p p !# m ~ r T T b r T , rT , e K N pT , + 2 , N pT , + 2 T " p p ! ~b) rpT pT !# r m ~ T T (2 m ~ , , S (0) N pT + + 2 , N pT + + 2 r m~ rpT pT ! + exp ,rT + 2m ~ ,2 N p + , 2 , T ! p p ~ N (2m~p, b) + r T , 2 T ; T where
~b = 1 log K ; m~ = 1 log L : S (0) S (0)
CHAPTER 20. Pricing Exotic Options
215
v(t,L) = 0
L
v(T,x) = (x  K)+
v(t,0) = 0
T
Figure 20.4: Initial and boundary conditions. If we let L!1 we obtain the classical BlackScholes formula
"
~b
p
p !#
v(0; S (0)) = S (0) 1 , N p , r T , 2 T T " ~b rpT pT !# , rT ,e K 1,N p , + 2 T p p ! 1 T S (0) r = S (0)N p log K + + 2 T T p p ! 1 r T S (0) , rT , e KN p log K + , 2 T : T
If we replace T by T , t and replace S (0) by x in the formula for v (0; S (0)), we obtain a formula for v (t; x), the value of the option at the time t if S (t) = x. We have actually derived the formula under the assumption x K L, but a similar albeit longer formula can also be derived for K < x L. We consider the function
h i v(t; x) = IE t;x e,r(T ,t)(S (T ) , K )+ 1fS (T )
This function satisfies the terminal condition
v(T; x) = (x , K )+ ; 0 x < L and the boundary conditions
v(t; 0) = 0; 0 t T; v(t; L) = 0; 0 t T: We show that v satisfies the BlackScholes equation
,rv + vt + rxvx + 21 2x2vxx; 0 t < T; 0 x L:
216 Let S (0) > 0 be given and define the stopping time
= minft 0; S (t) = Lg: Theorem 2.61 The process
e,r(t^ )v (t ^ ; S (t ^ )); 0 t T; is a martingale. Proof: First note that
S (T ) < L () > T: Let ! 2 be given, and choose t 2 [0; T ]. If (!) t, then
IE e,rT (S (T ) , K )+1fS (T )
But when (!) t, we have
v(t ^ (!); S (t ^ (!); !)) = v (t ^ (!); L) = 0; so we may write
IE
e,rT (S (T ) , K )+ 1fS (T )
On the other hand, if (!) > t, then the Markov property implies
IE e,rT (S (T ) , K )+ 1fS (T )
In both cases, we have
e,r(t^ )v(t ^ ; S (t ^ )) = IE e,rT (S (T ) , K )+ 1fS (T )
Suppose 0 u t T . Then
IE , rT + = IE IE e (S (T ) , K ) 1fS (T )
e,r(t^ )v(t ^ ; S (t ^ )) F (u)
= e,r(u^ )v (u ^ ; S (u ^ )) :
CHAPTER 20. Pricing Exotic Options
217
For 0 t T , we compute the differential
d e,rt v(t; S (t)) = e,rt(,rv + vt + rSvx + 21 2 S 2vxx ) dt + e,rt Svx dB:
Integrate from 0 to t ^ :
e,r(t^ )v (t ^ ; S (t ^ )) = v(0; S (0)) Z t^ + e,ru (,rv + vt + rSvx + 12 2 S 2vxx ) du 0
Z t^
+
0
e,ru Svx dB:
{z
}
A stopped martingale is still a martingale Because e,r(t^ )v (t ^ ; S (t ^ )) is also a martingale, the Riemann integral
Z t^ 0
e,ru (,rv + vt + rSvx + 21 2 S 2vxx ) du
is a martingale. Therefore,
,rv(u; S (u))+ vt(u; S (u)) + rS (u)vx(u; S (u))+ 12 2S 2(u)vxx(u; S (u)) = 0; 0 u t ^ : The PDE
,rv + vt + rxvx + 21 2x2vxx = 0; 0 t T; 0 x L;
then follows.
d e,rtv (t; S (t)) = e,rt S (t)vx(t; S (t)) dB(t); 0 t : Let X (t) be the wealth process corresponding to some portfolio (t). Then The Hedge
d(e,rtX (t)) = e,rt (t)S (t) dB(t): We should take and
X (0) = v (0; S (0)) (t) = vx (t; S (t)); 0 t T ^ :
Then
X (T ^ ) = v(T ^ ; S (T ^ )) ( v(T; S (T )) = (S (T ) , K )+ = v(; L) = 0
if if
>T T.
218 v(T, x)
0
K
L
x
L
x
v(t, x)
0
K
Figure 20.5: Practial issue.
20.3 A practical issue For t < T but t near T , v (t; x) has the form shown in the bottom part of Fig. 20.5. In particular, the hedging portfolio
(t) = vx (t; S (t))
can become very negative near the knockout boundary. The hedger is in an unstable situation. He should take a large short position in the stock. If the stock does not cross the barrier L, he covers this short position with funds from the money market, pays off the option, and is left with zero. If the stock moves across the barrier, he is now in a region of (t) = vx(t; S (t)) near zero. He should cover his short position with the money market. This is more expensive than before, because the stock price has risen, and consequently he is left with no money. However, the option has “knocked out”, so no money is needed to pay it off. Because a large short position is being taken, a small error in hedging can create a significant effect. Here is a possible resolution. Rather than using the boundary condition
v(t; L) = 0; 0 t T; solve the PDE with the boundary condition
v(t; L) + Lvx (t; L) = 0; 0 t T; where is a “tolerance parameter”, say 1%. At the boundary, Lvx (t; L) is the dollar size of the short position. The new boundary condition guarantees: 1.
Lvx(t; L) remains bounded;
2. The value of the portfolio is always sufficient to cover a hedging error of size of the short position.
times the dollar
Chapter 21
Asian Options Stock:
dS (t) = rS (t) dt + S (t) dB (t):
Payoff:
V =h Value of the payoff at time zero:
X (0) = IE
ZT 0
"
e,rT h
!
S (t) dt
ZT 0
!#
S (t) dt :
Introduce an auxiliary process Y (t) by specifying
dY (t) = S (t) dt: With the initial conditions
S (t) = x; Y (t) = y; we have the solutions
n
o
S (T ) = x exp (B(T ) , B (t)) + (r , 21 2 )(T , t) ; Y (T ) = y +
ZT t
S (u) du:
Define the undiscounted expected payoff
u(t; x; y) = IE t;x;y h(Y (T )); 0 t T; x 0; y 2 IR: 219
220
21.1 FeynmanKac Theorem The function u satisfies the PDE
ut + rxux + 12 2x2 uxx + xuy = 0; 0 t T; x 0; y 2 IR;
the terminal condition
u(T; x; y ) = h(y ); x 0; y 2 IR;
and the boundary condition
u(t; 0; y) = h(y ); 0 t T; y 2 IR: One can solve this equation. Then
v t; S (t);
Zt
is the option value at time t, where
0
The PDE for v is
S (u) du
v(t; x; y ) = e,r(T ,t) u(t; x; y ):
,rv + vt + rxvx + 12 2x2vxx + xvy = 0;
v (T; x; y ) = h(y ); v(t; 0; y) = e,r(T ,t)h(y ): One can solve this equation rather than the equation for u. 21.2 Constructing the hedge
Start with the stock price S (0). The differential of the value X (t) of a portfolio (t) is
dX = dS + r(X , S ) dt = S (r dt + dB ) + rX dt , rS dt = S dB + rX dt:
We want to have
X (t) = v t; S (t);
Zt
so that
X (T ) = v T; S (0); =h
ZT 0
0
S (u) du ;
ZT 0
!
S (u) du ;
!
S (u) du :
(1.1)
CHAPTER 21. Asian Options
221
The differential of the value of the option is
dv t; S (t);
Zt 0
S (u) du = vtdt + vx dS + vy S dt + 12 vxx dS dS = (vt + rSvx + Svy + 12 2 S 2vxx ) dt + Svx dB = rv (t; S (t)) dt + vx (t; S (t)) S (t) dB (t): (From Eq. 1.1)
Compare this with
dX (t) = rX (t) dt + (t) S (t) dB(t): Take (t) = vx (t; S (t)): If X (0) = v (0; S (0); 0), then
X (t) = v t; S (t);
Zt 0
S (u) du ; 0 t T;
because both these processes satisfy the same stochastic differential equation, starting from the same initial condition.
21.3 Partial average payoff Asian option Now suppose the payoff is
V =h where 0 <
ZT
!
S (t) dt ;
< T . We compute v(; x; y ) = IE ;x;y e,r(T , )h(Y (T )) just as before. For 0 t , we compute next the value of a derivative security which pays off v (; S ( ); 0) at time . This value is w(t; x) = IE t;xe,r( ,t) v (; S ( ); 0): The function w satisfies the BlackScholes PDE ,rw + wt + rxwx + 12 2x2wxx = 0; 0 t ; x 0; with terminal condition
w(; x) = v (; x; 0); x 0;
and boundary condition
w(t; 0) = e,r(T ,t)h(0); 0 t T:
The hedge is given by
8
0 t ;
< t T:
222 Remark 21.1 While no closedform for the Asian option price is known, the Laplace transform (in 2 the variable 4 (T , t)) has been computed. See H. Geman and M. Yor, Bessel processes, Asian options, and perpetuities, Math. Finance 3 (1993), 349–375.
Chapter 22
Summary of Arbitrage Pricing Theory A simple European derivative security makes a random payment at a time fixed in advance. The value at time t of such a security is the amount of wealth needed at time t in order to replicate the security by trading in the market. The hedging portfolio is a specification of how to do this trading.
22.1 Binomial model, Hedging Portfolio Let be the set of all possible sequences of n cointosses. We have no probabilities at this point. Let r 0; u > r + 1; d = 1=u be given. (See Fig. 2.1) Evolution of the value of a portfolio:
Xk+1 = k Sk+1 + (1 + r)(Xk , k Sk ): Given a simple European derivative security V (! 1; ! 2), we want to start with a nonrandom X0 and use a portfolio processes
0 ; 1(H ); 1 (T ) so that
X2 (!1 ; !2 ) = V (!1 ; !2 ) 8! 1; !2 :
(four equations)
There are four unknowns: X0; 0; 1(H ); 1(T ). Solving the equations, we obtain: 223
224
2 3 6 u , (1 + r) X (! ; T )77 ; ( ! ; H ) X1 (!1 ) = 1 +1 r 64 1 +u ,r ,d d X + 2 1  {z } u , d  2 {z1 }5 V (! ;H ) V (! ;T ) 1 + r , d u , (1 + r ) 1 X0 = 1 + r u , d X1(H ) + u , d X1(T ) ; ) , X2(! 1 ; T ) ; 1 (!1 ) = XS2 ((!!1 ;; H 2 1 H ) , S2 (!1 ; T ) X ) , X1(T ) : 0 = S1 ((H 1 H ) , S1 (T ) 1
1
The probabilities of the stock price paths are irrelevant, because we have a hedge which works on every path. From a practical point of view, what matters is that the paths in the model include all the possibilities. We want to find a description of the paths in the model. They all have the property
(log Sk+1 , log Sk
)2 =
Sk+1 2 log Sk
= ( log u)2 = (log u)2 : Let
= log u > 0. Then
nX ,1 k=0
(log Sk+1 , log Sk )2 = 2 n:
The paths of log Sk accumulate quadratic variation at rate 2 per unit time.
If we change u, then we change , and the pricing and hedging formulas on the previous page will give different results. We reiterate that the probabilities are only introduced as an aid to understanding and computation. Recall:
Xk+1 = k Sk+1 + (1 + r)(Xk , k Sk ):
Define
k = (1 + r)k :
Then
Xk+1 = Sk+1 + Xk , Sk ; k k k+1 k+1 k k
i.e.,
Xk+1 , Xk = Sk+1 , Sk : k k+1 k k+1 k
In continuous time, we will have the analogous equation
d X ((tt)) = (t) d S ((tt)) :
CHAPTER 22. Summary of Arbitrage Pricing Theory
225
P under which S kk is a martingale, then X kk will also be a If we introduce a probability measure If martingale, regardless of the portfolio used. Indeed,
If E X k+1 F k = If E X k + k S k+1 , S k F k k+1 k k+1 k X S S k +1 k k f = + k IE F k , : k k+1 {z k }  =0
Suppose we want to have must have
X2 = V , where V
is some
F 2measurable random variable.
Then we
X2 V 1 X = X1 = If f F 1 ; E F = I E 1 1 + r 1 1 2 2 E X 1 = If E V : X0 = X 0 = If 0
To find the riskneutral probability measure
1
2
If P under which S kk
is a martingale, we denote
If P f! k = H g, q~ = If P f!k = T g, and compute Sk+1 f IE F k = p~u Sk + q~d Sk k+1 k+1 k+1 1 = 1 + r [~pu + q~d] S k :
p~ =
k
We need to choose p~ and q~ so that
p~u + q~d = 1 + r; p~ + q~ = 1: The solution of these equations is
p~ = 1 +u ,r ,d d ; q~ = u ,u(1, +d r) : 22.2 Setting up the continuous model Now the stock price S (t); 0 t T , is a continuous function of t. We would like to hedge along every possible path of S (t), but that is impossible. Using the binomial model as a guide, we choose > 0 and try to hedge along every path S (t) for which the quadratic variation of log S (t) accumulates at rate 2 per unit time. These are the paths with volatility 2. To generate these paths, we use Brownian motion, rather than cointossing. To introduce Brownian motion, we need a probability measure. However, the only thing about this probability measure which ultimately matters is the set of paths to which it assigns probability zero.
226
B (t); 0 t T , be a Brownian motion defined on a probability space ( ; F ; P). 2 IR, the paths of t + B(t) accumulate quadratic variation at rate 2 per unit time. We want to define Let
For any
S (t) = S (0) expft + B(t)g; so that the paths of
log S (t) = log S (0) + t + B (t)
accumulate quadratic variation at rate 2 per unit time. Surprisingly, the choice of in this definition is irrelevant. Roughly, the reason for this is the following: Choose ! 1 2 . Then, for 1 2 IR,
1t + B(t; !1); 0 t T; is a continuous function of t. If we replace 1 by 2, then 2 t + B (t; ! 1 ) is a different function. However, there is an ! 2 2 such that
1t + B (t; !1 ) = 2t + B(t; !2); 0 t T: In other words, regardless of whether we use 1 or 2 in the definition of S (t), we will see the same paths. The mathematically precise statement is the following: If a set of stock price paths has a positive probability when S (t) is defined by
S (t) = S (0) expf1t + B(t)g; then this set of paths has positive probability when S (t) is defined by
S (t) = S (0) expf2t + B(t)g: Since we are interested in hedging along every path, except possibly for a set of paths which has probability zero, the choice of is irrelevant. The most convenient choice of is so
= r , 21 2 ;
S (t) = S (0) expfrt + B(t) , 12 2 tg;
and
e,rt S (t) = S (0) expfB(t) , 12 2tg is a martingale under IP . With this choice of , dS (t) = rS (t) dt + S (t) dB (t)
CHAPTER 22. Summary of Arbitrage Pricing Theory
227
and IP is the riskneutral measure. If a different choice of is made, we have
S (t) = S (0) expft + B(t)g; dS (t) = ( +{z12 2 )} S (t) dt + S (t) dB(t):
i h = rS (t) dt + , r dt + dB (t) : 
Be has the same paths as B.
{z
dBe(t)
}
We can change to the riskneutral measure If P , under which Brownian motion, and then proceed as if had been chosen to be equal to r , 21 2 .
Be is a
22.3 Riskneutral pricing and hedging Let If P denote the riskneutral measure. Then
dS (t) = rS (t) dt + S (t) dBe (t); e is a Brownian motion under IfP . Set where B (t) = ert :
S (t)
Then
d (t) = S ((tt)) dBe (t);
P. so S ((tt)) is a martingale under If Evolution of the value of a portfolio:
dX (t) = (t)dS (t) + r(X (t) , (t)S (t)) dt; which is equivalent to
X (t)
S (t)
d (t) = (t)d (t) = (t) S ((tt)) dBe (t):
(3.1)
(3.2)
(t) f Regardless of the portfolio used, X (t) is a martingale under IP . Now suppose V is a given F (T )measurable random variable, the payoff of a simple European derivative security. We want to find the portfolio process (T ); 0 t T , and initial portfolio (t) value X (0) so that X (T ) = V . Because X (t) must be a martingale, we must have
V X (t) = If E (t) (T ) F (t) ; 0 t T:
This is the riskneutral pricing formula. We have the following sequence:
(3.3)
228 1.
V
is given,
2. Define X (t); 0 t T , by (3.3) (not by (3.1) or (3.2), because we do not yet have (t)).
3. Construct (t) so that (3.2) (or equivalently, (3.1)) is satisfied by the defined in step 2.
X (t); 0 t T ,
(t) To carry out step 3, we first use the tower property to show that X (t) defined by (3.3) is a martingale under If P . We next use the corollary to the Martingale Representation Theorem (Homework Problem 4.5) to show that
X (t)
d (t) = (t) dBe (t)
(3.4)
(t) = (t) (t) : S (t)
(3.5)
for some proecss . Comparing (3.4), which we know, and (3.2), which we want, we decide to define
Then (3.4) implies (3.2), which implies (3.1), which implies that X (t); 0 the portfolio process (t); 0 t T .
t T , is the value of
From (3.3), the definition of X , we see that the hedging portfolio must begin with value
V f X (0) = IE (T ) ;
and it will end with value
V f X (T ) = (T )IE (T ) F (T ) = (T ) (VT ) = V:
Remark 22.1 Although we have taken r and to be constant, the riskneutral pricing formula is still “valid” when r and are processes adapted to the filtration generated by B . If they depend on e or on S , they are adapted to the filtration generated by B. The “validity” of the riskneutral either B pricing formula means:
V f X (0) = IE (T ) ; then there is a hedging portfolio (t); 0 t T , such that X (T ) = V ;
1. If you start with
2. At each time t, the value X (t) of the hedging portfolio in 1 satisfies
V X (t) = If (t) E (T ) F (t) :
Remark 22.2 In general, when there are multiple assets and/or multiple Brownian motions, the riskneutral pricing formula is valid provided there is a unique riskneutral measure. A probability measure is said to be riskneutral provided
CHAPTER 22. Summary of Arbitrage Pricing Theory
229
it has the same probabilityzero sets as the original measure; it makes all the discounted asset prices be martingales.
To see if the riskneutral measure is unique, compute the differential of all discounted asset prices e so that all these differentials have only dBe and check if there is more than one way to define B terms.
22.4 Implementation of riskneutral pricing and hedging To get a computable result from the general riskneutral pricing formula
V X (t) = If E (t) (T ) F (t) ;
one uses the Markov property. We need to identify some state variables, the stock price and possibly other variables, so that
X (t) = (t)If E (VT ) F (t)
is a function of these variables. Example 22.1 Assume r and are constant, and V variable. Define
= h(S(T)). We can take the stock price to be the state
h i v(t; x) = IEe t;x e,r(T ,t) h(S(T)) :
Then
, rT e X(t) = e h(S(T)) F (t) ert IE
= v(t; S(t));
(t) and X (t)
= e,rt v(t; S(t)) is a martingale under IPe .
Example 22.2 Assume r and are constant.
V =h Take S(t) and Y (t) =
ZT 0
!
S(u) du :
R t S(u) du to be the state variables. Define h i v(t; x; y) = IEe t;x;y e,r T ,t h(Y (T)) ; 0
(
where
Y (T) = y +
ZT t
)
S(u) du:
230 Then
X(t)
= ert
eIE e,rT h(S(T)) F (t)
= v(t; S(t); Y (t)) and
e
is a martingale under IP .
X(t) = e,rt v(t; S(t); Y (t)) (t)
Example 22.3 (Homework problem 4.2)
e dS(t) = r(t; Y (t)) S(t)dt + (t; Y (t))S(t) dB(t); e dY (t) = (t; Y (t)) dt + (t; Y (t)) dB(t); V = h(S(T)):
Take S(t) and Y (t) to be the state variables. Define
2 3 77 66 ( Z ) T 7 t;x;y 6 v(t; x; y) = IEe 66exp , r(u; Y (u)) du h(S(T ))77 : t 75 64 {z } (t) (T )
Then
)) F (t) X(t) = (t)IEe h(S(T " ( (T ) = IEe exp ,
ZT t
)
# r(u; Y (u)) du h(S(T)) F (t)
= v(t; S(t); Y (t)); and
e
X(t) = exp , Z t r(u; Y (u)) du v(t; S(t); Y (t)) (t) 0
is a martingale under IP .
In every case, we get an expression involving v to be a martingale. We take the differential and set the dt term to zero. This gives us a partial differential equation for v , and this equation must e term in the differential of the equation is the hold wherever the state processes can be. The dB differential of a martingale, and since the martingale is
X (t) = X (0) + Z t (u) S (u) dBe (u) (t) (u) 0
we can solve for (t). This is the argument which uses (3.4) to obtain (3.5).
CHAPTER 22. Summary of Arbitrage Pricing Theory
231
Example 22.4 (Continuation of Example 22.3)
X(t) = exp , Z t r(u; Y (u)) du v(t; S(t); Y (t)) (t)  0 {z } 1= (t)
e X(t) 1 d (t) = (t) ,r(t; Y (t))v(t; S(t); Y (t)) dt
is a martingale under IP . We have
+ vt dt + vx dS + vy dY
+ vxx dS dS + vxy dS dY + vyy dY dY 1 2
1 2
1 = (t) (,rv + vt + rSvx + vy + S vxx + Svxy + vyy ) dt e + (Svx + vy ) dB 1 2
2
2
1 2
2
The partial differential equation satisfied by v is
,rv + vt + rxvx + vy + 12 2x2 vxx + xvxy + 12 2 vyy = 0 where it should be noted that v
= v(t; x; y), and all other variables are functions of (t; y). We have X(t) 1 e d (t) = (t) [Svx + vy ] dB(t); where = (t; Y (t)), = (t; Y (t)), v = v(t; S(t); Y (t)), and S = S(t). We want to choose (t) so that (see (3.2))
X(t)
e d (t) = (t)(t; Y (t)) S(t) (t) dB(t):
Therefore, we should take (t) to be
Y (t)) (t) = vx (t; S(t); Y (t)) + (t; (t; Y (t)) S(t) vy (t; S(t); Y (t)):
232
Chapter 23
Recognizing a Brownian Motion Theorem 0.62 (Levy) Let B (t); 0 F (t); 0 t T , such that:
t T; be a process on ( ; F ; P), adapted to a filtration
1. the paths of B (t) are continuous, 2. 3.
B is a martingale, hBi(t) = t; 0 t T , (i.e., informally dB(t) dB(t) = dt).
Then B is a Brownian motion. Proof: (Idea) Let 0 s < t T be given. We need to show that B (t) , B (s) is normal, with mean zero and variance t , s, and B (t) , B (s) is independent of F (s). We shall show that the conditional moment generating function of B (t) , B (s) is
IE
eu(B (t),B(s)) F (s)
1
= e 2 u (t,s) : 2
Since the moment generating function characterizes the distribution, this shows that B (t) , B (s) is normal with mean 0 and variance t , s, and conditioning on F (s) does not affect this, i.e., B (t) , B(s) is independent of F (s). We compute (this uses the continuity condition (1) of the theorem)
deuB(t) = ueuB(t)dB(t) + 12 u2 euB(t)dB (t) dB (t); so
euB (t) = euB (s) +
Zt s
ueuB(v) dB(v ) + 12 u2 233
Zt s
euB(v)
dv: {z}
uses cond. 3
234
R Now 0t ueuB (v)dB (v ) is a martingale (by condition 2), and so IE
= 0:
IE
euB(t) F (s)
ueuB(v) dB(v ) F (s)
s
=,
It follows that
Z t
Zs 0
Z t
ueuB(v) dB(v ) + IE
0
Zt 2 uB ( v ) uB ( s ) 1 F (s) =e + 2 u IE e s
We define
'(v) = IE
euB(v) F (s)
;
so that
'(s) = euB(s) and
'(t) = euB(s) + 21 u2 '0(t) = 12 u2 '(t); 1 '(t) = ke 2 u t :
Zt s
'(v ) dv;
2
Plugging in s, we get
1
1
euB (s) = ke 2 u s =)k = euB(s), 2 u s : 2
Therefore,
1 IE eu(B(t),B(s)) F (s) = e 2 u (t,s):
1
IE euB(t) F (s) = '(t) = euB(s)+ 2 u (t,s) ; 2
2
2
ueuB (v)dB(v ) F (s)
dv:
CHAPTER 23. Recognizing a Brownian Motion
23.1 Identifying volatility and correlation Let B1 and B2 be independent Brownian motions and
dS1 = r dt + dB + dB ; 11 1 12 2 S1 dS2 = r dt + dB + dB ; 21 1 22 2 S 2
Define
q
Define processes W1 and W2 by
1 = 211 + 212 ; q 2 = 221 + 222 ; = 1121 + 1222 : 1 2
dW1 = 11 dB1 + 12 dB2 1 dB + dW2 = 21 1 22 dB2 : 2
Then W1 and W2 have continuous paths, are martingales, and
dW1 dW1 = 12 (11dB1 + 12 dB2 )2 1
= 12 (211dB1 dB1 + 212dB2 dB2 ) 1
= dt; and similarly
dW2 dW2 = dt:
Therefore, W1 and W2 are Brownian motions. The stock prices have the representation
dS1 = r dt + dW ; 1 1 S1 dS2 = r dt + dW : 2 2 S 2
The Brownian motions W1 and W2 are correlated. Indeed,
dW1 dW2 = 1 (11dB1 + 12 dB2 )(21dB1 + 22dB2) 1 2
= 1 (11 21 + 12 22 ) dt 1 2 = dt:
235
236
23.2 Reversing the process Suppose we are given that
dS1 = r dt + dW ; 1 1 S1 dS2 = r dt + dW ; 2 2 S 2
where W1 and W2 are Brownian motions with correlation coefficient . We want to find
"
#
#"
#
= 11 12 21 22 so that
"
11 12 11 21 " 21 2 22 2 12 22 # + + 11 21 12 22 11 12 = + 221 + 222 11 21 12 22 " 2 # 1 2 1 = 22 1 2
0 =
A simple (but not unique) solution is (see Chapter 19)
11 = 1; 12 = 0; q 21 = 2 ; 22 = 1 , 2 2 : This corresponds to
1 dW1 = 1dB1 =)dB1 = dW1; q 2 dW2 = 2 dB1 + 1 , 2 2 dB2 1 ; ( 6= 1) =) dB2 = dWp2 , dW 2 1, If = 1, then there is no B2 and dW2 = dB1 = dW1: Continuing in the case 6= 1, we have dB1 dB1 = dW1 dW1 = dt; dB2 dB2 = 1 ,1 2 dW2 dW2 , 2 dW1 dW2 + 2dW2 dW2
= 1 ,1 2 dt , 22 dt + 2 dt = dt;
CHAPTER 23. Recognizing a Brownian Motion
237
so both B1 and B2 are Brownian motions. Furthermore,
dB1 dB2 = p 1
(dW1 dW2 , dW1 dW1 ) 1 , 2 = p 1 2 ( dt , dt) = 0: 1,
We can now apply an Extension of Levy’s Theorem that says that Brownian motions with zero crossvariation are independent, to conclude that B1 ; B2 are independent Brownians.
238
Chapter 24
An outside barrier option Barrier process:
dY (t) = dt + dB (t): 1 1 Y (t) Stock process:
dS (t) = dt + dB (t) + q1 , 2 dB (t); 2 1 2 2 S (t)
> 0; 2 > 0; ,1 < < 1, and B1 and B2 are independent Brownian motions on some ( ; F ; P). The option pays off: (S (T ) , K )+ 1fY (T )
at time T , where
0 < S (0) < K; 0 < Y (0) < L;
Y (T ) = 0max Y (t): tT Remark 24.1 The option payoff depends on both the Y and S processes. In order to hedge it, we will need the money market and two other assets, which we take to be Y and S . The riskneutral measure must make the discounted value of every traded asset be a martingale, which in this case means the discounted Y and S processes. We want to find 1 and 2 and define
dBe1 = 1 dt + dB1; dBe2 = 2 dt + dB2 ; 239
240 so that
dY = r dt + dBe 1 1 Y = r dt + 1 1 dt + 1 dB1 ; dS = r dt + dBe + q1 , 2 dBe 2 1 2 2 S
q
= r dt + 2 1 dt + 1 , 2 2 2 dt q + 2 dB1 + 1 , 2 2 dB2 :
We must have
= r + 1 1 ; q = r + 21 + 1 , 2 2 2:
(0.1) (0.2)
We solve to get
1 = , r ; 1 , 2 1 : 2 = p r , 2 1 , 2 We shall see that the formulas for 1 and 2 do not matter. What matters is that (0.1) and (0.2) uniquely determine 1 and 2 . This implies the existence and uniqueness of the riskneutral measure. We define
n
o
Z (T ) = exp ,1 B1(T ) , 2B2 (T ) , 21 (12 + 22 )T ; Z If P (A) = Z (T ) dIP; 8A 2 F : A
Under If P , Be1 and Be2 are independent Brownian motions (Girsanov’s Theorem). riskneutral measure. Remark 24.2 Under both IP and If P , Y has volatility 1, S has volatility 2 and
i.e., the correlation between dY Y
dY dS = dt; 1 2 YS and dS S is .
The value of the option at time zero is
h i v(0; S (0); Y (0)) = If E e,rT (S (T ) , K )+ 1fY (T )
We need to work out a density which permits us to compute the righthand side.
If P is the unique
CHAPTER 24. An outside barrier option
241
Recall that the barrier process is
dY = r dt + dBe ; 1 1 Y so
o
n
Y (t) = Y (0) exp rt + 1 Be1 (t) , 21 21 t : Set
b = r=1 , 1=2; Bb (t) = bt + Be1 (t); c(T ) = max Bb (t): M 0tT Then
Y (t) = Y (0) expf1Bb (t)g; c(T )g: Y (T ) = Y (0) expf1M
b (T ) and M c(T ), appearing in Chapter 20, is The joint density of B
c(T ) 2 dm^ g If P fBb (T ) 2 d^b; M ^b)2 b^ 1 b2 ) ^ ^b) ( (2m ^ , 2(2 m ^ , exp , 2T + b , 2 T db dm; = p ^ T 2T m^ > 0; ^b < m: ^
The stock process.
dS = r dt + dBe + q1 , 2 dBe ; 2 1 2 2 S
so
q
S (T ) = S (0) expfrT + 2 Be1 (T ) , 21 222 T + 1 , 2 2Be2(T ) , 21 (1 , 2)22 T g q = S (0) expfrT , 12 22 T + 2 Be1 (T ) + 1 , 2 2 Be2 (T )g From the above paragraph we have
Be1 (T ) = ,bT + Bb (T ); so
q 2 1 b b S (T ) = S (0) expfrT + 2 B (T ) , 2 2 T , 2T + 1 , 2 2Be2 (T )g
242
24.1 Computing the option value
h i v (0; S (0); Y (0)) = If E e,rT (S (T ) , K )+ 1fY (T )
q
= e,rT If E S (0) exp (r , 21 22 , 2 b)T + 2 Bb (T ) + 1 , 2 2 Be2(T ) , K
:1fY (0)exp[ Mb (T )]
1
b (T ); M c(T )). The density of Be2(T ) is We know the joint density of (B
( ~2 ) 1 f e ~ IP fB2 (T ) 2 dbg = p exp , 2bT d~b; ~b 2 IR: 2T b (T ); Mc(T )) is independent of Be (T ) because Be1 and Furthermore, the pair of random variables (B eB2 are independent under IfP . Therefore, the joint density of the random vector (2Be2(T ); Bb (T ); M c(T )) is
c(T ) 2 dm; c(T ) 2 dm^ g If P fBe2 (T ) 2 d~b; Bb (T ) 2 d^b; M ^ g = If P fBe2 (T ) 2 d~bg:If P fBb (T ) 2 d^b; M The option value at time zero is
v (0; S (0); Y (0)) L log + Z Y Zm^ Z1 q , rT 2 1 b 2 ^ ~ =e S (0) exp (r , 2 2 , 2 )T + 2b + 1 , 2 b , K 1 1
(0)
,1 ( ,1 ) 2 ~ : p 1 exp , 2bT 2T ^b) ( (2m^ , ^b)2 b^ 1 b2 ) 2(2 m ^ , : p exp , + b , 2 T 2T 0
T 2T :d~b d^b dm: ^
The answer depends on T; S (0) and Y (0). It also depends on 1 ; 2 ; ; r; K and L. It does not depend on ; ; 1; nor 2 . The parameter b appearing in the answer is b = r1 , 21 : Remark 24.3 If we had not regarded Y as a traded asset, then we would not have tried to set its mean return equal to r. We would have had only one equation (see Eqs (0.1),(0.2))
q
= r + 21 + 1 , 2 22
(1.1)
to determine 1 and 2 . The nonuniqueness of the solution alerts us that some options cannot be hedged. Indeed, any option whose payoff depends on Y cannot be hedged when we are allowed to trade only in the stock.
+
CHAPTER 24. An outside barrier option
243
If we have an option whose payoff depends only on original equation for S ,
S , then Y
is superfluous. Returning to the
dS = dt + dB + q1 , 2 dB ; 2 2 2 1 S
q
we should set
dW = dB1 + 1 , 2dB2 ; so W is a Brownian motion under IP (Levy’s theorem), and dS = dt + dW: 2 S Now we have only Brownian motion, there will be only one , namely, = , r ; 2 f = dt + dW; we have so with dW dS = r dt + dW 2 f; S and we are on our way.
24.2 The PDE for the outside barrier option Returning to the case of the option with payoff
(S (T ) , K )+ 1fY (T )
h
v(t; x; y) = e,r(T ,t)If E t;x;y (S (T ) , K )+1fmaxtuT Y (u) < Lg ;
i
by replacing T , S (0) and Y (0) by T , t, x and y respectively in the formula for v (0; S (0); Y (0)). Now start at time 0 at S (0) and Y (0). Using the Markov property, we can show that the stochastic process is a martingale under If P . We compute
h i d e,rtv(t; S (t); Y (t))
e,rt v (t; S (t); Y (t))
,rv + vt + rSvx + rY vy + 21 22S 2vxx + 12SY vxy + 12 21Y 2vyy dt q 2 e e e + 2 Svx dB1 + 1 , 2Svx dB2 + 1Y vy dB1
= e,rt
244
y v(t, x, L) = 0, x >= 0
L
x v(t, 0, 0) = 0 Figure 24.1: Boundary conditions for barrier option. Note that t 2 [0; T ] is fixed.
Setting the dt term equal to 0, we obtain the PDE
, rv + vt + rxvx + ryvy + 12 22x2vxx
+ 1 2 xyvxy + 12 21 y 2 vyy = 0; 0 t < T; x 0; 0 y L:
The terminal condition is
v(T; x; y) = (x , K )+; x 0; 0 y < L;
and the boundary conditions are
v(t; 0; 0) = 0; 0 t T; v (t; x; L) = 0; 0 t T; x 0:
CHAPTER 24. An outside barrier option
245
,rv + vt + ryvy + 12 21y2vyy = 0
x=0
y=0 ,rv + vt + rxvx + 12 22x2vxx = 0
This is the usual BlackScholes formula in y .
This is the usual BlackScholes formula in x.
The boundary conditions are
The boundary condition is
the terminal condition is
v (T; 0; y) = (0 , K )+ = 0; y 0:
the terminal condition is
On the x = 0 boundary, the option value is v (t; 0; y ) = 0; 0 y L:
On the y = 0 boundary, the barrier is irrelevant, and the option value is given by the usual BlackScholes formula for a European call.
v (t; 0; 0) = e,r(T ,t)(0 , K )+ = 0;
v (t; 0; L) = 0; v (t; 0; 0) = 0;
v (T; x; 0) = (x , K )+ ; x 0:
24.3 The hedge After setting the dt term to 0, we have the equation
h i d e,rtv(t; S (t); Y (t))
q
= e,rt 2Svx dBe1 + 1 , 2 2 Svx dBe2 + 1 Y vy dBe1 ;
where vx that
= vx (t; S (t); Y (t)), vy = vy (t; S (t); Y (t)), and Be1 ; Be2 ; S; Y
are functions of t. Note
h i d e,rtS (t) = e,rt [,rS (t) dt + dS (t)]
q e 2 e 2S (t) dB1 (t) + 1 , 2S (t) dB2 (t) : h ,rt i ,rt d e Y (t) = e [,rY (t) dt + dY (t)] = e,rt 1Y (t) dBe1 (t): = e,rt
Therefore,
h i d e,rtv (t; S (t); Y (t)) = vx d[e,rtS ] + vy d[e,rtY ]:
Let 2 (t) denote the number of shares of stock held at time t, and let 1(t) denote the number of “shares” of the barrier process Y . The value X (t) of the portfolio has the differential
dX = 2dS + 1 dY + r[X , 2S , 1Y ] dt:
246 This is equivalent to
d[e,rtX (t)] = 2 (t)d[e,rtS (t)] + 1(t)d[e,rtY (t)]: To get X (t) = v (t; S (t); Y (t)) for all t, we must have
X (0) = v (0; S (0); Y (0)) and
2 (t) = vx (t; S (t); Y (t)); 1 (t) = vy (t; S (t); Y (t)):
Chapter 25
American Options This and the following chapters form part of the course Stochastic Differential Equations for Finance II.
25.1 Preview of perpetual American put
dS = rS dt + S dB Intrinsic value at time t : (K , S (t))+ :
Let L 2 [0; K ] be given. Suppose we exercise the first time the stock price is L or lower. We define
L = minft 0; S (t) Lg; vL (x) = IEe,rL (K , S (L ))+ ( K,x if x L, = , r L (K , L)IEe if x > L: The plan is to comute vL (x) and then maximize over L to find the optimal exercise price. We need to know the distribution of L .
25.2 First passage times for Brownian motion: first method (Based on the reflection principle) Let B be a Brownian motion under IP , let x > 0 be given, and define
= minft 0; B(t) = xg: is called the first passage time to x. We compute the distribution of . 247
Intrinsic value
248
K
K
Stock price
x
Figure 25.1: Intrinsic value of perpetual American put Define
M (t) = 0max B(u): ut
From the first section of Chapter 20 we have
)
(
2 IP fM (t) 2 dm; B (t) 2 dbg = 2(2pm , b) exp , (2m2,t b) dm db; m > 0; b < m: t 2t Therefore,
Z 1 Z m 2(2m , b) ( (2m , b)2 ) p IP fM (t) xg = exp , 2t db dm x ,1 t 2t ) ( Z1 2 2 b=m (2 m , b ) dm p exp , 2t = 2t x b=,1 ( 2) Z1 2 p exp , m2t dm: = 2t x
We make the change of variable z
= pmt in the integral to get
Z1
(
)
2 = p p2 exp , z dz: 2 x= t 2 Now
t()M (t) x;
CHAPTER 25. American Options
249
so
@ IP f tg dt IP f 2 dtg = @t @ IP fM (t) xg dt = @t " Z1 ( 2) # @ 2 = @t p p exp , z2 dz dt x= t 2 ( 2) 2 @ px dt = , p exp , x2t : @t t 2 ( 2) = px exp , x dt: 2t t 2t We also have the Laplace transform formula
IEe, =
Z1 0
e,t IP f 2 dtg
p = e,x 2;
> 0:
(See Homework)
Reference: Karatzas and Shreve, Brownian Motion and Stochastic Calculus, pp 9596.
25.3 Drift adjustment Reference: Karatzas/Shreve, Brownian motion and Stochastic Calculus, pp 196–197. For 0 t < 1, define
Be (t) = t + B (t); Z (t) = expf,B (t) , 12 2 tg; = expf,Be (t) + 12 2 tg;
Define
~ = minft 0; Be (t) = xg:
We fix a finite time T and change the probability measure “only up to T ”. More specifically, with T fixed, define Z
If P (A) =
A
Z (T ) dP; A 2 F (T ):
Under If P , the process Be (t); 0 t T , is a (nondrifted) Brownian motion, so
If P f~ 2 dtg = IP f 2 dtg ( 2) x = p exp , x dt; 0 < t T: 2t t 2t
250 For 0 < t T we have
h
i
IP f~ tg = IE 1f~tg = If E 1f~tg Z (1T )
h
i
= If E 1f~tg expfBe (T ) , 12 2 T g = If E 1f~tgIf E expfBe (T ) , 12 2 T g F (~ ^ t)
h
= If E 1f~tg expfBe (~ ^ t) , 21 2 (~ ^ t)g i h = If E 1f~tg expfx , 12 2 ~g =
Zt
i
expfx , 12 2 sgIf P f~ 2 dsg
( Z0t x 2) x 2 1 p exp x , 2 s , 2s ds = 0 s 2s ) ( Zt x 2 ( x , s ) p exp , 2s = ds: s 2s 0
Therefore,
(
)
2 IP f~ 2 dtg = px exp , (x ,2tt) dt; 0 < t T: t 2t Since T is arbitrary, this must in fact be the correct formula for all t > 0.
25.4 Driftadjusted Laplace transform Recall the Laplace transform formula for
= minft 0; B (t) = xg for nondrifted Brownian motion:
IEe, = For
( Z1 x 2) p x p exp ,t , 2t dt = e,x 2; > 0; x > 0: 0 t 2t ~ = minft 0; t + B (t) = xg;
CHAPTER 25. American Options the Laplace transform is
251
) ( Z1 x 2 ( x , t ) p exp ,t , 2t dt 0 t 2t ( ) Z1 x 2 x 2 1 p exp ,t , 2t + x , 2 t dt = 0 t 2t ( Z1 x 2) x 1 x 2 p exp ,( + 2 )t , 2t dt =e t 2t
IEe,~ =
0 p x , x 2+2 ; =e
> 0; x > 0;
where in the last step we have used the formula for IEe, with replaced by + 12 2 .
If ~(!) < 1, then
if ~(! ) = 1, then e,~(!)
lim e,~(! ) = 1; #0
= 0 for every > 0, so lim e,~(! ) = 0: #0
Therefore,
lim e,~(! ) = 1~<1 : #0
Letting #0 and using the Monotone Convergence Theorem in the Laplace transform formula
p
IEe,~ = ex,x we obtain If
0, then
If
< 0, then
p
IP f~ < 1g = ex,x
2+2 ;
2
= ex,xjj :
IP f~ < 1g = 1: IP f~ < 1g = e2x < 1:
(Recall that x > 0).
25.5 First passage times: Second method (Based on martingales) Let
> 0 be given. Then
Y (t) = expfB(t) , 21 2tg
252 is a martingale, so Y (t ^ ) is also a martingale. We have
1 = Y (0 ^ ) = IEY (t ^ ) = IE expfB (t ^ ) , 21 2 (t ^ )g: = t! lim1 IE expfB (t ^ ) , 12 2 (t ^ )g:
We want to take the limit inside the expectation. Since
0 expfB (t ^ ) , 12 2 (t ^ )g ex ; this is justified by the Bounded Convergence Theorem. Therefore,
1 = IE t! lim1 expfB (t ^ ) , 12 2 (t ^ )g: There are two possibilities. For those ! for which (! ) < 1, 1
lim expfB (t ^ ) , 12 2 (t ^ )g = ex, 2 : t!1 For those ! for which (! ) = 1, lim expfB (t ^ ) , 12 2 (t ^ )g t! lim1 expfx , 21 2tg = 0: t!1 2
Therefore,
1 = IE t! lim1 expfB (t ^ ) , 21 2(t ^ )g
1 = IE e x, 2 1<1 1 = IEex, 2 ; 2
2
1 2 where we understand ex, 2 to be zero if Let = 21 2 , so
p
= 1.
= 2. We have again derived the Laplace transform formula p e,x 2 = IEe, ; > 0; x > 0;
for the first passage time for nondrifted Brownian motion.
25.6 Perpetual American put
dS = rS dt + S dB S (0) = x S (t) = x expf(r , 12 2 )t + B(t)g
8 2 39 > > > < 66 r 77> = = x exp > 64 , t + B (t)75> : > > ; :  {z 2 }
CHAPTER 25. American Options Intrinsic value of the put at time t:
253
(K , S (t))+.
Let L 2 [0; K ] be given. Define for x L,
L = minft 0; S (t) = Lg = minft 0; t + B (t) = 1 log Lx g = minft 0; ,t , B (t) = 1 log Lx g Define
vL = (K , L)IEe,rL p = (K , L) exp , log Lx , 1 log Lx 2r + 2 x , , p2r+ : = (K , L) L 1
2
We compute the exponent
s
2 r p 1 1 r 1 2 , , 2r + = , 2 + 2 , 2r + , =2 s 2 = , r2 + 12 , 1 2r + r 2 , r + 2 =4 s 2 r 1 = , 2 + 12 , r 2 + r + 2=4 s 2 r 1 1 = , 2 + 2 , r + =2 r 1 r 1 = , 2 + 2 , + =2 = , 2r2 : 8 <(K , x); 0 x L; vL (x) = : , , 2 r= x (K , L) L ; x L: , ,2r= ; are all of the form Cx,2r= . The curves (K , L) x Therefore,
2
L
2
2
We want to choose the largest possible constant. The constant is
C = (K , L)L2r= ; 2
value
254
Kx K
σ2 (K  L) (x/L)2r/
K
Stock price
x
value
Figure 25.2: Value of perpetual American put
C3 x
2r/ σ 2
C2 x C1 x
2r/ σ 2
Stock price
Figure 25.3: Curves.
2r/ σ 2
x
CHAPTER 25. American Options
255
and
@C = ,L r + 2r (K , L)L r ,1 @L 2 r = L ,1 + 2r2 (K , L) L1 r = L , 1 + 2r2 + 2r2 K L : 2
2
2
We solve
2
2
2
2
2
2r K 2 r , 1 + 2 + 2 L = 0
to get
L = 22rK + 2r : Since 0 < 2r < 2 + 2r; we have
0 < L < K:
Solution to the perpetual American put pricing problem (see Fig. 25.4):
8 < v(x) = :(K , x); , x ,2r= (K , L ) L ;
0 x L ;
x L;
2
where
Note that
v0(x) =
L = 22rK + 2r :
(
,1; , 2r (K , L)(L)2r= x,2r= ,1; 2
2
2
0 x < L ;
x > L:
We have
lim v 0(x) = ,2 r2 (K , L) 1
x#L
L 2 + 2r r 2 rK = ,2 2 K , 2 + 2r 2rK 2 + 2r , 2r ! 2 + 2r r = ,2 2 2 + 2r 2r = ,1 = lim v 0 (x): x"L
value
256
Kx K 2r/ σ (K  L* )(x/L* )
L*
K
Stock price
2
x
Figure 25.4: Solution to perpetual American put.
25.7 Value of the perpetual American put Set
K:
= 2r2 ; L = 22rK = + 2r + 1 If 0 x < L , then v (x) = K , x. If L x < 1, then v(x) = (K , L{z)(L) } x,
h ,Cr x = IE e (K , L )+ 1
f<1g
i
;
(7.1) (7.2)
where
S (0) = x = minft 0; S (t) = L g:
If 0 x < L, then
,rv(x) + rxv0(x) + 12 2x2v00(x) = ,r(K , x) + rx(,1) = ,rK: If L x < 1, then ,rv(x) + rxv0(x) + 12 2x2v00(x) = C [,rx, , rx x, ,1 , 12 2 x2 (, , 1)x, ,2 ] = Cx, [,r , r , 21 2 (, , 1)] 2r ,
2 1 = C (, , 1)x r , 2 2 = 0:
In other words, v solves the linear complementarity problem: (See Fig. 25.5).
(7.3) (7.4)
CHAPTER 25. American Options
257
6 K@
v
@@
@@ @@
x
K
L
Figure 25.5: Linear complementarity For all x 2 IR, x 6= L ,
rv , rxv 0 , 21 2 x2v 00 0; v (K , x)+ ; One of the inequalities (a) or (b) is an equality.
(a) (b) (c)
The halfline [0; 1) is divided into two regions:
C = fx; v(x) > (K , x)+ g; S = fx; rv , rxv0 , 12 2x2v00 > 0g; and L is the boundary between them. If the stock price is in C , the owner of the put should not exercise (should “continue”). If the stock price is in S or at L , the owner of the put should exercise (should “stop”).
25.8 Hedging the put Let S (0) be given. Sell the put at time zero for v (S (0)). Invest the money, holding (t) shares of stock and consuming at rate C (t) at time t. The value X (t) of this portfolio is governed by
dX (t) = (t) dS (t) + r(X (t) , (t)S (t)) dt , C (t) dt; or equivalently,
d(e,rtX (t)) = ,e,rt C (t) dt + e,rt (t)S (t) dB(t):
258 The discounted value of the put satisfies
h i d e,rtv(S (t)) = e,rt ,rv (S (t)) + rS (t)v 0(S (t)) + 12 2 S 2(t)v 00 (S (t)) dt + e,rt S (t)v 0(S (t)) dB (t) = ,rKe,rt 1fS (t)
We should set
C (t) = rK 1fS (t)
v(S (t)) = K , S (t); (t) = v 0(S (t)) = ,1: To hedge the put when S (t) < L , short one share of stock and hold K in the money market. As long as the owner does not exercise, you can consume the interest from the money market position, i.e.,
C (t) = rK 1fS(t)
Properties of e,rt v (S (t)): 1. e,rt v (S (t)) is a supermartingale (see its differential above). 2. e,rt v (S (t)) e,rt (K , S (t))+ ,
0 t < 1;
3. e,rt v (S (t)) is the smallest process with properties 1 and 2. Explanation of property 3. Let Y be a supermartingale satisfying
Y (t) e,rt (K , S (t))+; 0 t < 1:
(8.1)
Then property 3 says that
Y (t) e,rt v (S (t)); 0 t < 1: (8.2) We use (8.1) to prove (8.2) for t = 0, i.e., Y (0) v (S (0)): (8.3) If t is not zero, we can take t to be the initial time and S (t) to be the initial stock price, and then adapt the argument below to prove property (8.2). Proof of (8.3), assuming Y is a supermartingale satisfying (8.1):
Case I: S (0) L : We have
Y (0) {z} (K , S (0))+ = v(S (0)): (8:1)
CHAPTER 25. American Options Case II: S (0) > L: For T
259
> 0, we have
Y (0) IEY ( ^ T ) (Stopped supermartingale is a supermartingale) h i IE Y ( ^ T )1f<1g : (Since Y 0) Now let T !1 to get
h
i
Y (0) Tlim !h1 IE Y ( ^ Ti )1f<1g IE Y ( )1f<1g (Fatou’s Lemma)
2 3 IE 64e,r (K , S {z( )})+1f<1g75 = v (S (0)):
(by 8.1)
L
(See eq. 7.2)
25.9 Perpetual American contingent claim Intinsic value: h(S (t)). Value of the American contingent claim:
x e,r h(S ( )) ; v(x) = sup IE
where the supremum is over all stopping times. Optimal exercise rule: Any stopping time which attains the supremum.
Characterization of v :
1. e,rt v (S (t)) is a supermartingale; 2. e,rt v (S (t)) e,rt h(S (t));
0 < t < 1;
3. e,rt v (S (t)) is the smallest process with properties 1 and 2.
25.10 Perpetual American call
v(x) = sup IE x e,r (S ( ) , K )+
Theorem 10.63
v (x) = x 8x 0:
260 Proof: For every t,
h
i
v(x) IE x e,rt (S (t) , K )+ h i IE x e,rt(S (t) , K ) h i = IE x e,rt S (t) , e,rt K = x , e,rt K: Let t!1 to get v (x) x.
Now start with S (0) = x and define
Y (t) = e,rt S (t): Then: 1. 2.
Y is a supermartingale (in fact, Y is a martingale); Y (t) e,rt(S (t) , K )+ ; 0 t < 1.
Therefore, Y (0) v (S (0)), i.e.,
x v (x):
Remark 25.2 No matter what we choose,
IE x e,r (S ( ) , K )+ < IE x e,r S ( ) x = v (x):
There is no optimal exercise time.
25.11 Put with expiration
T > 0. Intrinsic value: (K , S (t))+. Expiration time:
Value of the put:
v(t; x) = (value of the put at time t if S (t) = x) = sup IE xe,r( ,t) (K , S ( ))+: t T
 {z }
:stopping time See Fig. 25.6. It can be shown that jump. Let S (0) be given. Then
v; vt; vx are continuous across the boundary, while vxx has a
CHAPTER 25. American Options
261
x
6
v (T; x) = 0; x K
v>K,x ,rv + vt + rxvx + 12 2x2vxx = 0 K
v (T; x) = K , x; 0 x K
v=K,x vt = 0; vx = ,1; vxx = 0 ,rv + vt + rxvx + 12 2x2vxx = ,rK
L
T
t
Figure 25.6: Value of put with expiration 1. e,rt v (t; S (t));
0 t T; is a supermartingale;
2. e,rt v (t; S (t)) e,rt (K , S (t))+;
0 t T;
3. e,rt v (t; S (t)) is the smallest process with properties 1 and 2.
25.12 American contingent claim with expiration
T > 0. Intrinsic value: h(S (t)). Expiration time:
Value of the contingent claim:
v(t; x) = sup IE xe,r( ,t)h(S ( )): t T
Then
rv , vt , rxvx , 12 2 x2vxx 0; v h(x);
At every point (t; x) 2 [0; T ] [0; 1), either (a) or (b) is an equality. Characterization of v : Let S (0) be given. Then
(a) (b) (c)
262
0 t T; is a supermartingale; e,rtv (t; S (t)) e,rt h(S (t)); e,rtv (t; S (t)) is the smallest process with properties 1 and 2.
1. e,rt v (t; S (t)); 2. 3.
The optimal exercise time is
= minft 0; v (t; S (t)) = h(S (t))g If (!) = 1, then there is no optimal exercise time along the particular path ! .
Chapter 26
Options on dividendpaying stocks 26.1 American option with convex payoff function Theorem 1.64 Consider the stock price process
dS (t) = r(t)S (t) dt + (t)S (t) dB(t); where r and are processes and r(t) 0; 0 t T; a.s. This stock pays no dividends. Let h(x) be a convex function of x 0, and assume h(0) = 0. (E.g., h(x) = (x , K )+ ). An American contingent claim paying h(S (t)) if exercised at time t does not need to be exercised before expiration, i.e., waiting until expiration to decide whether to exercise entails no loss of value. Proof: For 0 1 and x 0, we have
h(x) = h((1 , )0 + x) (1 , )h(0) + h(x) = h(x): Let T be the time of expiration of the contingent claim. For 0 t T , ( ZT ) ( t ) 0 (T ) = exp , r(u) du 1 t and S (T ) 0, so
(t)
(t) h (T ) S (T ) (T ) h(S (T )):
(*)
Consider a European contingent claim paying h(S (T )) at time T . The value of this claim at time t 2 [0; T ] is
X (t) = (t) IE (1T ) h(S (T )) F (t) : 263
264
6
. ... . . . ... .... . . .r . . . .... .. .... ...... (x; h(x)) . . . .. . h(x) .... .... .... .... .... .... .... .... .... .... .... .... .... ..... ..... ..... ..... ..... ..... ......r ............. ....... .. .. ... .. . ... . .. h(x) .... .... .... .... .... .... .... .... .... .... .... .... .... ..... ..... ...........................................r... h .. . . . .. . . . . .. . . . . . . . . . . . ... . . .... . . . . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . .. ..............
x

Figure 26.1: Convex payoff function Therefore,
X (t) = 1 IE (t) h(S (T )) F (t) (t) (t) (T ) (1t) IE h ((Tt)) S (T ) F (t) (by (*)) S (T ) 1 (t) h (t) IE (T ) F (t) (Jensen’s inequality) = 1 h (t) S (t) ( S is a martingale) (t) (t) = (1t) h(S (t)):
This shows that the value X (t) of the European contingent claim dominates the intrinsic value h(S (t)) of the American claim. In fact, except in degenerate cases, the inequality
X (t) h(S (t)); 0 t T; is strict, i.e., the American claim should not be exercised prior to expiration.
26.2 Dividend paying stock Let r and be constant, let be a “dividend coefficient” satisfying
0 < < 1:
CHAPTER 26. Options on dividend paying stocks Let T > 0 be an expiration time, and let t1 price is given by
265
2 (0; T ) be the time of dividend payment. The stock
(
1 2 0 t t1 ; S (t) = S (0) expf(r , 2 )t +1B2 (t)g; (1 , )S (t1) expf(r , 2 )(t , t1 ) + (B (t) , B (t1 ))g; t1 < t T: Consider an American call on this stock. At times t 2 (t1 ; T ), it is not optimal to exercise, so the value of the call is given by the usual BlackScholes formula
v(t; x) = xN (d+(T , t; x)) , Ke,r(T ,t) N (d,(T , t; x)); t1 < t T;
x 2 log K + (T , t)(r =2) : T ,t
where
d (T , t; x) = p 1
At time t1 , immediately after payment of the dividend, the value of the call is
v (t1 ; (1 , )S (t1)): At time t1 , immediately before payment of the dividend, the value of the call is
w(t1 ; S (t1));
where
w(t1 ; x) = max (x , K )+ ; v (t1; (1 , )x :
Theorem 2.65 For 0 t t1 , the value of the American call is w(t; S (t)), where
h i w(t; x) = IE t;x e,r(t ,t) w(t1 ; S (t1)) : 1
This function satisfies the usual BlackScholes equation
,rw + wt + rxwx + 12 2x2wxx = 0; 0 t t1; x 0; (where w = w(t; x)) with terminal condition
w(t1; x) = max (x , K )+; v (t1 ; (1 , )x) ; x 0; and boundary condition
w(t; 0) = 0; 0 t T:
The hedging portfolio is
(t) =
(
wx(t; S (t)); vx(t; S (t));
0 t t1 ;
t1 < t T:
Proof: We only need to show that an American contingent claim with payoff w(t1; S (t1)) at time t1 need not be exercised before time t1 . According to Theorem 1.64, it suffices to prove 1.
w(t1; 0) = 0,
266 2.
w(t1; x) is convex in x.
Since v (t1 ; 0) = 0, we have immediately that
w(t1; 0) = max (0 , K )+ ; v(t1; (1 , )0) = 0: To prove that w(t1; x) is convex in x, we need to show that v (t1; (1 , )x) is convex is x. Obviously, (x , K )+ is convex in x, and the maximum of two convex functions is convex. The proof of the convexity of v (t1; (1 , )x) in x is left as a homework problem.
26.3 Hedging at time t1 Let x = S (t1).
Case I: v (t1; (1 , )x) (x , K )+ . The option need not be exercised at time t1 (should not be exercised if the inequality is strict). We have
w(t1 ; x) = v(t1; (1 , )x); (t1) = wx (t1 ; x) = (1 , )vx (t1 ; (1 , )x) = (1 , )(t1+); where
(t1 +) = lim (t) t#t1
is the number of shares of stock held by the hedge immediately after payment of the dividend. The postdividend position can be achieved by reinvesting in stock the dividends received on the stock held in the hedge. Indeed,
(t1 +) = 1 (t1) = (t1) + (t1) 1, 1, ( t ) S ( t ) 1 1 = (t1 ) + (1 , )S (t ) 1
dividends received = # of shares held when dividend is paid + price per share when dividend is reinvested Case II: v (t1; (1 , )x) < (x , K )+ . The owner of the option should exercise before the dividend payment at time t1 and receive (x , K ). The hedge has been constructed so the seller of the option has x , K before the dividend payment at time t1 . If the option is not exercised, its value drops from x , K to v (t1; (1 , )x), and the seller of the option can pocket the difference and continue the hedge.
Chapter 27
Bonds, forward contracts and futures Let fW (t); F (t); 0 t T g be a Brownian motion (Wiener process) on some ( ; F ; P). Consider an asset, which we call a stock, whose price satisfies
dS (t) = r(t)S (t) dt + (t)S (t) dW (t): Here, r and are adapted processes, and we have already switched to the riskneutral measure, which we call IP . Assume that every martingale under IP can be represented as an integral with respect to W . Define the accumulation factor
(t) = exp
Z t 0
r(u) du :
A zerocoupon bond, maturing at time T , pays 1 at time T and nothing before time T . According to the riskneutral pricing formula, its value at time t 2 [0; T ] is
B (t; T ) = (t) IE (1T ) F (t) (t) = IE (T ) F (t) " ( = IE exp ,
ZT t
) # r(u) du F (t) :
Given B (t; T ) dollars at time t, one can construct a portfolio of investment in the stock and money 267
268 market so that the portfolio value at time T is 1 almost surely. Indeed, for some process ,
B(t; T ) = (t) IE (1T ) F (t)  {z }
martingale Zt 1 = (t) IE (T ) + (u) dW (u) Zt 0 = (t) B (0; T ) +
0
(u) dW (u) ;
Zt
dB (t; T ) = r(t) (t) B (0; T ) + (u) dW (u) dt + (t) (t) dW (t) 0 = r(t)B (t; T ) dt + (t) (t) dW (t): The value of a portfolio satisfies
dX (t) = (t) dS (t) + r(t)[X (t) , (t)S (t)]dt = r(t)X (t) dt + (t)(t)S (t) dW (t): (*) We set
(t) = ((tt))S ((tt)) :
If, at any time t, X (t) = B (t; T ) and we use the portfolio (u);
t u T , then we will have
X (T ) = B(T; T ) = 1: If r(t) is nonrandom for all t, then
( ZT ) B (t; T ) = exp , r(u) du ; t
dB (t; T ) = r(t)B (t; T ) dt; i.e., = 0. Then given above is zero. If, at time t, you are given B (t; T ) dollars and you always invest only in the money market, then at time T you will have
B(t; T ) exp
(Z T t
)
r(u) du = 1:
If r(t) is random for all t, then is not zero. One generally has three different instruments: the stock, the money market, and the zero coupon bond. Any two of them are sufficient for hedging, and the two which are most convenient can depend on the instrument being hedged.
CHAPTER 27. Bonds, forward contracts and futures
269
27.1 Forward contracts We continue with the setup for zerocoupon bonds. The T forward price of the stock at time t 2 [0; T ] is the F (t)measurable price, agreed upon at time t, for purchase of a share of stock at time T , chosen so the forward contract has value zero at time t. In other words,
We solve for F (t):
1 IE (T ) (S (T ) , F (t)) F (t) = 0; 0 t T: 1 (S (T ) , F (t)) F (t) 0 = IE (T ) = IE S (T ) F (t) , F (t) IE (t) F (t) (T ) (t) (T ) = S (t) , F (t) B (t; T ): (t) (t)
This implies that
F (t) = BS(t;(tT) ) : Remark 27.1 (Value vs. Forward price) The T forward price F (t) is not the value at time t of the forward contract. The value of the contract at time t is zero. F (t) is the price agreed upon at time t which will be paid for the stock at time T .
27.2 Hedging a forward contract Enter a forward contract at time 0, i.e., agree to pay F (0) = BS(0(0) ;T ) for a share of stock at time T . At time zero, this contract has value 0. At later times, however, it does not. In fact, its value at time t 2 [0; T ] is
1 V (t) = (t) IE (T ) (S (T ) , F (0)) F (t) = (t) IE S ((TT )) F (t) , F (0) IE ((Tt)) F (t) = (t) S ((tt)) , F (0)B (t; T ) = S (t) , F (0)B (t; T ):
This suggests the following hedge of a short position in the forward contract. At time 0, short F (0) T maturity zerocoupon bonds. This generates income
F (0)B (0; T ) = BS(0(0) ; T ) B(0; T ) = S (0):
270 Buy one share of stock. This portfolio requires no initial investment. Maintain this position until time T , when the portfolio is worth
S (T ) , F (0)B (T; T ) = S (T ) , F (0): Deliver the share of stock and receive payment F (0). A short position in the forward could also be hedged using the stock and money market, but the implementation of this hedge would require a termstructure model.
27.3 Future contracts Future contracts are designed to remove the risk of default inherent in forward contracts. Through the device of marking to market, the value of the future contract is maintained at zero at all times. Thus, either party can close out his/her position at any time. Let us first consider the situation with discrete trading dates
0 = t0 < t1 < : : : < tn = T: On each [tj ; tj +1 ), r is constant, so
(tk+1 ) = exp
Z tk
+1
r(u) du
8 0k 9
is F (tk )measurable.
Enter a future contract at time tk , taking the long position, when the future price is (tk ). At time tk+1 , when the future price is (tk+1 ), you receive a payment (tk+1 ) , (tk ). (If the price has fallen, you make the payment ,((tk+1 ) , (tk )). ) The mechanism for receiving and making these payments is the margin account held by the broker. By time T
= tn , you have received the sequence of payments (tk+1 ) , (tk ); (tk+2 ) , (tk+1 ); : : : ; (tn ) , (tn,1 )
at times tk+1 ; tk+2 ; : : : ; tn . The value at time t = t0 of this sequence is
2n,1 X (t) IE 4 j =k
3 1 5 (tj +1) ((tj+1 ) , (tj )) F (t) :
Because it costs nothing to enter the future contract at time t, this expression must be zero almost surely.
CHAPTER 27. Bonds, forward contracts and futures The continuoustime version of this condition is
271
#
"Z T
1 (t) IE d(u) F (t) = 0; 0 t T: t (u) Note that (tj +1 ) appearing in the discretetime version is F (tj )measurable, as it should be when
approximating a stochastic integral.
Definition 27.1 The T future price of the stock is any F (t)adapted stochastic process
f(t); 0 t T g ; satisfying
IE
"Z T t
(T ) = S (T ) a.s., and
# 1 d(u) F (t) = 0; 0 t T: (u)
(a) (b)
Theorem 3.66 The unique process satisfying (a) and (b) is
(t) = IE S (T ) F (t) ; 0 t T:
Proof: first show that (b) holds if and only if R t 1 dWe 0 (u) (u) is also a martingale, so
"Z T
is a martingale.
If
is a martingale, then
#
1 d(u) F (t) = IE Z t 1 d(u) F (t) , Z t 1 d(u) IE t (u) 0 (u) 0 (u) = 0:
On the other hand, if (b) holds, then the martingale
M (t) = IE satisfies
"Z T 0
#
1 d(u) F (t) (u)
"Z T # Zt 1 1 M (t) = (u) d(u) + IE d(u) F (t) 0 t (u) Zt 1 = (u) d(u); 0 t T: 0
this implies
dM (t) = (1t) d(t); d(t) = (t) dM (t);
272 and so is a martingale (its differential has no dt term).
(t) = IE S (T ) F (t) ; 0 t T:
Now define
Clearly (a) is satisfied. By the tower property, is a martingale, so (b) is also satisfied. Indeed, this is the only martingale satisfying (a).
27.4 Cash flow from a future contract With a forward contract, entered at time 0, the buyer agrees to pay F (0) for an asset valued at S (T ). The only payment is at time T . With a future contract, entered at time 0, the buyer receives a cash flow (which may at times be negative) between times 0 and T . If he still holds the contract at time T , then he pays S (T ) at time T for an asset valued at S (T ). The cash flow received between times 0 and T sums to
ZT 0
d(u) = (T ) , (0) = S (T ) , (0):
Thus, if the future contract holder takes delivery at time T , he has paid a total of
((0) , S (T )) + S (T ) = (0) for an asset valued at S (T ).
27.5 Forwardfuture spread
Future price: (t) = IE S (T ) F (t) .
Forward price:
F (t) = BS(t;(tT) ) =
S (t) : (t)IE (1T ) F (t)
Forwardfuture spread:
(0) , F (0) = IE [S (T )] , Sh (0)1 i
IE
(T ) S (T ) 1 1 = 1 IE (T ) IE (S (T )) , IE (T ) :
IE
If (1T ) and S (T ) are uncorrelated,
(T )
(0) = F (0):
CHAPTER 27. Bonds, forward contracts and futures
273
If (1T ) and S (T ) are positively correlated, then
(0) F (0): This is the case that a rise in stock price tends to occur with a fall in the interest rate. The owner of the future tends to receive income when the stock price rises, but invests it at a declining interest rate. If the stock price falls, the owner usually must make payments on the future contract. He withdraws from the money market to do this just as the interest rate rises. In short, the long position in the future is hurt by positive correlation between (1T ) and S (T ). The buyer of the future is compensated by a reduction of the future price below the forward price.
27.6 Backwardation and contango Suppose
dS (t) = S (t) dt + S (t) dW (t): f (t) = t + W (t), Define = W Z (T ) = expf,W (T ) , 12 2 T g Z f IP (A) = Z (T ) dIP; 8A 2 F (T ): ,r ;
A
f is a Brownian motion under IfP , and Then W f (t): dS (t) = rS (t) dt + S (t) dW We have
(t) = ert S (t) = S (0) expf( , 21 2)t + W (t)g f (t)g = S (0) expf(r , 12 2 )t + W
Because (1T )
= e,rT is nonrandom, S (T ) and (1T ) are uncorrelated under If P . Therefore,
f (t) = IE [S (T ) F (t)]
= F (t) = BS(t;(tT) ) = er(T ,t)S (t):
The expected future spot price of the stock under IP is
h
n
IES (T ) = S (0)eT IE exp , 12 2T + W (T ) = eT S (0):
oi
274 The future price at time 0 is
(0) = erT S (0):
If > r, then (0) < IES (T ): This situation is called normal backwardation (see Hull). If then (0) > IES (T ). This is called contango.
< r,
Chapter 28
Termstructure models Throughout this discussion, fW (t); 0 t T g is a Brownian motion on some probability space ( ; F ; P), and fF (t); 0 t T g is the filtration generated by W . Suppose we are given an adapted interest rate process fr(t); lation factor Z
(t) = exp
t
0
0 t T g. We define the accumu
r(u) du ; 0 t T :
In a termstructure model, we take the zerocoupon bonds (“zeroes”) of various maturities to be the primitive assets. We assume these bonds are defaultfree and pay $1 at maturity. For 0 t T T , let B(t; T ) = price at time t of the zerocoupon bond paying $1 at time T . Theorem 0.67 (Fundamental Theorem of Asset Pricing) A term structure model is free of arbitrage if and only if there is a probability measure If P on (a riskneutral measure) with the same probabilityzero sets as IP (i.e., equivalent to IP ), such that for each T 2 (0; T ], the process
B(t; T ) ; 0 t T; (t) is a martingale under If P. Remark 28.1 We shall always have
dB (t; T ) = (t; T )B(t; T ) dt + (t; T )B (t; T ) dW (t); 0 t T; for some functions (t; T ) and (t; T ). Therefore
d
B(t; T ) (t)
1 1 = B (t; T ) d (t) + (t) dB (t; T ) = [(t; T ) , r(t)] B (t;(tT) ) dt + (t; T ) B (t;(tT) ) dW (t); 275
276 so IP is a riskneutral measure if and only if (t; T ), the mean rate of return of B (t; T ) under IP , is the interest rate r(t). If the mean rate of return of B (t; T ) under IP is not r(t) at each time t and for each maturity T , we should change to a measure If P under which the mean rate of return is r(t). If such a measure does not exist, then the model admits an arbitrage by trading in zerocoupon bonds.
28.1 Computing arbitragefree bond prices: first method Begin with a stochastic differential equation (SDE)
dX (t) = a(t; X (t)) dt + b(t; X (t)) dW (t): The solution X (t) is the factor. If we want to have nfactors, we let W be an ndimensional Brownian motion and let X be an ndimensional process. We let the interest rate r(t) be a function of X (t). In the usual onefactor models, we take r(t) to be X (t) (e.g., CoxIngersollRoss, HullWhite). Now that we have an interest rate process prices to be
fr(t); 0 t T g, we define the zerocoupon bond
1 B (t; T ) = (t) IE (T ) F (t) " ( ZT ) # = IE exp , r(u) du F (t) ; 0 t T T : t
We showed in Chapter 27 that
dB (t; T ) = r(t)B(t; T ) dt + (t) (t) dW (t) for some process . Since B (t; T ) has mean rate of return r(t) under IP , IP is a riskneutral measure and there is no arbitrage.
28.2 Some interestrate dependent assets Couponpaying bond: Payments P1 ; P2; : : : ; Pn at times T1 ; T2; : : : ; Tn. Price at time t is
X
fk:t
Pk B (t; Tk ):
Call option on a zerocoupon bond: Bond matures at time Price at time t is
T.
Option expires at time T1
(t) IE (1T ) (B (T1 ; T ) , K )+ F (t) ; 0 t T1: 1
< T.
CHAPTER 28. Termstructure models
277
28.3 Terminology Definition 28.1 (Termstructure model) Any mathematical model which determines, at least theoretically, the stochastic processes
B(t; T ); 0 t T; for all T
2 (0; T ].
Definition 28.2 (Yield to maturity) For 0 F (t)measurable randomvariable satisfying
t T T , the yield to maturity Y (t; T ) is the
B(t; T ) exp f(T , t)Y (t; T )g = 1; or equivalently,
Y (t; T ) = , T 1, t log B(t; T ):
Determining
B (t; T ); 0 t T T ;
is equivalent to determining
Y (t; T ); 0 t T T :
28.4 Forward rate agreement Let 0 t T < T + T be given. Suppose you want to borrow $1 at time T with repayment (plus interest) at time T + , at an interest rate agreed upon at time t. To synthesize a forwardrate (t;T ) (T + )maturity zeroes. agreement to do this, at time t buy a T maturity zero and short BB(t;T +) The value of this portfolio at time t is
B(t; T ) , B(Bt;(Tt; T+)) B(t; T + ) = 0: (t;T ) . The At time T , you receive $1 from the T maturity zero. At time T + , you pay $ BB(t;T +) effective interest rate on the dollar you receive at time T is R(t; T; T + ) given by
B (t; T ) = expf R(t; T; T + )g; B(t; T + ) or equivalently,
R(t; T; T + ) = , log B (t; T + ) , log B (t; T ) :
The forward rate is
@ log B(t; T ): f (t; T ) = lim R(t; T; T + ) = , @T #0
(4.1)
278 This is the instantaneous interest rate, agreed upon at time t, for money borrowed at time T . Integrating the above equation, we obtain
ZT t
ZT @ log B (t; u) du t @u u=T = , log B (t; u) u=t = , log B (t; T );
f (t; u) du = ,
so
( ZT ) B(t; T ) = exp , f (t; u) du : t
You can agree at time t to receive interest rate f (t; u) at each time u 2 [t; T ]. If you invest $ B (t; T ) at time t and receive interest rate f (t; u) at each time u between t and T , this will grow to
B(t; T ) exp
(Z T t
)
f (t; u) du = 1
at time T .
28.5 Recovering the interest r (t) from the forward rate
"
( ZT ) # B (t; T ) = IE exp , r(u) du F (t) ; t " ( ) # @ B(t; T ) = IE ,r(T ) exp , Z T r(u) du F (t) ; @T t @ B (t; T ) = IE ,r(t) F (t) = ,r(t):
@T On the other hand,
T =t
( ZT ) B (t; T ) = exp , f (t; u) du ; t ( ) @ B(t; T ) = ,f (t; T ) exp , Z T f (t; u) du ; @T t @ @T B (t; T ) T =t = ,f (t; t):
Conclusion: r(t) = f (t; t).
CHAPTER 28. Termstructure models
279
28.6 Computing arbitragefree bond prices: HeathJarrowMorton method For each T
2 (0; T ], let the forward rate be given by Zt Zt f (t; T ) = f (0; T ) +
Here f(u; T ); In other words, Recall that
(u; T ) du +
0
0
(u; T ) dW (u); 0 t T:
0 u T g and f(u; T ); 0 u T g are adapted processes.
df (t; T ) = (t; T ) dt + (t; T ) dW (t):
( ZT ) B(t; T ) = exp , f (t; u) du : t
Now
( ZT ) ZT d , f (t; u) du = f (t; t) dt , df (t; u) du t t ZT = r(t) dt , [(t; u) dt + (t; u) dW (t)] du "tZ T # "Z T # = r(t) dt , (t; u) du dt , (t; u) du dW (t) t t  {z }  {z } (t;T ) (t;T ) = r(t) dt , (t; T ) dt , (t; T ) dW (t):
Let
g(x) = ex; g 0(x) = ex; g 00(x) = ex:
Then
B (t; T ) = g ,
ZT t
and
dB (t; T ) = dg , = g0
, +
f (t; u) du ;
ZT
t ZT t
1 00 2g
h
!
!
f (t; u) du
!
f (t; u) du (r dt , dt , dW )
,
ZT t
!
f (t; u) du ( )2 dt
i
= B (t; T ) r(t) , (t; T ) + 12 ( (t; T ))2 dt , (t; T )B(t; T ) dW (t):
280
28.7 Checking for absence of arbitrage
IP is a riskneutral measure if and only if (t; T ) = 12 ( (t; T ))2 ; 0 t T T ; i.e.,
ZT t
(t; u) du =
Differentiating this w.r.t.
1 2
T , we obtain
(t; T ) = (t; T )
ZT t
ZT t
!2
(t; u) du ; 0 t T T :
(7.1)
(t; u) du; 0 t T T :
(7.2)
Not only does (7.1) imply (7.2), (7.2) also implies (7.1). This will be a homework problem. Suppose (7.1) does not hold. Then IP is not a riskneutral measure, but there might still be a riskneutral measure. Let f(t); 0 t T g be an adapted process, and define
f (t) = W
Zt 0
(u) du + W (t);
Zt
Z (t) = exp , If P (A) = Then
Z
A
0
(u) dW (u) ,
Z (T ) dIP
Zt 2 (u) du ; 1 2 0
8A 2 F (T ):
h
i
dB (t; T ) = B(t; T ) r(t) , (t; T ) + 12 ( (t; T ))2 dt , (ht; T )B(t; T ) dW (t) i = B (t; T ) r(t) , (t; T ) + 12 ( (t; T ))2 + (t; T )(t) dt f (t); 0 t T: , (t; T )B(t; T ) dW In order for B (t; T ) to have mean rate of return r(t) under If P , we must have (t; T ) = 21 ( (t; T ))2 + (t; T )(t); 0 t T T : Differentiation w.r.t. T yields the equivalent condition (t; T ) = (t; T )(t; T ) + (t; T )(t); 0 t T T :
(7.3)
(7.4)
Theorem 7.68 (HeathJarrowMorton) For each T 2 (0; T ], let (u; T ); 0 u T; and (u; T ); 0 u T , be adapted processes, and assume (u; T ) > 0 for all u and T . Let f (0; T ); 0 t T , be a deterministic function, and define
f (t; T ) = f (0; T ) +
Zt 0
(u; T ) du +
Zt 0
(u; T ) dW (u):
CHAPTER 28. Termstructure models
281
Then f (t; T ); 0 t T T is a family of forward rate processes for a termstructure model without arbitrage if and only if there is an adapted process (t); 0 t T , satisfying (7.3), or equivalently, satisfying (7.4). Remark 28.2 Under IP , the zerocoupon bond with maturity T has mean rate of return
r(t) , (t; T ) + 12 ((t; T ))2 and volatility (t; T ). The excess mean rate of return, above the interest rate, is
,(t; T ) + 21 ((t; T ))2; and when normalized by the volatility, this becomes the market price of risk
,(t; T ) + 21 ((t; T ))2 : (t; T )
The noarbitrage condition is that this market price of risk at time t does not depend on the maturity T of the bond. We can then set
"
(t; T ) + 1 ( (t; T ))2 # , 2 (t) = , ; (t; T )
and (7.3) is satisfied. (The remainder of this chapter was taught Mar 21) Suppose the market price of risk does not depend on the maturity T , so we can solve (7.3) for . Plugging this into the stochastic differential equation for B (t; T ), we obtain for every maturity T :
f (t): dB (t; T ) = r(t)B(t; T ) dt , (t; T )B(t; T ) dW Because (7.4) is equivalent to (7.3), we may plug (7.4) into the stochastic differential equation for f (t; T ) to obtain, for every maturity T :
df (t; T ) = [(t; T )(t; T ) + (t; T )(t)] dt + (t; T ) dW (t) f (t): = (t; T ) (t; T ) dt + (t; T ) dW 28.8 Implementation of the HeathJarrowMorton model Choose
(t; T ); 0 t T T ; (t); 0 t T :
282 These may be stochastic processes, but are usually taken to be deterministic functions. Define
(t; T ) = (t; T )(t; T ) + (t; T )(t); Zt f W (t) = (u) du + W (t); 0
Let f (0; T );
Zt Zt 2 1 Z (t) = exp , (u) dW (u) , 2 (u) du ; 0 0 Z If P (A) = Z (T ) dIP 8A 2 F (T ): A
0 T T ; be determined by the market; recall from equation (4.1):
@ log B(0; T ); 0 T T : f (0; T ) = , @T
Then f (t; T ) for 0 t T is determined by the equation
f (t); df (t; T ) = (t; T ) (t; T ) dt + (t; T ) dW
(8.1)
this determines the interest rate process
r(t) = f (t; t); 0 t T ; and then the zerocoupon bond prices are determined by the initial conditions B (0; T );
T , gotten from the market, combined with the stochastic differential equation f (t): dB (t; T ) = r(t)B(t; T ) dt , (t; T )B(t; T ) dW
(8.2)
0T (8.3)
P, Because all pricing of interest rate dependent assets will be done under the riskneutral measure If f is a Brownian motion, we have written (8.1) and (8.3) in terms of W f rather than under which W W . Written this way, it is apparent that neither (t) nor (t; T ) will enter subsequent computations. The only process which matters is (t; T ); 0 t T T , and the process (t; T ) = obtained from (t; T ).
ZT t
(t; u) du; 0 t T T ;
(8.4)
From (8.3) we see that (t; T ) is the volatility at time t of the zero coupon bond maturing at time T . Equation (8.4) implies
(T; T ) = 0; 0 T T : (8.5) This is because B (T; T ) = 1 and so as t approaches T (from below), the volatility in B (t; T ) must vanish. In conclusion, to implement the HJM model, it suffices to have the initial market data B (0; T ); T T ; and the volatilities
(t; T ); 0 t T T :
0
CHAPTER 28. Termstructure models
283
We require that (t; T ) be differentiable in T and satisfy (8.5). We can then define
@ (t; T ); (t; T ) = @T and (8.4) will be satisfied because
(t; T ) = (t; T ) , (t; t) =
ZT @ @u (t; u) du: t
f be a Brownian motion under a probability measure IfP , and we let B(t; T ); 0 t We then let W T T , be given by (8.3), where r(t) is given by (8.2) and f (t; T ) by (8.1). In (8.1) we use the initial conditions @ log B(0; T ); 0 T T : f (0; T ) = , @T
f and IP rather than IfP , Remark 28.3 It is customary in the literature to write W rather than W so that IP is the symbol used for the riskneutral measure and no reference is ever made to the market measure. The only parameter which must be estimated from the market is the bond volatility (t; T ), and volatility is unaffected by the change of measure.
284
Chapter 29
Gaussian processes Definition 29.1 (Gaussian Process) A Gaussian process X (t), t 0, is a stochastic process with the property that for every set of times 0 t1 t2 : : : tn , the set of random variables
X (t1); X (t2); : : : ; X (tn)
is jointly normally distributed. Remark 29.1 If X is a Gaussian process, then its distribution is determined by its mean function
m(t) = IEX (t) and its covariance function
(s; t) = IE [(X (s) , m(s)) (X (t) , m(t))]: Indeed, the joint density of X (t1); : : : ; X (tn) is IP fX (t1) 2 dx1; : : : ; X (tn) 2 dxn g n 1 o 1p ,1 (x , m(t))T dx1 : : : dxn ; = exp , ( x , m ( t )) 2 (2 )n=2 det where is the covariance matrix
2 3 (t1; t1) (t1; t2 ) : : : (t1; tn ) 6 7 = 664 (t:2:;:t1) (t:2:;:t2 ) :: :: :: (t:2:; :tn ) 775
(tn ; t1) (tn; t2 ) : : : (tn; tn) x is the row vector [x1; x2; : : : ; xn], t is the row vector [t1; t2; : : : ; tn], and m(t) = [m(t1); m(t2); : : : ; m(tn)]. The moment generating function is
IE exp where
(X n
k=1
)
n
o
uk X (tk ) = exp u m(t)T + 21 u uT ;
u = [u1; u2; : : : ; un].
285
286
29.1 An example: Brownian Motion Brownian motion W is a Gaussian process with m(t) = 0 and (s; t) = s ^ t. Indeed, if 0 s t, then
h
(s; t) = IE [W (s)W (t)] = IE W (s) (W (t) , W (s)) + W 2 (s) = IEW (s):IE (W (t) , W (s)) + IEW 2 (s) = IEW 2(s) = s ^ t:
i
To prove that a process is Gaussian, one must show that X (t1); : : : ; X (tn) has either a density or a moment generating function of the appropriate form. We shall use the m.g.f., and shall cheat a bit by considering only two times, which we usually call s and t. We will want to show that
(
IE exp fu1X (s) + u2X (t)g = exp u1 m1 + u2m2 + Theorem 1.69 (Integral w.r.t. a Brownian) Let dom function. Then Z
t
X (t) =
0
is a Gaussian process with m(t) = 0 and
(s; t) = Proof: (Sketch.) We have
1 [u1 u2 ] 2
"
# " #)
11 12 u1 21 22 u2
:
W (t) be a Brownian motion and (t) a nonran(u) dW (u)
Z s^t 0
2(u) du:
dX = dW:
Therefore,
deuX (s) = ueuX (s)(s) dW (s) + 21 u2 euX (s) 2(s) ds; Zs Zs euX (s) = euX (0) + u euX (v) (v ) dW (v ) + 21 u2 euX (v) 2 (v ) dv; IEeuX (s) = 1 + 12 u2
 0 Zs 0
{z
Martingale
}
0
2(v )IEeuX (v) dv;
d IEeuX (s) = 1 u2 2(s)IEeuX (s); 2 ds Zs uX ( s ) uX (0) 1 2 2 IEe =e exp 2 u (v ) dv = exp
Zs
0
1 u2 2(v ) dv 2 0
:
R This shows that X (s) is normal with mean 0 and variance 0s 2(v ) dv .
(1.1)
CHAPTER 29. Gaussian processes
287
Now let 0 s < t be given. Just as before,
deuX (t) = ueuX (t) (t) dW (t) + 12 u2euX (t) 2 (t) dt:
Integrate from s to t to get
Zt
Zt uX ( v ) 2 1 (v )e dW (v ) + 2 u 2 (v )euX (v) dv: s s
euX (t) = euX (s) + u
Take IE [: : : jF (s)] conditional expectations and use the martingale property
IE
Z t s
(v)euX (v) dW (v) F (s)
to get
= IE =0
Z t 0
Zs , (v)euX (v) dW (v)
(v )euX (v) dW (v ) F (s)
0
Zt 2 (v )IE euX (v) F (s) dv s d IE euX (t) F (s) = 1 u2 2(t)IE euX (t) F (s) ; t s: 2 dt
IE euX (t) F (s) = euX (s) + 12 u2
The solution to this ordinary differential equation with initial time s is
IE
euX (t) F (s)
Zt 2 2 uX ( s ) 1 =e exp 2 u (v ) dv ; s
t s:
(1.2)
We now compute the m.g.f. for (X (s); X (t)), where 0 s t:
IE
eu1 X (s)+u2X (t) F (s)
= eu1 X (s) IE
eu2X (t) F (s)
Z
t (1.2) (u1 +u2 )X (s) = e exp 21 u22 2 (v ) dv s h u1 X (s)+u2X (t)i IE e = IE IE eu1 X (s)+u2 X (t) F (s)
;
n (u +u )X (s)o 1 2 Z t 2 = IE e : exp 2 u2 (v ) dv s Z Zt s (1.1) 1
2
1 (u1 + u2 )2 2 (v ) dv + 21 u22 2(v ) dv 2 Z0 s Zs t 2 2 2 2 1 1 = exp 2 (u1 + 2u1u2 ) (v ) dv + 2 u2 (v ) dv 0 ( "R s 2 0R s 2# " #) = exp 12 [u1 u2 ] R0s 2 R0t 2 uu1 : 2 0 0
= exp
This shows that (X (s); X (t)) is jointly normal with IEX (s) = IEX (t) = 0,
IEX 2(s) =
Zs 0
2 (v ) dv;
IE [X (s)X (t)] =
IEX 2(t) =
Zs 0
Zt
2(v ) dv:
0
2(v ) dv;
288
Remark 29.2 The hard part of the above argument, and the reason we use moment generating functions, is to prove the normality. The computation of means and variances does not require the use of moment generating functions. Indeed,
X (t) = is a martingale and X (0) = 0, so For fixed s 0,
Zt 0
(u) dW (u)
m(t) = IEX (t) = 0 8t 0:
by the Itˆo isometry. For 0 s t,
IEX 2(s) =
Zs 0
2(v ) dv
IE [X (s)(X (t) , X (s))] = IE IE X (s)(X (t) , X (s)) F (s) 2 3 77 66 = IE 64X (s) IE X (t) F (s) , X (s) 75  {z }
0
= 0: Therefore,
IE [X (s)X (t)] = IE [X (s)(X (t) , X (s)) + X 2(s)] Zs = IEX 2(s) = 2 (v ) dv: If were a stochastic proess, the Itˆo isometry says
IEX 2(s) =
0
Zs
IE 2(v) dv and the same argument used above shows that for 0 s t, Zs 2 IE [X (s)X (t)] = IEX (s) = IE2(v ) dv: 0 However, when is stochastic, X is not necessarily a Gaussian process, so its distribution is not 0
determined from its mean and covariance functions. Remark 29.3 When is nonrandom,
X (t) =
Zt 0
(u) dW (u)
is also Markov. We proved this before, but note again that the Markov property follows immediately from (1.2). The equation (1.2) says that conditioned on F (s), the distribution of X (t) depends only R t on X (s); in fact, X (t) is normal with mean X (s) and variance s 2(v ) dv .
CHAPTER 29. Gaussian processes y=
v=
z
z
z
z
289
s
s y
v
t
(b)
(a) y=
z
z
s v y t (c)
Figure 29.1: Range of values of y; z; v for the integrals in the proof of Theorem 1.70.
Theorem 1.70 Let Define
W (t) be a Brownian motion, and let (t) and h(t) be nonrandom functions. X (t) =
Zt 0
(u) dW (u); Y (t) =
Zt 0
h(u)X (u) du:
Then Y is a Gaussian process with mean function mY (t) = 0 and covariance function
Y (s; t) =
Z s^t 0
2 (v )
Z s v
h(y ) dy
Z t v
h(y ) dy dv:
(1.3)
Proof: (Partial) Computation of Y (s; t): Let 0 s t be given. It is shown in a homework problem that (Y (s); Y (t)) is a jointly normal pair of random variables. Here we observe that
mY (t) = IEY (t) = and we verify that (1.3) holds.
Zt 0
h(u) IEX (u) du = 0;
290 We have
Y (s; t) = IE [Y (s)Y (t)] Z s Zt = IE h(y)X (y ) dy: h(z)X (z) dz = IE = = =
Z s0 Z t
Z s Z0 t 0
h(y )h(z )
0 Zz
sZ s
0
0 y
h(z )
+
=
h(y )h(z )
Z0s Z0t Zs
Zs
Z s Z 0z 0 Z0
+
h(y)h(z)X (y )X (z) dy dz
h(y )h(z )IE [X (y )X (z)] dy dz
Z0s Z0t
+
=
0
Z y^z Z0 0
h(y )h(z)
Z t z
h(y)
h(y ) dy
Z s y
h(z )2(v )
sZ y
z
2(v ) dv dy dz 2(v ) dv
Z y 0
Z t z
h(y ) 2(v )
dy dz
2(v ) dv
Z z
h(z ) dz
dz dy
(See Fig. 29.1(a))
2 (v) dv dz
0 Z y
0
2(v) dv dy
h(y ) dy dv dz
Z s
h(z ) dz dv dy
Z t y Z s Z 0s 0 = h(z) 2(v ) h(y ) dy dz dv 0 Zv Z z Z s s s + h(y ) 2(v ) h(z) dz dy dv (See Fig. 29.1(b)) y Z s 0 vZ s Z t 2 = (v ) h(y )h(z) dy dz dv 0 Z v Zz Z s s s 2 + (v ) h(y )h(z) dz dy dv 0 v y Z s Z t Zs = 2 (v ) h(y )h(z) dy dz dv (See Fig. 29.1(c)) Zvs v Z t Z0s = 2 (v ) h(y ) dy h(z ) dz dv Zvs Zvt Z0s = 2 (v ) h(y ) dy h(y ) dy dv 0 v v
Remark 29.4 Unlike the process
X (t) =
R t (u) dW (u), the process Y (t) = R t X (u) du is 0 0
CHAPTER 29. Gaussian processes
291
neither Markov nor a martingale. For 0 s < t,
IE [Y (t)jF (s)] =
Zs 0
h(u)X (u) du + IE
= Y (s) + = Y (s) +
Zt Zs t s
Z t
h(u)X (u) du F (s)
s h(u)IE [X (u) F (s)] du h(u)X (s) du
= Y (s) + X (s)
Zt s
h(u) du;
where we have used the fact that X is a martingale. The conditional expectation IE [Y (t)jF (s)] is not equal to Y (s), nor is it a function of Y (s) alone.
292
Chapter 30
Hull and White model Consider
dr(t) = ((t) , (t)r(t)) dt + (t) dW (t); where (t), (t) and (t) are nonrandom functions of t. We can solve the stochastic differential equation. Set
K (t) = Then
Zt 0
(u) du:
d eK (t)r(t) = eK(t) (t)r(t) dt + dr(t) = eK (t) ((t) dt + (t) dW (t)) : Integrating, we get
eK (t)r(t) = r(0) + so
Zt 0
eK (u) (u) du +
Zt 0
eK (u) (u) dW (u);
Zt Zt , K ( t ) K ( u ) K ( u ) r(t) = e r(0) + e (u) du + e (u) dW (u) : 0
0
From Theorem 1.69 in Chapter 29, we see that r(t) is a Gaussian process with mean function
mr (t) = e,K(t) r(0) + and covariance function
Zt 0
eK(u)(u) du
Z s^t , K ( s ) , K ( t ) r (s; t) = e e2K(u)2 (u) du:
The process r(t) is also Markov.
0
293
(0.1)
(0.2)
294
R We want to study 0T
r(t) dt. To do this, we define
X (t) =
Zt 0
Then
ZT 0
r(t) = e,K (t) r(0) + r(t) dt =
ZT 0
ZT 0
and its variance is
r(t) dt =
ZT
var
0
Zt 0
0
e,K (t)X (t) dt:
eK (u) (u) du + e,K (t)X (t);
Zt , K ( t ) K ( u ) e r(0) + e (u) du
R According to Theorem 1.70 in Chapter 29, 0T IE
ZT
eK (u) (u) dW (u); Y (T ) =
ZT 0
0
dt + Y (T ):
r(t) dt is normal. Its mean is
e,K (t) r(0) +
Zt 0
eK(u)(u) du dt;
(0.3)
!
r(t) dt = IEY 2 (T ) =
ZT 0
e2K (v) 2(v )
ZT v
e,K (y) dy
!2
dv:
The price at time 0 of a zerocoupon bond paying $1 at time T is
( ZT ) B (0; T ) = IE exp , r(t) dt 0 ( !) ZT ZT 2 1 = exp (,1)IE r(t) dt + 2 (,1) var r(t) dt 0 0 ZT ZTZ t = exp ,r(0) e,K (t) dt , e,K (t)+K (u) (u) du dt 0 0 0 !2 ZT ZT + 12
0
e2K (v)2 (v )
v
e,K (y) dy
dv
= expf,r(0)C (0; T ) , A(0; T )g; where
C (0; T ) = A(0; T ) =
ZT 0
e,K (t) dt;
Z TZt 0
0
e,K (t)+K (u)(u) du dt , 12
ZT 0
e2K (v) 2(v )
ZT v
e,K (y) dy
!2
dv:
CHAPTER 30. Hull and White model
295
u
=
t
t
T u Figure 30.1: Range of values of u; t for the integral.
30.1 Fiddling with the formulas Note that (see Fig 30.1)
ZTZ t 0
= (y = t; v = u) =
e,K (t)+K (u) (u) du dt
Z 0T Z T Z0T 0
e,K (t)+K (u)(u) dt du
u eK(v)(v )
ZT v
e,K(y) dy
!
dv:
Therefore,
! !23 Z T2 ZT ZT A(0; T ) = 4eK (v)(v ) e,K (y) dy , 12 e2K (v) 2(v ) e,K (y) dy 5 dv; v 0 v ZT
C (0; T ) = e,K(y) dy; 0 B (0; T ) = exp f,r(0)C (0; T ) , A(0; T )g :
Consider the price at time t 2 [0; T ] of the zerocoupon bond:
"
( ZT ) # B (t; T ) = IE exp , r(u) du F (t) : t
Because r is a Markov process, this should be random only through a dependence on r(t). In fact,
B(t; T ) = exp f,r(t)C (t; T ) , A(t; T )g ;
296 where
!23 ! ZT Z T2 ZT e,K (y) dy 5 dv; A(t; T ) = 4eK (v) (v) e,K (y) dy , 12 e2K (v)2 (v ) v t v ZT
C (t; T ) = eK(t)
t
e,K (y) dy:
The reason for these changes R is the following. We are now taking R the initial time to be t rather than zero, so it is plausible that 0T : : : dv should be replaced by tT : : : dv: Recall that
K (v ) =
Zv 0
and this should be replaced by
K (v ) , K (t) =
(u) du;
Zv t
(u) du:
Similarly, K (y ) should be replaced by K (y ) , K (t). Making these replacements in A(0; T ), we see that the K (t) terms cancel. In C (0; T ), however, the K (t) term does not cancel.
30.2 Dynamics of the bond price Let Ct(t; T ) and At (t; T ) denote the partial derivatives with respect to t. From the formula
B (t; T ) = exp f,r(t)C (t; T ) , A(t; T )g ; we have
i
h
dB (t; T ) = B (t; T ) ,C (t; T ) dr(t) , 21 C 2 (t; T ) dr(t) dr(t) , r(t)Ct(t; T ) dt , At (t; T ) dt = B (t; T ) , C (t; T ) ((t) , (t)r(t)) dt
, C (t; T )(t) dW (t) , 12 C 2(t;T )2(t) dt , r(t)Ct(t; T ) dt , At(t; T ) dt : Because we have used the riskneutral pricing formula
"
( ZT ) # B(t; T ) = IE exp , r(u) du F (t) t
to obtain the bond price, its differential must be of the form
dB (t; T ) = r(t)B(t; T ) dt + (: : : ) dW (t):
CHAPTER 30. Hull and White model
297
Therefore, we must have
,C (t; T ) ((t) , (t)r(t)) , 12 C 2(t; T )2(t) , r(t)Ct(t; T ) , At(t; T ) = r(t): We leave the verification of this equation to the homework. After this verification, we have the formula
dB (t; T ) = r(t)B (t; T ) dt , (t)C (t; T )B(t; T ) dW (t): In particular, the volatility of the bond price is (t)C (t; T ). 30.3 Calibration of the Hull & White model Recall:
dr(t) = ((t) , (t)r(t)) dt + (t) dB (t); Zt K (t) = (u) du;
0 ! !23 ZT2 ZT ZT e,K (y) dy 5 dv; A(t; T ) = 4eK (v)(v) e,K (y) dy , 12 e2K (v) 2 (v ) t v v ZT
C (t; T ) = eK(t) e,K(y) dy; t B (t; T ) = exp f,r(t)C (t; T ) , A(t; T )g :
Suppose we obtain B (0; T ) for all T 2 [0; T ] from market data (with some interpolation). Can we determine the functions (t), (t), and (t) for all t 2 [0; T ]? Not quite. Here is what we can do. We take the following input data for the calibration: 1.
B (0; T ); 0 T T ;
2.
r(0);
3.
(0);
4.
(t); 0 t T (usually assumed to be constant);
5.
(0)C (0; T ); 0 T T , i.e., the volatility at time zero of bonds of all maturities.
Step 1. From 4 and 5 we solve for
C (0; T ) =
ZT 0
e,K (y) dy:
298 We can then compute
@ C (0; T ) = e,K (T ) @T @ C (0; T ); =) K (T ) = , log @T @ K (T ) = @ Z T (u) du = (T ): @T @T 0 We now have (T ) for all T 2 [0; T ]. Step 2. From the formula
B (0; T ) = expf,r(0)C (0; T ) , A(0; T )g; we can solve for A(0; T ) for all T 2 [0; T ]. Recall that
! !23 Z T2 ZT ZT A(0; T ) = 4eK (v) (v) e,K (y) dy , 21 e2K (v)2 (v ) e,K (y) dy 5 dv: 0
v
v
We can use this formula to determine (T );
0 T T as follows:
"
!#
@ A(0; T ) = Z T eK(v) (v)e,K (T ) , e2K(v)2 (v )e,K (T ) Z T e,K(y) dy dv; @T 0 v " !# Z Z @ A(0; T ) = T eK(v) (v) , e2K(v) 2 (v ) T e,K (y) dy dv; eK (T ) @T 0 v Z T @ K (T ) @ K (T ) 2K (v) 2 ,K (T ) dv; @T e @T A(0; T ) = e (T ) , Z0 e (v ) e @ eK (T ) @ A(0; T ) = e2K (T )(T ) , T e2K(v)2 (v ) dv; eK(T ) @T @T 0 @ eK(T ) @ eK (T ) @ A(0; T ) = 0(T )e2K (T ) + 2(T ) (T )e2K(T ) , e2K (T )2(T ); 0 T T : @T @T @T This gives us an ordinary differential equation for , i.e., 0(t)e2K (t) + 2(t) (t)e2K(t) , e2K (t)2(t) = known function of t: From assumption 4 and step 1, we know all the coefficients in this equation. From assumption 3, we have the initial condition (0). We can solve the equation numerically to determine the function (t); 0 t T . Remark 30.1 The derivation of the ordinary differential equation for (t) requires three differentiations. Differentiation is an unstable procedure, i.e., functions which are close can have very different derivatives. Consider, for example,
f (x) = 0 8x 2 IR; x) 8x 2 IR: g (x) = sin(1000 100
CHAPTER 30. Hull and White model
299
Then
1 8x 2 IR; jf (x) , g(x)j 100 but because
g 0(x) = 10 cos(1000x); we have
jf 0(x) , g0(x)j = 10
for many values of x.
Assumption 5 for the calibration was that we know the volatility at time zero of bonds of all maturities. These volatilities can be implied by the prices of options on bonds. We consider now how the model prices options.
30.4 Option on a bond Consider a European call option on a zerocoupon bond with strike price K and expiration time T1 . The bond matures at time T2 > T1. The price of the option at time 0 is R T1 , r(u) du +
IE e
(B (T1 ; T2) , K )
0
R T r(u) du (expf,r(T1)C (T1; T2) , A(T1; T2)g , K )+ : + Z1Z1 = e,x expf,yC (T1; T2) , A(T1; T2)g , K f (x; y ) dx dy; ,1 ,1 R where f (x; y ) is the joint density of 0T r(u) du; r(T1) . R We observed at the beginning of this Chapter (equation (0.3)) that 0T r(u) du is normal with "Z T # ZT 4 = IEe,
0
1
1
1
1 = IE
2 =4 var 1
1
0
=
"Z T
1
0
r(u) du =
1
0
ZT 1
# Z0T
r(u) du =
IEr(u) du
1
0
Zv
r(0)e,K (v) + e,K (v)
e2K(v)2(v )
ZT
1
v
We also observed (equation (0.1)) that r(T1) is normal with
2 =4 IEr(T1) = r(0)e,K (T ) + e,K(T ) 1
22 =4 var (r(T1)) = e,2K (T1)
ZT
1
1
0
0
e,K(y) dy
ZT
1
eK(u)(u) du dv;
!2
dv:
eK (u) (u) du;
0 2 K ( u ) 2 e (u) du:
300 In fact,
R T 0 r(u) du; r(T1) is jointly normal, and the covariance is 1
12 = IE = =
"Z T
ZT
0
0
#
(r(u) , IEr(u)) du: (r(T1) , IEr(T1))
1
IE [(r(u) , IEr(u)) (r(T1) , IEr(T1))] du
1
r (u; T1) du;
0
ZT
1
where r (u; T1) is defined in Equation 0.2. The option on the bond has price at time zero of
Z1Z1 ,1 ,1
e,x expf,yC (T1; T2) , A(T1; T2)g , K
(
+ "
2 y2 2 1p1 , 2 exp , 2(1 ,1 2) x2 + 2xy + 1 2 22 1 2 1
#)
dx dy:
(4.1)
The price of the option at time t 2 [0; T1] is
RT IE e, t
1
r(u) du (B (T
+ 1; T2) , K ) F (t)
RT ,
= IE e
t r(u) du (expf,r(T1)C (T1; T2) , A(T1; T2)g , K )+ F (t) 1
(4.2)
Because of the Markov property, this is random only R T1through a dependence on r(t). To compute this option price, we need the joint distribution of t r(u) du; r(T1) conditioned on r(t). This
CHAPTER 30. Hull and White model
301
pair of random variables has a jointly normal conditional distribution, and
1 (t) = IE =
"Z T
1
t
ZT 1
t
# r(u) du F (t)
Zv
r(t)e,K(v)+K (t) + e,K (v)
eK(u)(u) du dv;
t 3 2 Z ! 2 T 21 (t) = IE 4 r(u) du , 1 (t) F (t)5 t !2 ZT ZT 1
=
1
t
e2K (v) 2(v )
2 (t) = IE r(T1) r(t)
1
v
e,K (y) dy
= r(t)e,K (T )+K (t) + e,K (T )
22 (t) = IE
1
(r(T1) , 2
1
(t))2 F (t)
ZT
1
t
dv;
eK (u)(u) du;
Z T1 , 2 K ( T 1) =e e2K (u) 2 (u) du; t " Z T1 ! # (t)1 (t)2 (t) = IE r(u) du , 1 (t) (r(T1) , 2 (t)) F (t) t Z T1 Zu = e,K (u),K(T1 ) e2K (v) 2(v ) dv du: t t The variances and covariances are not random. The means are random through a dependence on r(t). Advantages of the Hull & White model: 1. Leads to closedform pricing formulas. 2. Allows calibration to fit initial yield curve exactly. Shortcomings of the Hull & White model: 1. Onefactor, so only allows parallel shifts of the yield curve, i.e.,
B (t; T ) = exp f,r(t)C (t; T ) , A(t; T )g ; so bond prices of all maturities are perfectly correlated. 2. Interest rate is normally distributed, and hence can take negative values. Consequently, the bond price " ( ) #
B (t; T ) = IE exp ,
can exceed 1.
ZT t
r(u) du F (t)
302
Chapter 31
CoxIngersollRoss model In the Hull & White model, r(t) is a Gaussian process. Since, for each t, r(t) is normally distributed, there is a positive probability that r(t) < 0. The CoxIngersollRoss model is the simplest one which avoids negative interest rates. We begin with a ddimensional Brownian motion (W1 ; W2; : : : ; Wd). Let constants. For j = 1; : : : ; d, let Xj (0) 2 IR be given so that
> 0 and > 0 be
X12(0) + X22(0) + : : : + Xd2(0) 0; and let Xj be the solution to the stochastic differential equation
dXj (t) = , 12 Xj (t) dt + 12 dWj (t): Xj is called the OrsteinUhlenbeck process. It always has a drift toward the origin. The solution to
this stochastic differential equation is
1
Xj (t) = e, 2 t Xj (0) + 12
Zt 0
1
e 2 u dWj (u) :
This solution is a Gaussian process with mean function
1
mj (t) = e, 2 t Xj (0) and covariance function
1 (s+t) Z s^t u 1 , 2 (s; t) = 4 e 2 e du: 0
Define
r(t) =4 X12(t) + X22(t) + : : : + Xd2(t): If d = 1, we have r(t) = X12(t) and for each t, IP fr(t) > 0g = 1, but (see Fig. 31.1) IP
There are infinitely many values of t > 0 for which r(t) = 0 303
= 1
304 2
r(t) = X 1 (t)
t x2 ( X1 (t), X2 (t) )
x
1
Figure 31.1: r(t) can be zero. If d 2, (see Fig. 31.1)
IP fThere is at least one value of t > 0 for which r(t) = 0g = 0: Let f (x1; x2; : : : ; xd) = x21 + x22 + : : : + x2d . Then
(
fxi = 2xi ; fxi xj = 2
0
if i = j; if i 6= j:
Itˆo’s formula implies
dr(t) = =
d X i=1 d X i=1
fxi dXi + 12
i=1
fxi xi dXi dXi d 1 X
2Xi , 12 Xi dt + 12 dWi(t) + d X
= , r(t) dt + =
d X
!i=1
2
Xi dWi + d4 dt
2 dWi dWi 4 i=1
d X (t) q X pri (t) dWi(t): , r ( t ) dt + r ( t ) 4 i=1
d 2
Define
d Z t X (u) X pi dWi(u): W (t) = r(u) i=1 0
CHAPTER 31. CoxIngersollRoss model
305
Then W is a martingale,
d X X dW = pri dWi ; i=1
dW dW =
d X2 X i i=1
r dt = dt;
so W is a Brownian motion. We have
!
q 2 dr(t) = d4 , r(t) dt + r(t) dW (t):
The CoxIngersollRoss (CIR) process is given by
q
dr(t) = ( , r(t)) dt + r(t) dW (t); We define
d = 42 > 0:
If d happens to be an integer, then we have the representation
r(t) =
d X i=1
Xi2(t);
but we do not require d to be an integer. If d < 2 (i.e., < 12 2), then
IP fThere are infinitely many values of t > 0 for which r(t) = 0g = 1: This is not a good parameter choice. If d 2 (i.e., 12 2 ), then
IP fThere is at least one value of t > 0 for which r(t) = 0g = 0: With the CIR process, one can derive formulas under the assumption that integer, and they are still correct even when d is not an integer.
d = 4 2
is a positive
For example, here is the distribution of r(t) for fixed t > 0. Let r(0) 0 be given. Take
q
X1(0) = 0; X2(0) = 0; : : : ; Xd,1 (0) = 0; Xd(0) = r(0): For i = 1; 2; : : : ; d , 1, Xi (t) is normal with mean zero and variance 2 (t; t) = 4 (1 , e, t ):
306
Xd (t) is normal with mean and variance (t; t). Then
r(t) =
(t; t)

q
1
md(t) = e, 2 t r(0) dX ,1 i=1
!2
pXi(t) (t; t)
{z
2 Chisquare with d , 1 = ,2
4
2 X  d{z(t)}
+
}
(0.1)
Normal squared and independent of the other term
degrees of
freedom
Thus r(t) has a noncentral chisquare distribution.
31.1 Equilibrium distribution of r (t) As t!1, md (t)!0. We have
!2 d X X ( t ) i p r(t) = (t; t) : (t; t) i=1
As t!1, we have (t; t) = 4 , and so the limiting distribution of r(t) is 4 times a chisquare 4 degrees of freedom. The chisquare density with 4 degrees of freedom is with d = 2 2 2
f (y ) =
2
1 y , e,y=2 : 22= , 2 2
2
2
2
2
We make the change of variable r = 4 y . The limiting density for r(t) is 2
1 4 r e, r 2 22= , 2 2 , = 2 2 12 r e, r : , ,2
p(r) = 4 :
2
2
2
2
2
2
2
2
2
2
2
2
2
2
We computed the mean and variance of r(t) in Section 15.7.
31.2 Kolmogorov forward equation Consider a Markov process governed by the stochastic differential equation
dX (t) = b(X (t)) dt + (X (t)) dW (t):
CHAPTER 31. CoxIngersollRoss model
307
h
y
0
Figure 31.2: The function h(y ) Because we are going to apply the following analysis to the case X (t) 0 for all t.
X (t) = r(t), we assume that
We start at X (0) = x 0 at time 0. Then X (t) is random with density p(0; t; x; y ) (in the y variable). Since 0 and x will not change during the following, we omit them and write p(t; y ) rather than p(0; t; x; y ). We have
IEh(X (t)) =
Z1 0
h(y )p(t; y ) dy
for any function h. The Kolmogorov forward equation (KFE) is a partial differential equation in the “forward” variables t and y . We derive it below. Let h(y ) be a smooth function of y Fig. 31.2). Itˆo’s formula implies
0 which vanishes near y = 0 and for all large values of y (see
h i dh(X (t)) = h0(X (t))b(X (t)) + 21 h00(X (t))2 (X (t)) dt + h0 (X (t))(X (t)) dW (t); so
h(X (t)) = h(X (0)) +
Zt 0
Z th 0
i
h0 (X (s))b(X (s)) + 12 h00 (X (s))2 (X (s)) ds +
h0 (X (s))(X (s)) dW (s);
IEh(X (t)) = h(X (0)) + IE
Z th 0
i
h0 (X (s))b(X (s)) dt + 12 h00(X (s))2 (X (s)) ds;
308 or equivalently,
Z1 0
h(y)p(t; y ) dy = h(X (0)) +
Z tZ 1 Z0 0
h0 (y )b(y )p(s; y ) dy ds +
Zt 1 00 2 1 2 0 0 h (y ) (y )p(s; y ) dy ds:
Differentiate with respect to t to get
Z1 0
h(y)pt(t; y) dy =
Z1 0
h0(y )b(y )p(t; y ) dy + 12
Z1 0
h00(y )2 (y )p(t; y ) dy:
Integration by parts yields
y=1 Z 1 @ (b(y)p(t; y )) dy; , h(y) @y 0 {z y=0 } 0  =0 y=1 Z 1 Z1 @ 2(y )p(t; y ) dy 00 2 0 2 h (y) (y)p(t; y ) dy = h (y) (y )p(t; y ) , h0 (y ) @y 0  {z y=0 } 0 =0 y=1 Z 1 @ @ 2 2(y )p(t; y ) dy: = ,h(y ) @y 2(y )p(t; y ) + h(y ) @y 2 y=0 } 0  {z Z1
h0(y )b(y)p(t; y ) dy = h(y)b(y)p(t; y )
=0
Therefore,
Z1 0
h(y)pt(t; y) dy = ,
or equivalently,
Z1 0
Z1 0
@ (b(y)p(t; y )) dy + 1 Z 1 h(y ) @ 2 2(y )p(t; y ) dy; h(y ) @y 2 @y 2 0
"
#
@ (b(y )p(t; y )) , 1 @ 2 2(y )p(t; y ) dy = 0: h(y) pt(t; y ) + @y 2 @y 2
This last equation holds for every function h of the form in Figure 31.2. It implies that
@ ((b(y )p(t; y )) , 1 @ 2 2(y )p(t; y ) = 0: pt(t; y ) + @y 2 @y 2
If there were a place where (KFE) did not hold, then we could take points, but take h to be zero elsewhere, and we would obtain
Z1 " 0
#
(KFE)
h(y ) > 0 at that and nearby
@ (bp) , 1 @ 2 (2 p) dy 6= 0: h pt + @y 2 @y 2
CHAPTER 31. CoxIngersollRoss model
309
If the process X (t) has an equilibrium density, it will be
p(y ) = t! lim1 p(t; y ): In order for this limit to exist, we must have
0 = t! lim1 pt (t; y ):
Letting t!1 in (KFE), we obtain the equilibrium Kolmogorov forward equation
@ (b(y)p(y)) , 1 @ 2 2(y )p(y ) = 0: 2 @y 2 @y
When an equilibrium density exists, it is the unique solution to this equation satisfying
p(y ) 0 8y 0;
Z1 0
p(y ) dy = 1:
31.3 CoxIngersollRoss equilibrium density We computed this to be , p(r) = Cr e, r ; 2
2
2
2
2
where
2
C = 2
2
2
12 :
,
2
We compute
2 p0(r) = 2,2 : p(rr) , 2 2 p(r) = 22 , 12 2 , r p(r); r
p00(r) = , 22r2 , 21 2 , r p(r) + 22 r (, )p(r) + 22r , 12 2 , r p0 (r) = 22r , 1r ( , 12 2 , r) , + 22r ( , 12 2 , r)2 p(r)
We want to verify the equilibrium Kolmogorov forward equation for the CIR process:
@ (( , r)p(r)) , 1 @ 2 (2 rp(r)) = 0: 2 @r2 @r
(EKFE)
310 Now
@ (( , r)p(r)) = , p(r) + ( , r)p0(r); @r @ 2 (2 rp(r)) = @ (2 p(r) + 2 rp0(r)) @r2 @r = 22 p0(r) + 2rp00(r): The LHS of (EKFE) becomes
, p(r)+ ( , r)p0(r) , 2p0(r) , 12 2rp00(r) = p(r) , + ( , r , 2 ) 22 r ( , 12 2 , r)
2 1 2 2 2 1 1 + ( , 2 , r) + , 2r ( , 2 , r) r = p(r) ( , 1 2 , r) 2 ( , 1 2 , r) 2
2 2 r , 12 2 22r ( , 12 2 , r)
= 0;
+ r1 ( , 12 2 , r) , 22 r ( , 21 2 , r)2
as expected.
31.4 Bond prices in the CIR model The interest rate process r(t) is given by
q
dr(t) = ( , r(t)) dt + r(t) dW (t); where r(0) is given. The bond price process is
"
( ZT ) # B (t; T ) = IE exp , r(u) du F (t) : t
Because
" ( ZT ) # Zt exp , r(u) du B (t; T ) = IE exp , r(u) du F (t) ; 0
0
the tower property implies that this is a martingale. The Markov property implies that B (t; T ) is random only through a dependence on r(t). Thus, there is a function B (r; t; T ) of the three dummy variables r; t; T such that the process B (t; T ) is the function B (r; t; T ) evaluated at r(t); t; T , i.e.,
B(t; T ) = B(r(t); t; T ):
CHAPTER 31. CoxIngersollRoss model Because exp pute
311
n Rt o , 0 r(u) du B(r(t); t; T ) is a martingale, its differential has no dt term. We com
Zt
d exp ,
r(u) du B (r(t); t; T )
Z0t = exp , r(u) du ,r(t)B (r(t); t; T ) dt + Br (r(t); t; T ) dr(t) + 0 1 Brr (r(t); t; T ) dr(t) dr(t) + Bt (r(t); t; T ) dt 2
The expression in [: : : ] equals
:
p
= ,rB dt + Br ( , r) dt + Br r dW + 12 Brr 2 r dt + Bt dt: Setting the dt term to zero, we obtain the partial differential equation
, rB(r; t; T ) + Bt (r; t; T ) + ( , r)Br (r; t; T ) + 21 2rBrr (r; t; T ) = 0; 0 t < T; r 0: The terminal condition is
(4.1)
B(r; T; T ) = 1; r 0:
Surprisingly, this equation has a closed form solution. Using the Hull & White model as a guide, we look for a solution of the form
B (r; t; T ) = e,rC (t;T ),A(t;T ); where C (T; T ) = 0;
A(T; T ) = 0. Then we have Bt = (,rCt , At )B; Br = ,CB; Brr = C 2B;
and the partial differential equation becomes
0 = ,rB + (,rCt , At )B , ( , r)CB + 21 2 rC 2B = rB (,1 , Ct + C + 12 2 C 2 ) , B (At + C ) We first solve the ordinary differential equation
,1 , Ct(t; T ) + C (t; T ) + 12 2C 2(t; T ) = 0; C (T; T ) = 0; and then set
A(t; T ) =
ZT t
C (u; T ) du;
312 so A(T; T ) = 0 and
At (t; T ) = ,C (t; T ):
It is tedious but straightforward to check that the solutions are given by
sinh( (T , t)) ;
cosh( (T , t)) + 12 sinh( (T , t)) 2 3 1 2 (T ,t) 2
e 5; A(t; T ) = , 2 log 4
cosh( (T , t)) + 21 sinh( (T , t))
C (t; T ) =
where
=
1 2
q
u ,u u ,u 2 + 22; sinh u = e ,2 e ; cosh u = e +2 e :
Thus in the CIR model, we have
"
( ZT ) # IE exp , r(u) du F (t) = B(r(t); t; T ); t
where
B(r; t; T ) = exp f,rC (t; T ) , A(t; T )g ; 0 t < T; r 0; and C (t; T ) and A(t; T ) are given by the formulas above. Because the coefficients in
q
dr(t) = ( , r(t)) dt + r(t) dW (t) do not depend on t, the function B (r; t; T ) depends on t and T only through their difference = T , t. Similarly, C (t; T ) and A(t; T ) are functions of = T , t. We write B(r; ) instead of B (r; t; T ), and we have B (r; ) = exp f,rC ( ) , A( )g ; 0; r 0; where
) C ( ) = cosh( sinh( ; ) + 1 sinh( )
2 2 3 1 2
e 5; A( ) = , 22 log 4
cosh( ) + 12 sinh( ) q 1
= 2 2 + 22 :
We have
( ZT ) B(r(0); T ) = IE exp , r(u) du : 0
Now r(u) > 0 for each u, almost surely, so B (r(0); T ) is strictly decreasing in T . Moreover,
B(r(0); 0) = 1;
CHAPTER 31. CoxIngersollRoss model
313
Z1
lim B (r(0); T ) = IE exp , T !1 But also,
0
r(u) du = 0:
B(r(0); T ) = exp f,r(0)C (T ) , A(T )g ;
so
r(0)C (0) + A(0) = 0; lim [r(0)C (T ) + A(T )] = 1; T !1 and
r(0)C (T ) + A(T ) is strictly inreasing in T .
31.5 Option on a bond The value at time t of an option on a bond in the CIR model is
"
( ZT ) # + v(t; r(t)) = IE exp , r(u) du (B(T1; T2) , K ) F (t) ; t 1
where T1 is the expiration n R time ofothe option, T2 is the maturity time of the bond, and 0 t T1 T2. As usual, exp , 0t r(u) du v(t; r(t)) is a martingale, and this leads to the partial differential equation
(where v
,rv + vt + ( , r)vr + 12 2rvrr = 0; 0 t < T1; r 0:
= v (t; r).) The terminal condition is
v(T1; r) = (B(r; T1; T2) , K )+ ; r 0: Other European derivative securities on the bond are priced using the same partial differential equation with the terminal condition appropriate for the particular security.
31.6 Deterministic time change of CIR model Process time scale: In this time scale, the interest rate r(t) is given by the constant coefficient CIR equation q
dr(t) = ( , r(t)) dt + r(t) dW (t):
Real time scale: In this time scale, the interest rate r^(t^) is given by a timedependent CIR equation
q
dr^(t^) = (^(t^) , ^(t^)^r(t^)) dt^ + ^ (t^) r^(t^) dW^ (t^):
t: 314
6
Process time
.. ... .. ... ... .. . ..
. .. .. . .. ... . .. .. .. .. .. . .. .. ... ... ... ... .. . ..
t = '(t^)


t^ :
Real time
A period of high interest rate volatility
Figure 31.3: Time change function. There is a strictly increasing time change function t Fig. 31.3).
= '(t^) which relates the two time scales (See
^ (^r; ^t; T^) denote the price at real time t^ of a bond with maturity T^ when the interest rate at time Let B ^t is r^. We want to set things up so
B^ (^r; ^t; T^) = B (r; t; T ) = e,rC (t;T ),A(t;T ); = '(T^), and C (t; T ) and A(t; T ) are as defined previously. We need to determine the relationship between r^ and r. We have where t = '(t^); T
( ZT ) B(r(0); 0; T ) = IE exp , r(t) dt ; ( Z0T^ ) B(^r(0); 0; T^) = IE exp , r^(t^) dt^ : 0
With T
= '(T^), make the change of variable t = '(t^), dt = '0(t^) dt^ in the first integral to get
( Z T^ ) 0 B(r(0); 0; T ) = IE exp , r('(t^))' (t^) dt^ ; 0
and this will be B (^ r(0); 0; T^) if we set
r^(t^) = r('(t^)) '0(t^):
CHAPTER 31. CoxIngersollRoss model
315
31.7 Calibration
!
^ B^ (^r(t^); ^t; T^) = B r^0(t^) ; '(t^); '(T^) (' (t) ) ^)) ^ C ( ' ( t ) ; ' ( T ^ = exp ,r^(t^) , A('(t^); '(T )) '0 (t^) n o = exp ,r^(t^)C^ (t^; T^) , A^(t^; T^) ; where
^
^
C^ (t^; T^) = C ('(t)0 ;^'(T )) ' (t) A^(t^; T^) = A('(t^); '(T^)) do not depend on ^ t and T^ only through T^ , t^, since, in the real time scale, the model coefficients are time dependent.
^ (^r(0); 0; T^ ) for all T^ 2 [0; T^ ]. We calibrate by writing the equation Suppose we know r^(0) and B
n
o
B^ (^r(0); 0; T^) = exp ,r^(0)C^ (0; T^) , A^(0; T^) ; or equivalently,
^ ^ , log B^ (^r(0); 0; T^) = 'r^0(0) (0) C ('(0); '(T )) + A('(0); '(T )): Take ; and so the equilibrium distribution of r(t) seems reasonable. These values determine the functions C; A. Take '0(0) = 1 (we justify this in the next section). For each T^, solve the equation for '(T^):
, log B^ (^r(0); 0; T^) = r^(0)C (0; '(T^)) + A(0; '(T^)):
(*)
The righthand side of this equation is increasing in the '(T^) variable, starting at 0 at time 0 and having limit 1 at 1, i.e.,
r^(0)C (0; 0) + A(0; 0) = 0; lim [^r(0)C (0; T ) + A(0; T )] = 1: T !1
^ (^r(0); 0; T^) < 1; (*) has a unique solution for each T^. For T^ Since 0 , log B ^ is '(0) = 0. If T1 < T^2, then
= 0, this solution
, log B^ (r(0); 0; T^1) < , log B^ (r(0); 0; T^2); so '(T^1) < '(T^2). Thus ' is a strictly increasing timechangefunction with the right properties.
316
31.8 Tracking down '0 (0) in the time change of the CIR model Result for general term structure models:
@ log B(0; T ) , @T
T =0
Justification:
= r(0):
( ZT ) B(0; T ) = IE exp , r(u) du : 0 ( ZT ) , log B(0; T ) = , log IE exp , r(u) du 0 R T , r(u) du
@ log B (0; T ) = IE r(T )eR , @T T IEe, r(u) du @ log B(0; T ) = r(0): , @T 0
0
T =0
In the real time scale associated with the calibration of CIR by time change, we write the bond price as
B^ (^r(0); 0; T^);
thereby indicating explicitly the initial interest rate. The above says that
@ ^ ^ , ^ log B(^r(0); 0; T ) ^ = r^(0): @T T =0
The calibration of CIR by time change requires that we find a strictly increasing function '(0) = 0 such that
, log B^ (^r(0); 0; T^) = '01(0) r^(0)C ('(T^)) + A('(T^)); T^ 0;
' with
^ (^r(0); 0; T^), determined by market data, is strictly increasing in T^, starts at 1 when T^ where B ^ (^r(0); 0; T^) is as shown in Fig. 31.4. and goes to zero as T^!1. Therefore, , log B Consider the function
r^(0)C (T ) + A(T );
Here C (T ) and A(T ) are given by
sinh( T ) ;
cosh( T ) + 12 sinh( T ) 2 3 1 T 2 5; A(T ) = , 22 log 4 cosh( T e ) + 12 sinh( T )
C (T ) =
=
1 2
q
2 + 22 :
(cal)
= 0,
CHAPTER 31. CoxIngersollRoss model
317
6, log B^ (^r(0); 0; T^)
Goes to 1
Strictly increasing
 T^ Figure 31.4: Bond price in CIR model
6 r^(0)C (T ) + A(T )
, log B^ (^r(0); 0; T^)
.... .... ..... ..... ..... ..... ..... ..... .... .... .... . .. ... ... ... .. . .. ... .. .. .. ..
'(T^)
T
Figure 31.5: Calibration The function r^(0)C (T ) + A(T ) is zero at T = 0, is strictly increasing in T , and goes to 1 as T !1. This is because the interest rate is positive in the CIR model (see last paragraph of Section 31.4). To solve (cal), let us first consider the related equation
, log B^ (^r(0); 0; T^) = r^(0)C ('(T^)) + A('(T^)):
(cal’)
Fix T^ and define '(T^) to be the unique T for which (see Fig. 31.5)
, log B^ (^r(0); 0; T^) = r^(0)C (T ) + A(T ) If T^ = 0, then '(T^) = 0. If T^1 < T^2, then '(T^1) < '(T^2). As T^!1, '(T^)!1. We have thus
defined a timechange function ' which has all the right properties, except it satisfies (cal’) rather than (cal).
318 We conclude by showing that '0(0) = 1 so ' also satisfies (cal). From (cal’) we compute
@ ^ ^ r^(0) = , ^ log B (^r(0); 0; T ) ^ @T T =0 = r^(0)C 0('(0))'0(0) + A0('(0))'0(0) = r^(0)C 0(0)'0(0) + A0(0)'0(0):
We show in a moment that C 0 (0) = 1, A0(0) = 0, so we have
r^(0) = r^(0)'0(0): Note that r^(0) is the initial interest rate, observed in the market, and is striclty positive. Dividing by r^(0), we obtain
'0 (0) = 1:
Computation of C 0(0):
1 C 0 ( ) = 2 cosh( ) cosh( ) + 21 sinh( ) 1
cosh( ) + 2 sinh( )
h
, sinh( )
2 sinh( ) + 12 cosh( )
i
C 0(0) = 12 ( + 0) , 0(0 + 12 ) = 1: Computation of A0 (0):
"
cosh( ) + 12 sinh( ) 2
e =2
A0 ( ) = , 2
#
1 =2 cosh( ) + 1 sinh( ) e 2 2
cosh( ) + 12 sinh( ) 2
, e =2 2 sinh( ) + 21 cosh( ) ; 1 ) A0(0) = , 22 + 0 ( +1 0)2 (
+ 0) ,
(0 + 2 " 2 #2 = , 22 12 2 , 12 2 = 0:
Chapter 32
A twofactor model (Duffie & Kan) Let us define:
X1 (t) = Interest rate at time t X2(t) = Yield at time t on a bond maturing at time t + 0 Let X1(0) > 0, X2 (0) differential equations
> 0 be given, and let X1(t) and X2(t) be given by the coupled stochastic
q
dX1(t) = (a11X1(t) + a12X2(t) + b1) dt + 1 1 X1(t) + 2 X2(t) + dW1(t); (SDE1) q q dX2(t) = (a21X1(t) + a22X2(t) + b2) dt + 2 1 X1(t) + 2 X2(t) + ( dW1(t) + 1 , 2 dW2 (t)); (SDE2)
where W1 and W2 are independent Brownian motions. To simplify notation, we define
Y (t) =4 1X1(t) + 2X2 (t) + ; q W3(t) =4 W1(t) + 1 , 2W2 (t): Then W3 is a Brownian motion with
dW1(t) dW3(t) = dt; and
dX1 dX1 = 21Y dt; dX2 dX2 = 22 Y dt; dX1 dX2 = 1 2Y dt: 319
320
32.1 Nonnegativity of Y
dY = 1 dX1 + 2 dX2 = ( 1a11X1 + 1a12 X2 + 1b1) dt + ( 2a21 X1 + 2a22 X2 + 2 b2) dt q p + Y ( 1 1 dW1 + 22 dW1 + 2 1 , 22 dW2) = [( 1a11 + 2 a21)X1 + ( 1a12 + 2a22)X2 ] dt + ( 1 b1 + 2 b2) dt 1q + ( 12 21 + 2 1 21 2 + 22 22 ) 2 Y (t) dW4(t) where
p
2 W4(t) = ( 11 +q 2222 )W1(t) + 2 1 , 2 22 W2 (t) 1 1 + 2 1 212 + 2 2 is a Brownian motion. We shall choose the parameters so that: Assumption 1: For some ,
1 a11 + 2a21 = 1; 1 a12 + 2a22 = 2:
Then
dY = [ 1X1 + 2X2 + ] dt + ( 1b1 + 2b2 , ) dt 1p + ( 12 21 + 2 1 21 2 + 22 22 ) 2 Y dW4 1p = Y dt + ( 1b1 + 2 b2 , ) dt + ( 12 21 + 2 1 2 1 2 + 22 22) 2 Y dW4: From our discussion of the CIR process, we recall that Y will stay strictly positive provided that: Assumption 2: and Assumption 3:
Y (0) = 1X1 (0) + 2X2(0) + > 0;
1b1 + 2b2 , 12 ( 12 21 + 2 1 2 1 2 + 22 22):
Under Assumptions 1,2, and 3,
Y (t) > 0; 0 t < 1;
almost surely,
and (SDE1) and (SDE2) make sense. These can be rewritten as
q
dX1(t) = (a11X1 (t) + a12X2 (t) + b1) dt + 1 Y (t) dW1(t); q dX2(t) = (a21X1 (t) + a22X2 (t) + b2) dt + 2 Y (t) dW3(t):
(SDE1’) (SDE2’)
CHAPTER 32. A twofactor model (Duffie & Kan)
321
32.2 Zerocoupon bond prices The value at time t T of a zerocoupon bond paying $1 at time T is
"
( ZT ) # B(t; T ) = IE exp , X1(u) du F (t) : t
Since the pair (X1; X2) of processes is Markov, this is random only through a dependence on X1(t); X2(t). Since the coefficients in (SDE1) and (SDE2) do not depend on time, the bond price depends on t and T only through their difference = T , t. Thus, there is a function B (x1 ; x2; ) of the dummy variables x1; x2 and , so that
"
( ZT ) # B(X1(t); X2(t); T , t) = IE exp , X1(u) du F (t) : t
The usual tower property argument shows that
Zt exp , X1(u) du B (X1(t); X2(t); T , t) 0
is a martingale. We compute its stochastic differential and set the dt term equal to zero.
Zt d exp , X1(u) du B(X1(t); X2(t); T , t) Z0t = exp , X1(u) du ,X1 B dt + Bx dX1 + Bx dX2 , B dt 1
0
Zt
= exp ,
0
+
1 Bx x 2 1 1
2
dX1 dX1 + Bx1 x2 dX1 dX2 + 21 Bx2 x2
dX2 dX2
,X1B + (a11X1 + a12X2 + b1)Bx + (a21X1 + a22X2 + b2)Bx , B
X1(u) du
1
2
+ 12 21 Y Bx x + 1 2 Y Bx x + 12 22 Y Bx x dt
p
1 1
p
1 2
+ 1 Y Bx dW1 + 2 Y Bx dW3 1
2 2
2
The partial differential equation for B (x1 ; x2; ) is
, x1B , B +(a11x1 + a12x2 + b1)Bx +(a21x1 + a22x2 + b2)Bx + 21 21( 1x1 + 2x2 + )Bx x 1
2
+ 1 2( 1x1 + 2x2 + )Bx x + 21 22 ( 1x1 + 2x2 + )Bx x = 0: 1 2
2 2
1 1
(PDE)
We seek a solution of the form
B(x1 ; x2; ) = exp f,x1 C1 ( ) , x2C2 ( ) , A( )g ; valid for all 0 and all x1 ; x2 satisfying 1x1 + 2 x2 + > 0:
(*)
322 We must have
B(x1 ; x2; 0) = 1; 8x1 ; x2 satisfying (*); because = 0 corresponds to t = T . This implies the initial conditions C1(0) = C2(0) = A(0) = 0: We want to find C1( ); C2( ); A( ) for > 0. We have B (x1; x2; ) = ,x1C10 ( ) , x2C20 ( ) , A0 ( ) B (x1; x2; ); Bx (x1; x2; ) = ,C1( )B (x1; x2; ); Bx (x1; x2; ) = ,C2( )B (x1; x2; ); Bx x (x1; x2; ) = C12( )B(x1 ; x2; ); Bx x (x1; x2; ) = C1( )C2( )B(x1; x2; ); Bx x (x1; x2; ) = C22( )B(x1 ; x2; ):
(IC)
1 2
1 1 1 2 2 2
0 = B (x1 ; x2; ) ,x1 + x1C10 ( ) + x2 C20 ( ) + A0 ( ) , (a11x1 + a12 x2 + b1)C1( ) , (a21x1 + a22x2 + b2)C2( )
(PDE) becomes
+ 12 21 ( 1x1 + 2 x2 + )C12( ) + 1 2( 1x1 + 2x2 + )C1( )C2( ) 2 2 1 + 2 2 ( 1x1 + 2 x2 + )C2 ( )
= x1 B (x1 ; x2; ) , 1 + C10 ( ) , a11C1( ) , a21C2 ( ) +
1 2 1C 2 ( ) + 1 2 1C1( )C2( ) + 1 2 1C 2 ( ) 1 2 2 1 2 2
+ x2 B (x1 ; x2; ) C20 ( ) , a12C1( ) , a22C2 ( ) +
1 2 2C 2 ( ) + 1 2 2C1( )C2( ) + 1 2 2C 2 ( ) 1 2 2 1 2 2
+ B (x1 ; x2; ) A0( ) , b1 C1( ) , b2C2( ) + 12 21 C12( ) + 1 2C1 ( )C2( ) + 21 22C22 ( )
We get three equations:
C10 ( ) = 1 + a11C1 ( ) + a21C2( ) , 12 21 1C12 ( ) , 1 2 1C1 ( )C2( ) , 12 22 1C22 ( ); C1(0) = 0; C20 ( ) = a12C1( ) + a22 C2( ) , 21 21 2C12 ( ) , 12 2C1( )C2( ) , 12 22 2C22 ( ); C2(0) = 0; A0( ) = b1C1( ) + b2C2 ( ) , 21 21C12 ( ) , 1 2 C1( )C2( ) , 12 22 C22( ); A(0) = 0;
(1) (2) (3)
CHAPTER 32. A twofactor model (Duffie & Kan)
323
We first solve (1) and (2) simultaneously numerically, and then integrate (3) to obtain the function A( ).
32.3 Calibration Let 0
> 0 be given. The value at time t of a bond maturing at time t + 0 is B(X1(t); X2(t); 0) = expf,X1 (t)C1(0) , X2 (t)C2(0) , A(0 )g
and the yield is
, 1 log B(X1(t); X2(t); 0) = 1 [X1(t)C1(0) + X2(t)C2(0) + A(0)] : 0
0
But we have set up the model so that X2(t) is the yield at time t of a bond maturing at time t + 0 . Thus
X2(t) = 1 [X1(t)C1(0 ) + X2(t)C2(0) + A(0)] : 0
This equation must hold for every value of X1(t) and X2 (t), which implies that
C1 (0) = 0; C2 (0) = 0; A( ) = 0: We must choose the parameters
a11; a12; b1; a21 ; a22; b2; 1 ; 2; ; 1; ; 2; so that these three equations are satisfied.
324
Chapter 33
Change of num´eraire Consider a Brownian motion driven market model with time horizon T . For now, we will have one asset, which we call a “stock” even though in applications it will usually be an interest rate dependent claim. The price of the stock is modeled by
dS (t) = r(t) S (t) dt + (t)S (t) dW (t); where the interest rate process
(0.1)
r(t) and the volatility process (t) are adapted to some filtration
fF (t); 0 t T g. W is a Brownian motion relative to this filtration, but fF (t); 0 t T g may be larger than the filtration generated by W .
This is not a geometric Brownian motion model. We are particularly interested in the case that the interest rate is stochastic, given by a term structure model we have not yet specified. We shall work only under the riskneutral measure, which is reflected by the fact that the mean rate of return for the stock is r(t). We define the accumulation factor
(t) = exp
Z t 0
r(u) du ;
so that the discounted stock price S ((tt)) is a martingale. Indeed,
d S ((tt)) = S ((tt)) (t) dW (t):
The zerocoupon bond prices are given by
"
( ZT ) # B (t; T ) = IE exp , r(u) du F (t) (t) t = IE (T ) F (t) ; 325
326 so
B(t; T ) = IE 1 F (t) (t) (T )
is also a martingale (tower property). The T forward price F (t; T ) of the stock is the price set at time t for delivery of one share of stock at time T with payment at time T . The value of the forward contract at time t is zero, so
0 = IE (T ) (S (T ) , F (t; T )) F (t) = (t)IE S (T ) Ft , F (t; T )IE (t) F (t) (T ) (T ) (t)
= (t) S (t) , F (t; T )B (t; T ) (t) = S (t) , F (t; T )B (t; T )
Therefore,
F (t; T ) = BS(t;(tT) ) : Definition 33.1 (Num´eraire) Any asset in the model whose price is always strictly positive can be taken as the num´eraire. We then denominate all other assets in units of this num´eraire. Example 33.1 (Money market as num´eraire) The money market could be the num´eraire. At time t, the S (t) B(t;T ) stock is worth (t) units of money market and the T maturity bond is worth (t) units of money market. Example 33.2 (Bond as num´eraire) The T maturity bond could be the num´eraire. At time t T , the stock is worth F(t; T ) units of T maturity bond and the T maturity bond is worth 1 unit.
We will say that a probability measure IPN is riskneutral for the num´eraire N if every asset price, divided by N , is a martingale under IPN . The original probability measure IP is riskneutral for the num´eraire (Example 33.1). Theorem 0.71 Let N be a num´eraire, i.e., the price process for some asset whose price is always strictly positive. Then IPN defined by
IPN (A) = N1(0) is riskneutral for N .
Z N (T ) (T ) dIP; 8A 2 F (T ); A
CHAPTER 33. Change of num´eraire Note:
327
IP and IPN are equivalent, i.e., have the same probability zero sets, and Z IP (A) = N (0) N ((TT )) dIPN ; 8A 2 F (T ): A
Proof: Because N is the price process for some asset, N= is a martingale under IP . Therefore,
Z N (T ) 1 IPN ( ) = N (0) (T ) dIP
1 = N (0) :IE N ((TT)) (0) = N1(0) N (0) = 1;
and we see that IPN is a probability measure.
Let Y be an asset price. Under IP , Y= is a martingale. We must show that under IPN , Y=N is a martingale. For this, we need to recall how to combine conditional expectations with change of measure (Lemma 1.54). If 0 t T T and X is F (T )measurable, then
N (0) (t) N (T ) IEN X F (t) = N (t) IE N (0) (T ) X F (t) N (T ) ( t ) = N (t) IE (T ) X F (t) :
Therefore,
Y (T ) (t) N (T ) Y (T ) IEN N (T ) F (t) = N (t) IE (T ) N (T ) F (t) = N ((tt)) Y ((tt)) Y (t) ; =N (t)
which is the martingale property for Y=N under IPN .
33.1 Bond price as num´eraire Fix T
2 (0; T ] and let B(t; T ) be the num´eraire. The riskneutral measure for this num´eraire is 1 Z B (T; T ) IPT (A) = B(0; T ) =
ZA
1 B(0; T ) A
(T ) dIP 1 dIP 8A 2 F (T ): (T )
328 Because this bond is not defined after time T , we change the measure only “up to time ) using B (01;T ) B ((T;T T ) and only for A 2 F (T ).
IPT
is called the stock is
T forward measure.
Denominated in units of
T ”, i.e.,
T maturity bond, the value of the
F (t; T ) = BS(t;(tT) ) ; 0 t T:
This is a martingale under IPT , and so has a differential of the form
dF (t; T ) = F (t; T )F (t; T ) dWT (t); 0 t T;
(1.1)
i.e., a differential without a dt term. The process fWT ; 0 t T g is a Brownian motion under IPT . We may assume without loss of generality that F (t; T ) 0.
We write F (t) rather than F (t; T ) from now on.
33.2 Stock price as num´eraire Let S (t) be the num´eraire. In terms of this num´eraire, the stock price is identically 1. The riskneutral measure under this num´eraire is
1 IPS (A) = S (0)
Z S (T ) (T ) dIP; 8A 2 F (T ): A
Denominated in shares of stock, the value of the T maturity bond is
B(t; T ) = 1 : S (t) F (t) This is a martingale under IPS , and so has a differential of the form
d F 1(t) = (t; T ) F 1(t) dWS (t); where fWS (t); 0 t T g is a Brownian motion under generality that (t; T ) 0.
IPS .
(2.1)
We may assume without loss of
Theorem 2.72 The volatility (t; T ) in (2.1) is equal to the volatility F (t; T ) in (1.1). In other words, (2.1) can be rewritten as
d F 1(t) = F (t; T ) F 1(t) dWS (t);
(2.1’)
CHAPTER 33. Change of num´eraire Proof: Let g (x) = 1=x, so g 0(x) = ,1=x2 ;
329
g 00 (x) = 2=x3. Then
d F 1(t) = dg (F (t)) = g 0(F (t)) dF (t) + 21 g 00 (F (t)) dF (t) dF (t) = , F 21(t) F (t; T )F (t; T ) dWT (t) + F 31(t) 2F (t; T )F 2(t; T ) dt
h i = F 1(t) ,F (t; T ) dWT (t) + 2F (t; T ) dt = F (t; T ) F 1(t) [,dWT (t) + F (t; T ) dt]:
Under IPT ; ,WT is a Brownian motion. Under this measure, F 1(t) has volatility F (t; T ) and mean rate of return 2F (t; T ). The change of measure from IPT to IPS makes F 1(t) a martingale, i.e., it changes the mean return to zero, but the change of measure does not affect the volatility. Therefore,
(t; T ) in (2.1) must be F (t; T ) and WS must be
WS (t) = ,WT (t) +
Zt 0
F (u; T ) du:
33.3 Merton option pricing formula The price at time zero of a European call is
1 + V (0) = IE (T ) (S (T ) , K ) S (T ) 1 = IE (T ) 1fS (T )>K g , KIE (T ) 1fS(T )>K g Z S (T ) dIP , KB (0; T ) Z 1 = S (0) dIP S (0) ( T ) B (0 ; T ) (T ) fS (T )>K g fS (T )>K g = S (0)IPS fS (T ) > K g , KB (0; T )IPT fS (T ) > K g = S (0)IPS fF (T ) > K g , KB (0; T )IPT fF (T ) > K g = S (0)IPS 1 < 1 , KB (0; T )IPT fF (T ) > K g: F (T ) K
330 This is a completely general formula which permits computation as soon as we specify F (t; T ). If we assume that F (t; T ) is a constant F , we have the following:
1 = B (0; T ) exp n W (T ) , 1 2 T o ; F S 2 F 1 F (1T) S (0) S (0) 2 1 IPS F (T ) < K = IPS F WS (T ) , 2 F T < log KB (0; T ) (0) + 1 pT = IPS WpS (T ) < 1p log KBS(0 ; T) 2 F = N (1);
where
1 =
T
1p log
F T
F T
S (0) + 1 2 T : KB (0; T ) 2 F
Similarly,
o n 1 2 T ; F (T ) = BS(0(0) exp W ( T ) , F T F 2 ; T ) ; T ) IPT fF (T ) > K g = IPT F WT (T ) , 12 2F T > log KBS(0 (0) WT (T ) 1 KB (0 ; T ) 2 1 p p = IPT > log + 2 F T ,WTT (T ) F 1T S (0) S (0) 2 1 p < p log KB(0; T ) , 2 F T = IPT T F T = N (2); where
2 =
1p log S (0) , 1 2 T : KB (0; T ) 2 F F T
If r is constant, then B (0; T ) = e,rT ,
1p log S (0) + (r + 1 2 )T ; 2 F F T K 1 S (0) 2 1 2 = p log K + (r , 2 F )T ;
1 =
F T
and we have the usual BlackScholes formula. When formula
r is not constant, we still have the explicit
V (0) = S (0)N (1) , KB (0; T )N (2):
CHAPTER 33. Change of num´eraire
331
As this formula suggests, if F is constant, then for 0 t T , the value of a European call expiring at time T is
V (t) = S (t)N (1(t)) , KB (t; T )N (2(t));
where
F (t) 2 1 1(t) = log + (T , t) ; F T , t K 2 F 1 F ( t ) 2 1 2(t) = p log K , 2 F (T , t) : T ,t p1
F
This formula also suggests a hedge: at each time t, hold KN (2(t)) bonds.
N (1(t)) shares of stock and short
We want to verify that this hedge is selffinancing. Suppose we begin with $ V (0) and at each time t hold N (1(t)) shares of stock. We short bonds as necessary to finance this. Will the position in the bond always be ,KN (2(t))? If so, the value of the portfolio will always be
S (t)N (1(t)) , KB (t; T )N (2(t)) = V (t);
and we will have a hedge. Mathematically, this question takes the following form. Let
(t) = N (1(t)): At time t, hold (t) shares of stock. If X (t) is the value of the portfolio at time t, then X (t) , (t)S (t) will be invested in the bond, so the number of bonds owned is X B(t)(,t;T() t) S (t) and the portfolio value evolves according to
dX (t) = (t) dS (t) + X (Bt)(t;,T() t) S (t) dB(t; T ):
(3.1)
The value of the option evolves according to
dV (t) = N (1(t)) dS (t) + S (t) dN (1(t)) + dS (t) dN (1(t)) , KN (2(t)) dB(t; T ) , K dB(t; T ) dN (2(t)) , KB(t; T ) dN (2(t)): If X (0) = V (0), will X (t) = V (t) for 0 t T ?
(3.2)
Formulas (3.1) and (3.2) are difficult to compare, so we simplify them by a change of num´eraire. This change is justified by the following theorem. Theorem 3.73 Changes of num´eraire affect portfolio values in the way you would expect. Proof: Suppose we have a model with k assets with prices S1 ; S2; : : : ; Sk . At each time t, hold i (t) shares of asset i, i = 1; 2; : : : ; k , 1, and invest the remaining wealth in asset k. Begin with a nonrandom initial wealth X (0), and let X (t) be the value of the portfolio at time t. The number of shares of asset k held at time t is
,1 i (t)Si (t) X (t) , Pki=1 k (t) = ; Sk (t)
332 and X evolves according to the equation
dX = =
kX ,1 i=1 k X i=1
i dSi + X ,
kX ,1 i=1
!
k i Si dS S k
i dSi:
Note that
Xk (t) =
k X
i (t)Si (t);
i=1 and we only get to specify 1; : : : ; k,1, not k , in advance. Let N be a num´eraire, and define
X (t) ; Sc(t) = Si (t) ; i = 1; 2; : : : ; k: Xb (t) = N i (t) N (t) Then
dXb = N1 dX + X d N1 + dX d N1 !
X k k k X X = N1 i dSi + i Si d N1 + i dSi d N1 i=1 i=1 i=1 k X = 1 dS + S d 1 + dS d 1 =
Now
i=1 k X i=1
i
N
i
i
N
i
i dc Si:
,1 i Si X , Pki=1 k = Sk ,1 i Si =N X=N , Pki=1 = Sk =N ,1 i c bX , Pki=1 Si : = c Sk Therefore,
! c k kX ,1 X Sk dXb = i dc Si + Xb , iSci dc i=1
i=1
Sk
N
CHAPTER 33. Change of num´eraire
333
This is the formula for the evolution of a portfolio which holds i shares of asset i, i = 1; 2; : : : ; k , 1, and all assets and the portfolio are denominated in units of N . We return to the European call hedging problem (comparison of (3.1) and (3.2)), but we now use the zerocoupon bond as num´eraire. We still hold (t) = N (1(t)) shares of stock at each time t. In terms of the new num´eraire, the asset values are Stock: Bond:
S (t) = F (t); B(t; T ) B(t; T ) = 1: B(t; T )
The portfolio value evolves according to
dXb (t) = (t) dF (t) + (Xb (t) , (t)) d(1) 1 = (t) dF (t):
(3.1’)
In the new num´eraire, the option value formula
V (t) = N (1(t))S (t) , KB (t; T )N (2(t)) becomes
Vb (t) = BV(t;(tT) ) = N (1(t))F (t) , KN (2(t));
and
dVb = N (1(t)) dF (t) + F (t) dN (1(t)) + dN (1(t)) dF (t) , K dN (2(t)): To show that the hedge works, we must show that
F (t) dN (1(t)) + dN (1(t)) dF (t) , K dN (2(t)) = 0: This is a homework problem.
(3.2’)
334
Chapter 34
BraceGatarekMusiela model 34.1 Review of HJM under riskneutral IP
f (t; T ) = Forward rate at time t for borrowing at time T: df (t; T ) = (t; T )(t; T ) dt + (t; T ) dW (t); where
(t; T ) =
ZT t
(t; u) du
The interest rate is r(t) = f (t; t). The bond prices
"
( ZT ) # B (t; T ) = IE exp , r(u) du F (t) ( ZT t ) = exp , f (t; u) du t
satisfy
dB (t; T ) = r(t) B(t; T ) dt ,
 ({zt; T })
volatility of T maturity bond.
B(t; T ) dW (t):
To implement HJM, you specify a function
(t; T ); 0 t T: A simple choice we would like to use is where
(t; T ) = f (t; T )
> 0 is the constant “volatility of the forward rate”. This is not possible because it leads to (t; T ) =
ZT t
f (t; u) du;
df (t; T ) = 2 f (t; T )
ZT t
!
f (t; u) du dt + f (t; T ) dW (t); 335
336 and Heath, Jarrow and Morton show that solutions to this equation explode before
T.
The problem with the above equation is that the dt term grows like the square of the forward rate. To see what problem this causes, consider the similar deterministic ordinary differential equation
f 0 (t) = f 2(t); where f (0) = c > 0. We have
f 0(t) = 1; f 2(t) , dtd f (1t) = 1;
Zt 1 1 , f (t) + f (0) = 1 du = t 0 1 1 = t , 1=c = ct , 1 ; , f (t) = t , f (0) c c f (t) = 1 , ct :
This solution explodes at t
= 1=c.
34.2 BraceGatarekMusiela model New variables: Current time t
Time to maturity
= T , t:
Forward rates:
r(t; ) = f (t; t + ); r(t; 0) = f (t; t) = r(t); @ r(t; ) = @ f (t; t + ) @ @T
(2.1) (2.2)
Bond prices:
D(t; ) = B(t; t + )
Z t+ = exp , f (t; v ) dv Zt (u = v , t; du = dv ) : = exp , f (t; t + u) du Z0 = exp , r(t; u) du
(2.3)
0
@ D(t; ) = @ B(t; t + ) = ,r(t; )D(t; ): @ @T
(2.4)
CHAPTER 34. BraceGatarekMusiela model We will now write (t; ) = (t; T
337
, t) rather than (t; T ). In this notation, the HJM model is
df (t; T ) = (t; ) (t; ) dt + (t; ) dW (t); dB (t; T ) = r(t)B(t; T ) dt , (t; )B (t; T ) dW (t);
(2.5) (2.6)
where
(t; ) =
Z 0
(t; u) du;
(2.7)
@ (t; ) = (t; ): @
(2.8)
We now derive the differentials of r(t; ) and D(t; ), analogous to (2.5) and (2.6) We have
dr(t; ) =
df  (t;{zt + )}
differential applies only to first argument
@ f (t; t + ) dt + @T
@ r(t; ) dt = (t; ) (t; ) dt + (t; ) dW (t) + @
(2.5),(2.2)
@ hr(t; ) + 1 ( (t; ))2i dt + (t; ) dW (t): = @ 2
(2.8)
(2.9)
Also,
dD(t; ) =
dB  (t;{zt + )}
differential applies only to first argument
@ B(t; t + ) dt + @T
= r(t) B (t; t + ) dt , (t; )B (t; t + ) dW (t) , r(t; )D(t; ) dt (2.1) = [r(t; 0) , r(t; )] D(t; ) dt , (t; )D(t; ) dW (t): (2.10) (2.6),(2.4)
34.3 LIBOR Fix > 0 (say, = 14 year). $ D(t; ) invested at time t in a (t + )maturity bond grows to $ 1 at time t + . L(t; 0) is defined to be the corresponding rate of simple interest:
D(t; )(1 + L(t; 0)) = 1;
(
)
Z@ 1 + L(t; 0) = D(1t; ) = exp r(t; u) du ; 0 nR o exp 0@ r(t; u) du , 1 L(t; 0) = :
338
34.4 Forward LIBOR
> 0 is still fixed. At time t, agree to invest $ DD(t;(t;+)) at time t + , with payback of $1 at time t + + . Can do this at time t by shorting DD(t;(t;+)) bonds maturing at time t + and going long one bond maturing at time t + + . The value of this portfolio at time t is (t; + ) D(t; ) + D(t; + ) = 0: , DD (t; ) The forward LIBOR L(t; ) is defined to be the simple (forward) interest rate for this investment: D(t; + ) (1 + L(t; )) = 1; D(t; ) R r(t; u) dug D ( t; ) exp f, n R 0+ o 1 + L(t; ) = D(t; + ) = exp , 0 r(t; u) du = exp
L(t; ) =
(Z +
)
r(t; u) du ;
nR + o exp r(t; u) du , 1
:
(4.1)
Connection with forward rates:
( ) ( ) @ exp Z + r(t; u) du = r(t; + ) exp Z + r(t; u) du =0 =0 @ = r(t; ); so
nR
o
exp + r(t; u) du , 1 f (t; t + ) = r(t; ) = lim
L(t; ) =
nR +#0
exp
o
r(t; u) du , 1 ; >0
fixed: (4.2)
r(t; ) is the continuously compounded rate. L(t; ) is the simple rate over a period of duration . We cannot have a lognormal model for r(t; ) because solutions explode as we saw in Section 34.1. For fixed positive , we can have a lognormal model for L(t; ). 34.5 The dynamics of L(t; )
t 0; 0, appearing in (2.5) so that dL(t; ) = (: : : ) dt + L(t; ) (t; ) dW (t)
We want to choose (t; );
CHAPTER 34. BraceGatarekMusiela model
339
for some (t; ); t 0; 0. This is the BGM model, and is a subclass of HJM models, corresponding to particular choices of (t; ). Recall (2.9):
@ hr(t; u) + 1 ( (t; u))2i dt + (t; u) dW (t): dr(t; ) = @u 2 Therefore,
d
Z +
! Z +
r(t; u) du =
dr(t; u) du
Z + @ h Z + i r (t; u) + 12 ( (t; u))2 du dt + (t; u) du dW (t) h @u i = r(t; + ) , r(t; ) + 21 ( (t; + ))2 , 12 ( (t; ))2 dt + [ (t; + ) , (t; )] dW (t)
(5.1)
=
and
o 3 2 nR + exp r ( t; u ) du , 15 dL(t; ) (4=:1) d 4 (Z + ) Z + 1 = exp
r(t; u) du d
(
)
Z + 1 + 2 exp r(t; u) du (4.1), (5.1) 1 = [1 + L(t; )]
r(t; u) du d
Z +
!2
r(t; u) du
[r(t; + ) , r(t; ) + 21 ((t; + ))2 , 12 ((t; ))2] dt + [ (t; + ) , (t; )] dW (t) + 12 [ (t; + ) , (t; )]2 dt = 1 [1 + L(t; )] [r(t; + ) , r(t; )] dt + (t; + )[ (t; + ) , (t; )] dt = +[ (t; + ) , (t; )] dW (t) :
(5.2)
340 But
o 3 2 nR + exp r ( t; u ) du , 15 @ L(t; ) = @ 4 @ @ (Z + ) = exp
r(t; u) du :[r(t; + ) , r(t; )]
= 1 [1 + L(t; )][r(t; + ) , r(t; )]: Therefore,
@ L(t; ) dt + 1 [1 + L(t; )][ (t; + ) , (t; )]:[ (t; + ) dt + dW (t)]: dL(t; ) = @ Take (t; ) to be given by
(t; )L(t; ) = 1 [1 + L(t; )][ (t; + ) , (t; )]:
(5.3)
Then
@ L(t; ) + (t; )L(t; )(t; + )] dt + (t; )L(t; ) dW (t): dL(t; ) = [ @
(5.4)
Note that (5.3) is equivalent to
) (t; ) : (t; + ) = (t; ) + L1 (+t;L (t; ) Plugging this into (5.4) yields
"
(5.3’)
#
@ L(t; ) + (t; )L(t; )(t; ) + L2(t; ) 2(t; ) dt dL(t; ) = @ 1 + L(t; ) + (t; )L(t; ) dW (t): 34.6 Implementation of BGM Obtain the initial forward LIBOR curve
L(0; ); 0; from market data. Choose a forward LIBOR volatility function (usually nonrandom)
(t; ); t 0; 0:
(5.4’)
CHAPTER 34. BraceGatarekMusiela model
341
Because LIBOR gives no rate information on time periods smaller than , we must also choose a partial bond volatility function
(t; ); t 0; 0 < for maturities less than from the current time variable t.
2 [0; ) solve (5.4’) to obtain L(t; ); t 0; 0 < : Plugging the solution into (5.3’), we obtain (t; ) for < 2 . We then solve (5.4’) to obtain L(t; ); t 0; < 2;
With these functions, we can for each
and we continue recursively. Remark 34.1 BGM is a special case of HJM with HJM’s (t; ) generated recursively by (5.3’). In BGM, (t; ) is usually taken to be nonrandom; the resulting (t; ) is random. Remark 34.2 (5.4) (equivalently, (5.4’)) is a stochastic partial differential equation because of the @ @ L(t; ) term. This is not as terrible as it first appears. Returning to the HJM variables t and T , set
K (t; T ) = L(t; T , t):
Then
@ L(t; T , t) dt dK (t; T ) = dL(t; T , t) , @
and (5.4) and (5.4’) become
dK (t; T ) = (t; T , t)K (t; T ) [ (t; T , t + ) dt + dW (t)] K ( t; T )
( t; T , t ) = (t; T , t)K (t; T ) (t; T , t) dt + dt + dW (t) : 1 + K (t; T ) Remark 34.3 From (5.3) we have
If we let #0, then
(t; )L(t; ) = [1 + L(t; )] (t; + ) , (t; ) :
@
(t; )L(t; )!@ (t; + ) = (t; ); =0
and so
(t; T , t)K (t; T )! (t; T , t): We saw before (eq. 4.2) that as #0, L(t; )!r(t; ) = f (t; t + );
(6.1)
342 so
K (t; T )!f (t; T ):
Therefore, the limit as #0 of (6.1) is given by equation (2.5):
df (t; T ) = (t; T , t) [ (t; T , t) dt + dW (t)] : Remark 34.4 Although the dt term in (6.1) has the term to this equation do not explode because
2 (t;T ,t)K 2 (t;T ) involving K 2 , solutions 1+K (t;T )
2(t; T , t)K 2 (t; T ) 2(t; T , t)K 2(t; T ) 1 + K (t; T ) K (t; T ) 2 (t; T , t)K (t; T ): 34.7 Bond prices Let (t) = exp
nR t
o
0 r(u) du : From (2.6) we have
d
B(t; T ) (t)
= 1(t) [,r(t)B (t; T ) dt + dB (t; T )] = , B (t; T ) (t; T , t) dW (t): (t)
) The solution B (t;T (t) to this stochastic differential equation is given by
B (t; T ) = exp , Z t (u; T , u) dW (u) , 1 Z t( (u; T , u))2 du : 2 0 (t)B (0; T ) 0
This is a martingale, and we can use it to switch to the forward measure
Z 1 1 IPT (A) = B (0; T ) (T ) dIP Z B(T;AT ) = (T )B (0; T ) dIP 8A 2 F (T ): A
Girsanov’s Theorem implies that
WT (t) = W (t) + is a Brownian motion under IPT .
Zt 0
(u; T , u) du; 0 t T;
CHAPTER 34. BraceGatarekMusiela model
343
34.8 Forward LIBOR under more forward measure From (6.1) we have
dK (t; T ) = (t; T , t)K (t; T ) [ (t; T , t + ) dt + dW (t)] = (t; T , t)K (t; T ) dWT + (t); so
K (t; T ) = K (0; T ) exp and
K (T; T ) = K (0; T ) exp = K (t; T ) exp
Z t 0
(u; T , u) dWT + (u) ,
(Z T 0
(Z T t
(u; T , u) dWT + (u) ,
is normal with variance
ZT t
(u; T , u) dWT +(u) , 12 2(t) =
and mean , 12 2(t).
ZT t
ZT t
ZT t
2(u; T , u) du
ZT
1 2 0
(u; T , u) dWT + (u) , 12
We assume that is nonrandom. Then
X (t) =
Zt
1 2 0
)
2(u; T , u) du
)
(8.1)
2(u; T , u) du :
2(u; T , u) du
(8.2)
2(u; T , u) du
34.9 Pricing an interest rate caplet Consider a floating rate interest payment settled in arrears. At time T + , the floating rate interest payment due is L(T; 0) = K (T; T ); the LIBOR at time T . A caplet protects its owner by requiring him to pay only the cap c if K (T; T ) > c. Thus, the value of the caplet at time T + is (K (T; T ) , c)+ . We determine its value at times 0 t T + .
Case I: T
t T + .
CT + (t) = IE (T (+t) ) (K (T; T ) , c)+ F (t) (t) + = (K (T; T ) , c) IE (T + ) F (t) = (K (T; T ) , c)+ B (t; T + ):
(9.1)
344 Case II: 0 t T . Recall that
IPT + (A) =
Z
A
Z (T + ) dIP; 8A 2 F (T + );
where
) : Z (t) = (Bt)(Bt;(0T; + T + ) We have
(t) + CT + (t) = IE (T + ) (K (T; T ) , c) F (t) 2 3 66 B(T + ; T + ) 77 ( t ) B (0 ; T + ) + 6 = B (t; T + ) B (t; T + ) IE 6 (T + )B (0; T + ) (K (T; T ) , c) F (t)77 5  {z } 4 {z } Z (T +) Z t = B (t; T + )IET + (K (T; T ) , c)+ F (t) 1 ( )
From (8.1) and (8.2) we have
K (T; T ) = K (t; T ) expfX (t)g; R where X (t) is normal under IPT + with variance 2(t) = tT 2(u; T , u) du and mean , 21 2(t). Furthermore, X (t) is independent of F (t). CT + (t) = B(t; T + )IET + Set
h
(K (t; T ) expfX (t)g , c)+ F (t)
:
i
g(y) = IET + (y expfX (t)g , c)+ = y N 1 log y + 12 (t) , c N 1 log y , 12 (t) : (t) c (t) c Then
CT + (t) = B(t; T + ) g (K (t; T )); 0 t T , : In the case of constant , we have p (t) = T , t; and (9.2) is called the Black caplet formula.
(9.2)
CHAPTER 34. BraceGatarekMusiela model
345
34.10 Pricing an interest rate cap Let
T0 = 0; T1 = ; T2 = 2; : : : ; Tn = n:
A cap is a series of payments
(K (Tk; Tk) , c)+
k = 0; 1; : : : ; n , 1:
at time Tk+1 ;
The value at time t of the cap is the value of all remaining caplets, i.e.,
C (t) =
X
k:tTk
CTk (t):
34.11 Calibration of BGM The interest rate caplet c on L(0; T ) at time T
+ has timezero value
CT + (0) = B (0; T + ) g (K (0; T )); where g (defined in the last section) depends on
ZT 0
2(u; T , u) du:
Let us suppose is a deterministic function of its second argument, i.e.,
(t; ) = ( ): Then g depends on
ZT 0
2(T , u) du =
ZT
2(v ) dv:
0
R If we know the caplet price CT + (0), we can “back out” the squared volatility 0T 2(v ) dv . If we know caplet prices CT + (0); CT + (0); : : : ; CTn+ (0); 0
where T0
ZT
0
0
1
< T1 < : : : < Tn , we can “back out”
2(v) dv;
ZT
1
T0
2(v) dv =
ZT
1
0
2(v ) dv ,
ZT
0
0
2(v ) dv; ::: ;
In this case, we may assume that is constant on each of the intervals
(0; T0); (T0; T1); : : : ; (Tn,1 ; Tn);
Z Tn Tn,1
2(v ) dv:
(11.1)
346 and choose these constants to make the above integrals have the values implied by the caplet prices.
R If we know caplet prices CT + (0) for all T 0, we can “back out” 0T 2(v ) dv and then differenp tiate to discover 2( ) and ( ) = 2( ) for all 0. To implement BGM, we need both ( );
0, and
(t; ); t 0; 0 < : Now (t; ) is the volatility at time t of a zero coupon bond maturing at time t + (see (2.6)). Since is small (say 14 year), and 0 < , it is reasonable to set (t; ) = 0; t 0; 0 < : We can now solve (or simulate) to get
L(t; ); t 0; 0; or equivalently,
K (t; T ); t 0; T 0;
using the recursive procedure outlined at the start of Section 34.6.
34.12 Long rates The long rate is determined by long maturity bond prices. Let n be a large fixed positive integer, so that n is 20 or 30 years. Then
1
D(t; n) = exp = =
n Y
(Z n 0
exp
k=1 n Y
)
r(t; u) du
(Z k
(k,1)
)
r(t; u) du
[1 + L(t; (k , 1) )];
k=1
where the last equality follows from (4.1). The long rate is
n 1 log 1 = 1 X n D(t; n ) n k=1 log[1 + L(t; (k , 1))]:
34.13 Pricing a swap Let T0
0 be given, and set T1 = T0 + ; T2 = T0 + 2; : : : ; Tn = T0 + n:
CHAPTER 34. BraceGatarekMusiela model The swap is the series of payments
(L(Tk; 0) , c)
For 0 t T0, the value of the swap is
347
at time Tk+1 ; k = 0; 1; : : : ; n , 1:
nX ,1
IE ( T(t) ) (L(Tk ; 0) , c) F (t) : k+1 k=0
Now
1 + L(Tk ; 0) = B (T ;1T k
so
k+1)
;
1 1 L(Tk; 0) = B(T ; T ) , 1 : k k+1
We compute
IE ( T(t) ) (L(Tk; 0) , c) F (t) k+1 1 ( t ) = IE (T ) B (T ; T ) , 1 , c F (t)
k k+1 2 k+1 3 66 (Tk ) 77 ( t ) 6 = IE 6 (T )B (T ; T ) IE (T ) F (Tk ) F (t)77 , (1 + c)B (t; Tk+1) k+1 4 k k k+1  5 {z } B (Tk ;Tk ) (t) = IE (T ) F (t) , (1 + c)B (t; Tk+1) k+1 = B (t; Tk ) , (1 + c)B (t; Tk+1): +1
The value of the swap at time t is
nX ,1
(t) IE (T ) (L(Tk ; 0) , c) F (t)
k=0 nX ,1
=
k=0
k+1
[B (t; Tk ) , (1 + c)B (t; Tk+1)]
= B (t; T0) , (1 + c)B (t; T1) + B (t; T1) , (1 + c)B (t; T2) + : : : + B (t; Tn,1 ) , (1 + c)B (t; Tn ) = B (t; T0) , cB (t; T1) , cB (t; T2) , : : : , cB (t; Tn ) , B (t; Tn ): The forward swap rate wT (t) at time t for maturity T0 is the value of c which makes the timet 0
value of the swap equal to zero:
wT (t) = [B (Bt;(Tt; T) 0+) ,: : B: +(t;BT(nt;)T )] : 0
1
n
In contrast to the cap formula, which depends on the term structure model and requires estimation of , the swap formula is generic.