Linear Approximation to Policy Function in IRIS Toolbox

Transcription

1 Linear Approximation to Policy Function in IRIS Toolbox Michal Andrle July 19, Introduction This short note documents the solution of linear approximation to policy function of nonlinear dynamic stochastic general equilibrium model in the IRIS Toolobx by Jaromir Benes. There are currently two options one with unanticipated stochastic shocks with zero mean expectations and perfect foresight policy function. 2 Problem Formulation A linear approximation to nonlinear model can be written as F 1 E t x t+1 + F 2 x t + F 3 ε t + c =0, (1) where the vector x stores transition variables of the model, ε denotes vector of stochastic shocks with E t ε t+k =0fork>0 unless perfect foresight solution is assumed. In this note we abstract from formulation of the problem that allows for solution of non-stationary model and the determination of the constant or, in the sequel, of trends in series. Thus we impose for convenience c = 0 and vector x can be interpreted as deviation from a balanced growth path (BGP). The vector x may be partitioned in the following form x t+1 [ x P t x N t+1, (2) where x P t denotes predetermined and x N t+1 non-predetermined transition variables. Following Klein(2000), the policy function is solved using generalized Schur decomposition. However, definition of auxilliary processes is somewhat different than usual. Let us define matrices Q, Z, A, B such that QF 1 Z = A QF 2 Z = B, (3) 1

2 where A, B are quasi-trianglular. We define new auxilliary variables as follows [ [ Z11 Z 12 St+1 Z 22 U t+1 Z 21 [ x P t x N t+1. (4) Using (4), we can rewrite (1) as follows [ [ [ A11 A 12 St+1 B11 E 0 A t + 22 U t+1 0 [ B 12 St + Dε B 22 U t =0. t (5) Due to properly reordered variables and Schur decomposition the lower part of (5) implies unstable dynamics. 2.1 Policy Function in Case of Unanticipated Shocks Forward Solution of Unstable Part Fist, we solve the unstable system forward and impose transversality condition. Thus we have A 22 E t U t+1 + B 22 U t + D 3 ε t =0, (6) implying that U t = lim k ( B 1 22 A 22) k E t U t+k ( B22 1 A 22) k B22 1 D 2E t ε t+k. (7) Either directly from transversality conditiosn derived from optimization principles of the model or imposed directly no-bubble solution is required, hence we postulate lim k ( B 1 22 A 22) k E t U t+k =0, (8) implying that U t = ( B22 1 A 22) k B22 1 D 2E t ε t+k. (9) Under our original assumption that E t ε t+k = 0 for k > 0, the solution specializes to U t = B22 1 D 2ε t = R U ε t. (10) Solution to Stable Path Having solution to U t we can solve upper part of (5), hence we solve A 11 E t S t+1 + A 12 U t+1 + B 11 S t + B 12 U t + D 1 ε t =0. (11) Since U t = R U ε t,thenu t+1 = R U ε t+1 and E t U t+1 =0. After plugging these realations into (11), we obtain A 11 E t S t+1 = B 11 S t B 12 R U ε t + D 1 ε t. (12) 2

3 One has to realize that in general due to stochastic nature of the problem E t S t+1 S t+1, but we require solution in terms of S t+1 and S t. Using the fact that for predetermined variables we have E t x P t xp t =0,we can use (4) and write E t (Z 11 S t+1 + Z 12 U t+1 ) (Z 11 S t+1 + Z 12 U t+1 )=0, (13) implying that E t S t+1 S t+1 = Z11 1 Z 12R U ε t+1. (14) Plugging (14) into (12) we obtain the solution S t+1 = A 1 11 B 11S t A 1 11 B 12R U ε t A 1 11 D 1ε t Z 1 11 Z 12R U ε t+1. (15) Using (4) we know that S t+1 = Z 1 11 xp t Z 1 11 Z 12U t+1 = Z 1 11 xp t Z 1 11 Z 12R U ε t+1 (16) Plugging (16) into solution (15) we obtain final stable solution Z 1 11 xp t = A 1 11 xp t 1 A 1 11 D 1ε t + ( A 1 11 Z 12 A 1 R U ε t. (17) We define a new auxilliary variable α as α t S t + Z 1 11 Z 12U t Z 1 11 xp t (18) and rewrite the stable transition equation (17) in terms of new variable α, i.e. α t = A 1 11 B 11α t 1 A 1 11 D 1ε t ( A 1 11 B 11G + A 1 R U ε t, (19) where G = Z 1 11 Z 12. To solve for x N t in terms of α, one must realize that x N t = Z 21 S t + Z 22 U t. We can thus rewrite hence or x N t x N t = Z 21 [ Z 1 11 xp t Z 1 11 Z 12U t + Z22 R U ε t. (20) = Z 21 α t 1 + [ Z 21 Z 1 11 Z 12 + Z 22 Ut (21) x N t = Z 21 α t 1 +[Z 21 G + Z 22 U t (22) 3

4 2.1.3 State Space System with No Anticipations The state-space model for transition equations thus may be written as follows [ [ [ x N t T F R F = α t T A α t 1 + R A ε t (23) We define x P t = Uα t. (24) T F Z 21 (25) T A A 1 11 B 11 (26) R F (Z 21 G + Z 22 ) R U (27) R A [ A 1 11 D 1 ( A 1 11 B 11G + A 1 R U (28) G Z (29) R U B (30) U Z 11 (31) 4

5 2.2 Policy Function with Perfect Foresight In case of full perfect foresight we solve the model (1) using assumption that E t ε t+k = ε t+k, k > 0. (32) There are multiple ways how to solve for this case. Even an algorithm calculated under assumption of no foresight may be accomodated for inclusion of perfect foresight by appropriate use of auxilliary variables. One can then combine foresight with surprises. For convenience we focus initially on pure foresight case, where there are no surprises and all shocks are perfectly anticipated. Furthermore, it is assummed that a steady-state of ε t+k is zero. The solution used bellow cannot form an infinite sum of geometric series, i.e. proper permament shock Forward Solution of Unstable Part Starting from identical quasi-triangular decoupled system (5) we solve for the lower unstable part. Imposing no-bubble equlibrium the solution collapses into U t = ( B22 1 A 22) k B22 1 D 2E t ε t+k. (33) For convenience let us define J ( B22 1 A ) 22 R U ( B 22 D 2 ), (34) hence simplifying the notation using U t = J k R U E t ε t+k. (35) It is clear that agents discount future shocks at the rate of J, which is formed using matrices of unstable dynamics. Intuitively, the model operates on unstable trajectory until the occurence of the shock anticipated. There are thus twoeffects ofanticipated shocks (i) announcement effect and (ii) implementation effect. For better intuition we shall use illustrative example, when agents anticipate only two periods ahead shocks and then expect zero realisations of shocks, i.e Solving Stable Part U t = R U ε t + JR U ε t+1 + J 2 R U ε t+2. (36) Since we assume pure perfect foresight, it is clear that E t S t+1 = S t+1 and we can thus write the stable part of the system as A 11 S t+1 + A 12 U t+1 + B 11 S t + B 12 U t + D 1 ε t =0. (37) 5

6 Using the definition of α S t + Z 1 11 Z 12U t we can rewrite the solution to the stable part of system as α t = A 1 11 B 11α t 1 + ( Z 1 11 Z 12 + A 1 11 A 12) Ut+1 + ( A 1 11 Z 12 A 1 Ut A 1 11 D 1ε t. It should be clear that we can write a recursion for determination of U t as and rewrite (38) as (38) U t = JU t+1 + R U ε t, (39) α t = A 1 11 B 11α t 1 + ( Z 1 11 Z 12 + A 1 11 A 12) Ut+1 + ( A 1 11 Z 12 A 1 (JUt+1 + R U ε t ) A 1 11 D 1ε t, which is the final solution of the problem. As in previous case we need to solve for x N t identical, giving a solution (40) in terms of α. The procedure is x N t = Z 21 α t 1 + [ Z 21 Z 1 11 Z 12 + Z 22 Ut. (41) Practical Implementation of the Solution As in the previous case the solution for transition variables will take a statespace form. The solution can be written down in a form ε t [ x N t ε t+1 = Tα α t 1 + R t., (42) ε t+n where R is expanded matrix of period t impact of current and anticipated shocks. The first left-block of R consists of stacked matrices [R F ; R A capturing the effect of current period shock. To write down the matrix R it is convenient to define auxilliary matrices. Thus, let X A0 = ( A 1 11 B 11Z11 1 Z 12 A 1 11 B ) 12 (43) X A1 = ( Z11 1 Z 12 + A 1 11 A ) 12 (44) X A = X A1 + X A0 J (45) 6

7 for the use with lower-part of the state-space and X F = (Z 21 G + Z 22 ) (46) for the upper part. The expanded matrix R canthenbewrittenas [ R F X R = F R U X F JR U X F J 2 R U... X F J N 1 R U R A X A R U X A JR U X A J 2 R U... X A J N 1 R U (47) 7