RBC model: 

max Et   t  ln Ct   ln Lt  t 0

ut ( Ct , Lt )

Subject to:

Yt  Ct  It

Nt  Lt  1 Kt 1  It  (1   ) Kt Yt  At Kt Nt1

ln At 1   ln At   t 1

Step 1: derive for the equilibrium conditions by Lagrangian

Yt 1     1  L  E0   ln Ct   ln(1  Nt )  t [Ct  Kt 1  (1   ) Kt  At Kt Nt ] t 0     

t

Take derivatives with respect to Ct , Nt , Kt 1 , and t :

Ct :

L 1 1 0  t  t  Ct Ct Ct Yt

L  1 0  t At K t N t1 Nt : Nt 1  Nt Nt Wt Yt 1

L   0  t   Et [t 1 (1    At 1 Kt1 Nt11 )] Kt 1 : Kt 1 Kt 1 Rt 1 Yt 1

L  0  Ct  Kt 1  (1   ) K t  At K t N t1  0 t : t

Step 1’: derive for the equilibrium conditions by Bellman equation The problem can be restated in the recursive form by Bellman equation.

V (kt )  max[u(Ct , Lt )   Et [V (kt 1 )]] Substitute Ct and Lt by constraints, and the resulting Bellman equation will have only two endogenous control variables Nt (intratemporal) and Kt 1 (intertemporal), but two state variables At (exogenous state variable) and Kt (endogenous state variable). Take F.O.C. of Bellman equation, and apply envelope theorem to get the same sets of equilibrium conditions. The equations can be restated by defining several intermediate variables like Wt , Rt , and Yt . This revision does not change the problem, but it makes future steps easier. The whole dynamic stochastic system can be characterized by both the equilibrium conditions (E1-E6) as well as the law of motion for the state variable At (E7):

(E1)





1  Nt

Wt Ct

Yt Nt

(E2) Wt  (1   )

(E3) Yt  At Kt Nt1 (E4) 1  Et [ 

Ct Rt 1 ] Ct 1

Yt Kt

(E5) Rt  1    

(E6) Ct  Kt 1  (1   ) Kt  Yt  0 (E7) ln At 1   ln At   t 1 Step 2: obtain the steady state The steady state of the system can be obtained by dropping the time subscripts, because variables will not change in steady state. (S1)

 1 N



W C

(S2) W  (1   )

Y N

(S3) Y  K  N 1 (S4) 1   R (S5) R  1    

Y K

(S6) C   K  Y  0 (S7) A  1

Step 3: linearize the system (1) Direct Approach: use Taylor expansion directly (1.1) Any univariate nonlinear equation f ( X t )  0 : In steady state, the equation becomes f ( X )  0 . Take first order Taylor expansion around the steady state:

f  X t   f  X   f   X  X t  X   f   X  X

(1.2) Any multivariate nonlinear equation f ( X1t , In steady state, the equation becomes f ( X1 ,

Xt  X  f   X  Xxˆt X

, X nt )  0 :

, X n )  0 , and so:

n

f  X 1t ,..., X nt     f X i ( X i )  X i  xˆit   0 i 1

(2) Indirect Approach: approximation rules implied by Taylor expansion

ln(1  x)  x  e x  1  x (basic result from Taylor expansion)  Xt  X 

Xt  X  X  X  (1  xˆt ) X

 X t  X  e  xˆt  X  (1   xˆt )  X t Yt   X  Y  e xˆt e  yˆt  X  Y  e xˆt   yˆt  X  Y  (1   xˆt   yˆt )

Linearize the Euler’s equations (S1-S6) as well as the law of motion for the exogenous state variable around the steady state. (L1) 0  (1  N )Wwˆ t  WNnˆt   Ccˆt

ˆt (L2) 0  yˆt  nˆt  w (L3) 0   yˆt  aˆt   kˆt  1    nˆt

(L4)  Et  cˆt 1   Et [rˆt 1 ]  cˆt





Y yˆt  kˆt  rˆt K K C K (L6)  kˆt 1  cˆt  (1   ) kˆt  yˆt Y Y Y (L5) 0  

(L7) aˆt 1   aˆt   t 1 Step 4: restate the linear system in matrix form Define the vector of variables in period t for this system:

 xt  exogenous/endogenous state variables  y    , endogenous control variables   t 

 zt   shocks

Note that expectation is needed for controls if we bring the vector to period t  1 :

 xt 1  exogenous/endogenous state variables  E y    , endogenous control variables   t t 1  

 zt 1   shocks 

The state vector xt has a dimension of N  Nexogenous  Nendogenous , the control vector yt has a dimension of M . In our case, N  1  1  2 and M  5 , and the vector zt has a dimension of 1. The 7 linear equations can then be rewritten in the matrix form:

 0  0   0  0  0  0  0

0 0 K 0 Y 0 0 0 1 0 0 0 0 0 0 A

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

1  0  0  1  0 0 0

  ˆt 1   a    ˆ   kt 1     E cˆ  t t  1   E n ˆ   t t 1    ˆt 1   Et y   E w ˆ   t t 1    Et rˆt 1   



   0    0   0  1 0   0 

0 1  K Y Y  K 0



0 0

0 C Y

0

0

0

0

1

0

0

0

Y

0

1 0 0

0 1 1

 C WN B

Hence, it can be written as:

K 0 1 1 0

0 0 1   1 N W  

0  

 0    1  0  0  0   0 

 ˆt  a    kˆ   t    cˆt    ˆt  n   y ˆ   t w ˆ   t    rˆt 



  1   0  0    0    0    0    0 C

     t 1  zt 1

 x  x  A  t 1   B  t   Czt 1  Et yt 1   yt  If the matrix A is nonsingular, we can multiply A1 on both sides directly. If the matrix A is singular, we can apply Uhlig (1997). The system becomes: x  xt 1  1  t  1  A B E y   y   A C zt 1 F  t  G  t t 1  x   F  t   Gzt 1  yt 

Apply Jordan decomposition to the matrix F :

1 F  P P  P  0  N  M  N  M   0 1

0

1

0

0  0  P N  M 

Note that the diagonal elements of  are the eigenvalues of F , while the column vectors of P 1 are the corresponding eigenvectors, based on:

det  F   I   0 Denote the number of unstable eigenvalues (   1 ) as u , and the number of stable eigenvalues (   1 ) as s . Note that u  s  M  N . Blanchard-Kahn condition is: (i)

If u  M , there is unique solution to the system (saddle path);

(ii)

If u  M , there is no solution to the system;

(iii)

If u  M , there are infinite solutions to the system.

We only focus on the saddle path case. Models that do not satisfy this condition cannot be solved. Hence, check this condition always! Now we can rewrite  in two blocks, with a stable block and an unstable block:

  s 0

0  u 

Similarly, we can partition P and G similarly:

P P   11  P21

P12   Gs  , G   P22  Gu 

The original system becomes:

 xt 1   P11 P12  E y   P P   t t 1   21 22   P11 P12   xt 1    s P P  E y    0  21 22   t t 1  

1

  s 0   P11 P12   xt   Gs   0    P P   y   G  zt 1 u   21 22   t    u 0   P11 P12   xt   P11 P12   Gs   z u   P21 P22   yt   P21 P22  Gu  t 1

 x    t 1   Et yt 1 

x   t   yt 

 x     t 1    s  Et yt 1   0

G   s  Gu 

0   xt  Gs   z u   yt  Gu  t 1

(1) The unstable block (red part) of the system can be solved forward by iteration:

Et yt 1  u yt  Gu zt 1  yt  u1Et yt 1  u1Gu zt 1  yt  u1Gu zt 1 Recover the solution for yt using the definition of yt :

yt   u1Gu zt 1  xt   P11  y   P  t   21  Gs   P11   Gu   P21

P12   xt    P22   yt    yt  P21 xt  P22 yt   P12   Gs   Gu  P21Gs  P22Gu P22  Gu  

P21 xt  P22 yt   u1  P21Gs  P22Gu  zt 1

yt   P221 P21 xt  P221 u1  P21Gs  P22Gu  zt 1

This is the solution (policy/transition function) for endogenous control variables yt in terms of both endogenous states and exogenous states as well as pure shocks. (2) The stable block (red part) of the system can be solved by substituting in the solu-

tion of yt , based on the original form of the system:

 xt 1   xt   E y   F  y   Gzt 1  t t 1   t  xt 1   F11 F12   xt   Gs  E y   F       zt 1  t t 1   21 F22   yt  Gu  xt 1  F11 xt  F12 yt  Gs zt 1





xt 1  F11  F12 P221P21 xt  Gs  P221 u1  P21Gs  P22Gu   zt 1 This is the solution (law of motion) for endogenous state variables xt 1 in terms of both endogenous states and exogenous states as well as pure shocks.