0 β t u(c t ), 0 <β<1,

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1 Part 2 1. Certainty-Equivalence Solution Methods Consider the model we dealt with previously, but now the production function is y t = f(k t,z t ), where z t is a stochastic exogenous variable. For example, it follows the AR(1) process log(z t )=ρlog(z t 1 )+ε t, ε t (0,σ 2 ) (useful for multiplicative specification). The problem we analyzed changes to Max {ct,k t+1 } = E P t=0 0 β t u(c t ), 0 <β<1, t=0 subject to c t = f(k t,z t )+(1 δ)k t k t+1, k 0 given, lim T βt E 0 u 1 (c T )k T =0. The Euler equation is now u1 (f(k u 1 (f(k t,z t )+(1 δ)k t k t+1 )+βe t+1,z t+1 )+(1 δ)k t+1 k t+2 ) t =0, (f 1 (k t+1,z t+1 )+1 δ) u 1 (c t )+βe t [u 1 (c t+1 )(f 1 (k t+1,z t+1 )+1 δ)] = 0. Denoting R t+1 = f 1 (k t+1,z t+1 )+1 δ, the Euler equation becomes u 1 (c t )+βe t [u 1 (c t+1 ) R t+1 ]=0, where R t+1 is the rate of return on saving at time t for one period Detour: A Certain Rate of Return Let s deviate for a moment from the model we started with, by assuming that income y t is exogenous and i.i.d, withmeanȳ. Households save on the asset a t+1, which yield R t+1 next period determined at time t. The problem now is Max {ct,k t+1 } = E P t=0 0 β t u(c t ), 0 <β<1, t=0 s.t. c t = y t + R t a t a t+1, a 0 given, lim T βt E 0 u 1 (c T )a T =0.

2 The Euler equation is u 1 (c t )+βr t+1 E t u 1 (c t+1 )=0. Note that this is similar to the Euler equation we had before, but now the rate of return is outside the expectation. A certainty-equivalence procedure is to solve u 1 (c t )+βr t+1 u 1 (E t c t+1 )=0, i.e., to treat future variables as if they are fully certain at their expected levels, given the current information. How different is the solution of a t+1 from this procedure from the correct decision? Graph of u 1 (c t+1 ) as a function of c t+1 If u 111 = 0: no approximation An application of certainty-equivalence is to solve a linearized version of the Euler equation u 1 (c t )+βr t+1 E t u 1 (c t+1 )=0 u 11 dc t + βu 1 dr t+1 + βru 11 E t dc t+1 where the derivatives are coefficients because are taken at a given point Return to the Main Model Now we go back to u 1 (c t )=βe t [u 1 (c t+1 ) R t+1 ], where R t+1 = f 1 (k t+1,z t+1 )+1 δ. Applying certainty-equivalence implies to solve u 1 (c t )=β [u 1 (E t (c t+1 )) E t (R t+1 )]. Here, certainty-equivalence not only ignores the curvature of u 1 ( ), which is likely to reduce the right-hand side, but also the covariance between u 1 ( ) and R t+1, which should be negative in general equilibrium. Ignoring the negative covariance, therefore, increases the right-hand side. Hence, the sign of the bias in the choice of k t+1 is unclear. 2

3 1.3. A Certainty-Equivalence Procedure If we decide that uncertainty is not an important issue, then linearization, or log linearization, is a good possibility if there are no kinks in the relevant area. For example, if there is a constraint which may bind or not, then the linearization, or any other order approximation ( perturbation methods ) will not work properly, unless the order of the approximation is high enough. Consider the case where investment has a minimum size. As we saw earlier, in this case the investment decision is not monotonic, and hence at the kink it cannot be approximated by a derivative. Here we ll solve the following problem: 1. Investment should have the minimum size Φ 2. The production function is y t = f(k t,z t ), where z t is a productivity shock following the process log z t+1 = ρ log z t + ε t, ε t (0,σ 2 ). The Lagrangean of this problem is: P L = E 0 β t {u(c t )+λ t [f(k t,z t )+(1 δ)k t k t+1 c t ]+ξ t [k t+1 (1 δ)k t Φ]}, t=0 subject to a given k 0 and E 0 is the expectation operator conditioned on information at time 0. Besides the usual assumptions about the functions u(.) and f(.), we assume that lim c u 1 (c) =0(i.e., there is no satiation). This implies that λ t > 0, or that the resource constraint always binds. The first-order conditions for maximization of L are: 0 = u 1 (c t ) λ t, 0 = λ t + βe t λ t+1 [f 1 (k t+1,z t+1 )+1 δ]+ξ t βe t ξ t+1 (1 δ), 0 = ξ t [k t+1 (1 δ)k t Φ], ξ t 0, 0 = f(k t,z t )+(1 δ)k t k t+1 c t, t =0, 1, 2,... 3

4 Substituting for λ t yields u 1 (c t ) = βe t u 1 (c t+1 )[f 1 (k t+1,z t+1 )+1 δ]+ξ t βe t ξ t+1 (1 δ), 0 = ξ t [k t+1 (1 δ)k t Φ], ξ t 0, 0 = f(k t,z t )+(1 δ)k t k t+1 c t. Steady State To allow for a standard steady state, we need to assume that Φ δk, where k solves [f 1 (k, 1) + 1 δ] =0. Let us solve the deterministic problem where z 0 is given, and z t = z ρt 0. For the computational solution, we define the dummy variable binding t = ½ 0 if kt+1 (1 δ)k t > Φ 1 if k t+1 (1 δ)k t = Φ t = 1, 2,... and then the system of equations for the solution can be written as 0 = (1 binding t ) { u 1 (c t )+βu 1 (c t+1 )[f 1 (k t+1,z t+1 )+1 δ]} +binding t {k t+1 (1 δ)k t Φ}, 0 = f(k t,z t )+(1 δ)k t k t+1 c t, t =0, 1, 2,... Note that the computation does not solve directly the Lagrange multipliers ξ t, but they can be calculated from the solution: From the Euler equation 0= u 1 (c t )+βu 1 (c t+1 )[f 1 (k t+1,z t+1 )+1 δ]+ξ t βξ t+1 (1 δ) we can compute ξ t βξ t+1 (1 δ), but not the two terms separately. However, given that (a) the process for z converges to it s mean, and (b) the constraint does not bind at the steady state, the constraint can bind only at the beginning of the path. In some period, the constraint ceases to bind, and stays slack till the end of the horizon. If the constraint ceases to bind in period τ, then, from the Euler equation for τ 1, we can compute ξ τ 1 because ξ τ =0. Then, we can use ξ τ 1 to compute ξ τ 2 and so on. 4

5 Structure of the program while changes>0 binding0 = binding; %initial and final capital values Kstate= ******; Kfinal= ******; [X,fval]=fsolve(@(X) fsystem_hmw2complete(x,binding,z,kstate,kfinal,params),x0,opt ; % "fsolve" solves the system of equations in the file "fsystem_hmw2", % given "Kstate" (K in the initial period) and Kfinal (K in the % final period), the parameters, an initial value for % the solution "X0" and puts the solution in "X" K(2:T+1,1) = X; K(1,1) = Kstate; % check whether the constraints are violated and when K1(T+1,1) = Kfinal; K1(1:T) = K(2:T+1,1); % the following generates a vector with ones where the constraint is violated, and % zeros elsewhere violations = (***** Euler equation investment **** - capphi < ); % now replace the original zeros by the corresponding elements in "violations" binding(binding==0) = violations(binding==0); changes=sum(abs(binding-binding0)); end function y=fsystem_hmw2complete(x,binding,z,kstate,kfinal,params);... % let s define two vectors: "current" (K) and "next period" (K1) % for the "current", the initial stock is given K(1,1) = Kstate; % for the "next period" the final stock is given K1(T+1,1) = Kfinal;... % the other variables 5

6 Y = Z.*K.^alpha; I = K1 - (1-delta)*K; C=Y-I; C1(1:T,1) = C(2:T+1,1);... % set of T+1 equations y(1:t+1) = (1-binding).* (****Euler expression =0****)... + binding.*( *** min investment constraint =0 ****); The Rolling Solution We solve for the path starting from period 0 with {k 0,z 0 }, and get k 1. Given the realization of ε 1 we have log (z 1 )=θ log(z 0 )+ε 1. Wesolveforthepathstartingfromperiod1with{k 1,z 1 }, and get k 2 Given the realization of ε 2 we get log (z 2 )=θ log(z 1 )+ε 2, and so on for an artificial sample of arbitraty length. In Matlab notation, the loop calculating "timeseries" observations is the following: We want to compute the vector k(timeseries,1) given k(1) = a starting value and the vector of the randomly produced productivity levels z(timeseries,1), whose values are known only period-by-period. % generating the entire productivity shock realization for the sample eps = RANDOM( norm,0,stdev,timeseries,1); logz(1,1)=eps(1,1); for i=2:timeseries logz(i,1)=rho*logz(i-1,1)+eps(i,1); end z=exp(logz); 6

7 for t=1:timeseries-1 Zexp(1,1) = z(t,1); for j=2:t; Zexp(j,1) = Zexp(1,1)^(rho^(j-1)); end; **** The solution for one path we already saw ***** end Estimation of ˆk t+1 = α 0 + α 1ˆkt + α 1 z t + residual to see how well this approximates the decision rule. timeseries = 100 No constraint CoeffTable = Coef StdErr tstat R2 = # of binding periods = 0 investment Φ =0.99 δk 7

8 CoeffTable = Coef StdErr tstat R2 = # of binding periods = 63 Investment/output ratio 8

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10 Investment/output ratio in the US INV/Y

11 Consumption/output ratio

12 Consumption/output ratio in the US CONS/Y

13 Annual output growth in the U.S D(LOG(Y))

14 Output growth from the model