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# problem3sol - Problem Set 3 Solution Econ 210 Core Macro...

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Unformatted text preview: Problem Set 3 Solution Econ 210 Core Macro Monika Piazzesi Stanford University Due Wednesday, October 13 in class Before you start this problem, read Chapter 5 of Krueger and Chapter 4 of SLP in its entirety. There is no need to go through proofs of the theorems. Here, we are interested in applying the theorems. 1. Famous example (a) Since there is no borrowing constraint, the consumer can borrow an in&nite amount in period 0 and roll his debt over (by borrow- ing more and more to repay his previous debt). The associated lifetime discounted utility is w ( x ) = 1 because the consumer has an in&nite amount of consumption. (b) The state variable is wealth x , the control variable is next period¡s wealth y and the recursive formulation is v ( x ) = sup y & x & f x & &y + &v ( y ) g ; where we substituted c = x & &y for consumption. (c) The function w ( ¡ ) = 1 satis&es this functional equation because if you plug in&nity into the RHS, the optimal choice of next period¡s wealth is y = &1 and so you will get in&nity for v ( ¡ ) on the LHS. (d) The value function e v ( x ) = x also satis&es the Bellman equation. To see why, plug it into the RHS of the Bellman equation e v ( x ) = sup y & x & f x & &y + &y g = sup y & x & x = x 1 which shows that e v ( x ) = x: Consider the feasible plan x n = x =& n . For this plan, we have lim n !1 & n e v ( x n ) = lim n !1 & n x & n = x > and so the last equation violates the limit condition of the Prin- ciple of Optimality: lim n !1 & n v ( x n ) = 0 and thus we cannot conclude that e v = w . (In fact, we know from ( a: ) that w & 1 for all x and so e v ( x ) = x is di/erent from w ( x ) = 1 : ). 2. Famous example (variation, with a borrowing constraint of zero) (a) The objective function is w ( x ) = ( x ¡ &x 1 ) + & ( x 1 ¡ &x 2 ) + & 2 ( x 2 ¡ &x 3 ) ::: = x : The solution b x = x and b x t = 0 for t > satis&es lim t !1 & t w ( b x t ) = lim t !1 & t 0 = 0 and so by Theorem 43, b x t is an optimal plan for the sequential problem. (b) The plan b x t = x =& t is feasible and satis&es w ( b x t ) = F ( b x t ; b x t +1 ) + &w ( b x t +1 ) but lim t !1 & t w ( b x t ) = lim t !1 & t & x & t ¡ = x and so Theorem 43 does not apply, and so we cannot conclude that b x t = x =& t is an optimal plan. (a) When A = A H , the planner¡s problem is given by the following functional equation v H ( k ) = max k f u ( A H F ( k ) + (1 ¡ ¡ ) k ¡ k ) + &v L ( k ) g when A = A L , the planner¡s problem is given by v L ( k ) = max k f u ( A L F ( k ) + (1 ¡ ¡ ) k ¡ k ) + &v H ( k ) g 2 (b) If u ( c ) = log c , & = 1 and F ( k ) = k & , then the above value functions become v H ( k ) = max k f log ( A H k & & k ) + ¡v L ( k ) g v L ( k ) = max k f log ( A L k & & k ) + ¡v H ( k ) g We guess that value functions are of the form v H = E + F log k v L = G + J log k In that case, the equation for v H becomes E + F log k = max k f log ( A H k & & k ) + ¡ ( G + J log k ) g The &rst-order conditions are & 1 A H k & & k + ¡...
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problem3sol - Problem Set 3 Solution Econ 210 Core Macro...

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