Sciences Po − Fall 2017 St´ ephane Guibaud

Financial Markets

Practice Problems on Derivatives: Suggested Solutions 1. (a) The aluminium producer is at risk if the price of aluminium falls. Rio Tinto’s CFO may consider selling forward aluminium or entering a short futures position. (b) When entering the contract, Rio Tinto’s CFO understood the possibility that, in case the spot price rises to $1,900 per ton, the company would not benefit from the price increase, having locked in a lower selling price of $1,615. But equally, had the spot price fallen to $1,350 per ton, the company would have benefited from the hedge. If the company ends up losing the upside, it does not mean that hedging was a bad decision ex ante. 2. (a) We use the forward pricing formula for a non-dividend paying underlying F = (1 + r)S0 = 1.05 × 50 = $52.5. (b) The forward price is too high. To take advantage of the mispricing, we take a short position in the forward contract, with payoff at expiration 55 − ST . We buy the underlying at cost 50 and hold it until maturity. We also borrow 50 with repayment 52.5 in one year. Today, we have zero net cashflow, and the total value of our combined position at maturity is (55 − ST ) + ST − 52.5 = $2.5. Alternatively, we can borrow 52.4 today, with repayment 55 in one year. Today we reap a profit of $2.4 (i.e., the present value of $2.5) and we break even at maturity. (c) Say we take a short forward position today at the fair forward price 52.5$, which we can do at no cost. At maturity, we will make a sure profit of 52.5 − 47.46 = $5.04. Discounting back to the present at the one-year risk free rate gives a value of 5.04/1.05 = $4.8. This is the marked-to-market value of our existing long position! 3. (a) The forward price should be lower. Namely, we now have F = (1 + r)(S0 − I), where I denotes the present value of the dividend received in six months. If the forward price remained equal to $52.5, there would be an arbitrage opportunity. Today, we would take a short forward position and borrow $50 to buy the underlying and hold it for a year (see question 2b). We would have zero net cash flow today and at maturity, but we would receive the dividend on the stock we own after 6 months—which would constitute an arbitrage. 1

(b) This is irrelevant because the dividend is paid after the maturity of the forward contract under consideration. (c) If one incurs a transaction cost of 50 cents per share when buying the stock, then the forward price should be higher. Namely, we now have F = (1 + r)(S0 + τ ), where τ denotes the transaction cost. For τ = 0.5, we obtain F = $53.0. In the presence of the transaction cost, a forward price of $53.0 does not create an arbitrage opportunity (contrary to what would happen under the frictionless assumption of question 2). This is because the effective cost of buying the underlying stock is S0 + τ . (d) In this case we have F ≤ (1 + r)S0 . Indeed, when F < (1 + r)S0 , constructing an arbitrage strategy would require to buy forward, sell the underlying and buy a ZC bond with the proceeds. This would tend to push the forward price up and the spot price down, bringing them back “in line” with the theoretical formula F = (1+r)S0 . However, if arbitrageurs are unable to implement the strategy because they do not own the underlying and cannot find someone to lend it to them temporarily for short sale, the forward price can remain below (1 + r)S0 . 4. (a) Payoff diagram: Call − Put

0 K

ST

−K Mathematically, max(0, ST − K) − max(0, K − ST ) = ST − K for all ST . (b) By selling one put and buying one call at strike price K = F , we obtain the same payoff as from a long forward position. Since we can get the latter at not cost, absence of arbitrage requires C − P = 0. Therefore P = C = $8.25. Using put-call parity would give the same result. 2

5. (a) At expiration, the stock price ST will be either 112 or 94. The put payoff is max [0, X − ST ], hence the put payoff will be either 0 or 9. In order to replicate the put payoff, we need to buy ∆ shares of the stock and n units of the bond such that the value of the portfolio matches the put payoff at expiration: ‘up’ state: ‘down’ state:

∆ × 112 + n × 100 ∆ × 94 + n × 100

= =

0 (1) 9 (2)

Equation (1) minus equation (2) implies 18∆ = −9 ⇒ ∆ = −1/2. Substituting ∆ = −1/2 into equation (1) gives −56 + 100n = 0 ⇒ n = 0.56. Hence by short selling 1/2 share of stock and buying 0.56 units of ZC bond, we can perfectly replicate the put payoff. In order to prevent arbitrage, the price of the put must coincide with the cost of replicating its payoff. Therefore, the fair put price is P0 = 0.56 × 97 − 0.5 × 100 = $4.32. (b) The payoff of the put is either Pu = 0 or Pd = 9, depending on which state (‘up’ or ‘down’) realizes. The rate of return on the put in state s ∈ {u, d} is equal to Rs = (Ps − P0 )/P0 . If the stock price goes up, the rate of return is −100%. If the stock price goes down, the rate of return is (9 − 4.32)/4.32 = 108.3%. Given that both states are equally likely, the expected rate of return on the put is 0.5 ∗ (−1) + 0.5 ∗ 1.083 = 0.0415 = 4.15%. 6. Parameters: u = 1.2, d = 0.8. Stock tree: u2 S=72 uS=60 



 H   HH S=50 udS=48 HH  H  HH   H  HH H HH dS=40 HH HH

Period 0

Period 1

d2 S=32 Period 2

3

(a) Call payoffs: 27 ?

 

  H  HH  ? 3  HH  H  HH    H  HH H HH H ? HH H

Period 0

Period 1

0 Period 2

First compute risk-neutral probability: q=

1 + 0.1 − 0.8 = 0.75. 1.2 − 0.8

Then go backward and use the risk-neutral valuation formula (as in Lecture 11) to compute Cu = 19.09, Cd = 2.05 and C = 13.48.

4

(b) Put payoffs: 0     H   HH ?  H0  HH  H   HH   H HH H HH ? H HH

?

Period 0

Period 1

13 Period 2

Again, using the risk-neutral valuation, we find that Pu =0, Pd = 2.95, P = 0.67. (c) Put-call parity: C = S + P − X dT . Period 0: Period 1, u-node: Period 1, d-node:

√ 1 2 (1 + 10%) √ 1 19.09 = 60 + 0 − 45 (1 + 10%) √ 1 2.05 = 40 + 2.95 − 45 (1 + 10%)

13.48 = 50 + 0.67 − 45

(d) See Lecture 11: Replication Method in Two-Step Binomial Tree. At u-node: Since Puu = Pud = 0, the option is worth 0. The replicating portfolio at this node consists of zero units of stock and zero units of Bond. At d-node: we need to hold ∆d units of stocks and Bd units of bond, so that 48∆d + 100Bd = 0 32∆d + 100Bd = 13

Solving this system, we have ∆d = −0.8125, Bd = 0.39. The cost of the replicating portfolio at node d is −0.8125 × 40 + 0.39

5

100 = 2.95. 1 + 10%

At initial node: we need to hold ∆0 units of stocks and B0 units of bond, so that 60∆0 + 100B0 = 0 40∆0 + 100B0 = 2.95

Solving this system, we have ∆0 = −0.1477, B0 = 0.0885. This characterizes the dynamic (self-financing) replicating strategy for the European put. The initial cost of entering the dynamic replicating strategy is −0.1477 × 50 + 0.0885

6

100 = 0.67. 1 + 10%