ECON608 Macroeconomic Theory I Fall 2014 Matthew Hoelle Purdue University Dynamic Programming Practice Problems 1. If T : X ! X is a contraction and X is a complete metric space, prove that there exists at most one …xed point. 2. Prove that if is upper hemicontinuous and single-valued, then it is a continuous function. For these results, you can use the sequential de…nition of continuity: A function f is continuous i¤ fxn g ! x implies ff (xn )g ! f (x): 3. Prove that the constraint correspondence is well-de…ned and continuous. 4. Prove part (i) of Berge’s Maximum Theorem, which is stated as follows: If u(c) + V (k 0 ) is continuous and is continuous, then (ii) for (k) = arg maxu(c) + V (k 0 ) 8k 2 A: is upper hemicontinuous c;k0 2 (k) 5. Prove part (ii) of Berge’s Maximum Theorem, which is stated as follows: If u(c) + V (k 0 ) is continuous and T V (k) = max u(c) + V (k 0 ) 8k 2 A: is continuous, then (ii) T V is continuous for c;k0 2 (k) 6. Recall that CB(A) is the set of continuous, bounded, and concave functions with compact domain A (a complete metric space). Show that T : CB(A) ! CB(A) ; where the mapping T is de…ned as before: max u(c) + V (k 0 ) (T V )(k) = : c; k 0 c + k 0 Rk + w 7. Assume that u is strictly concave, and prove that there exists a unique solution to max u(c) + V (k 0 ) : c; k 0 c + k 0 Rk + w 8. Write down the …rst order conditions of the Bellman equation. Use them to argue that the value function V is di¤erentiably strictly increasing. 9. Recall that the policy function is de…ned as g : A ! A such that g(k) = arg max u(Rk + w k0 2A k 0 ) + V (k 0 ): Assume that u is strictly concave. Then g is a function. From Berge’s Theorem, g is continuous. Prove that g satis…es the following properties: (a) g is continuously di¤erentiable (C 1 ) (b) g is strictly increasing (Dg(k) > 0 8k 2 A) Hint: Use the Implicit Function Theorem. You may assume that V is twice continuously di¤erentiable (C 2 ), a fact shown by Santos, Econometrica (1991). 1 Dynamic Programming Practice Problems (Solutions) 1. If T : X ! X is a contraction and X is a complete metric space, prove that there exists at most one …xed point. Solution: Suppose, in order to obtain a contradiction, that there exist two …xed points, x and x ; with x 6= x : By de…nition of a …xed point, both T (x ) = x and T (x ) = x : By the de…nition of a contraction, for some 2 (0; 1) : d (T (x ) ; T (x )) d (x ; x ) : By the de…nition of the …xed points, this implies d (x ; x ) d (x ; x ) ; which is only true if d (x ; x ) = 0: But this is ruled out by our initial supposition x 6= x : This contradiction completes the argument. 2. Prove that if is upper hemicontinuous and single-valued, then it is a continuous function. For these results, you can use the sequential de…nition of continuity: A function f is continuous i¤ fxn g ! x implies ff (xn )g ! f (x): Solution: If is upper hemicontinuous, then for any sequence fkn g ! k and (cn ; kn0 ) 2 (kn ) 8n 2 N with fcn ; kn0 g ! (c; k 0 ) ; the limit point (c; k 0 ) 2 (k) : Since is single-valued, then 8n; 9! (cn ; kn0 ) = (kn ) and 9! (c; k 0 ) = (k) : This states that (kn ) ! (k) ; which is exactly the de…nition of continuity. 3. Prove that the constraint correspondence Solution: that is well-de…ned and continuous. Recall that the contraint correspondence k 7! c; k 0 2 [0; c] A : c + k0 :A [0; c] A is de…ned such Rk + w : is well-de…ned as (k) is nonempty, since it contains the element (0; 0) for any k since R and w 0 by assumption (R = D1 f (k; 1) 0 and w = D2 f (k; 1) 0). 0 Continuity requires the veri…cation of two parts: upper hemicontinuity (uhc) and lower hemicontinuity (lhc). Claim: is upper hemicontinuous. Consider a sequence fkn g ! k and any sequence fcn ; kn0 g such that (cn ; kn0 ) 2 (kn ) 8n 2 N and fcn ; kn0 g ! (c; k 0 ) : We must show that (c; k 0 ) 2 (k) : Suppose otherwise, that either c < 0; c > c; k 0 < 0; or k 0 > k: Then there must exist n such that cn < 0; cn > c; kn < 0; or kn > k: This contradicts that (cn ; kn0 ) 2 (kn ) 8n 2 N: Claim: is lower hemicontinuous. Consider a sequence fkn g ! k and any (c; k 0 ) 2 following manner: cn ; kn0 where n = = 0 n c; k Rkn +w c+k0 1 8n 2 N; (k) : I de…ne the sequence fcn ; kn0 g in the if Rkn + w < c + k 0 and c + k 0 > 0 : otherwise 2 If c + k 0 = 0; then since R 0 and w 0 by assumption (R = D1 f (k; 1) 0 and w = D2 f (k; 1) 0), the terms (cn ; kn0 ) = n (c; k 0 ) = (0; 0) trivially satisfy the budget constraint cn + kn0 Rkn + w for any n 2 N: We then focus on the nontrivial case in which c + k 0 > 0: I need to verify that the sequence fcn ; kn0 g satis…es (i) fcn ; kn0 g ! (c; k 0 ) and (ii) (cn ; kn0 ) 2 (kn ) 8n 2 N: As fkn g ! k and increasing utility implies c + k 0 = Rk + w; then f n g ! 1; implying fcn ; kn0 g ! (c; k 0 ) : Since n 2 [0; 1] and 0 c c and 0 k0 k; then 0 cn c and 0 0 kn k 8n 2 N: The budget constraint for any n 2 N is given by: cn + kn0 = n c + k0 = Rkn +w c+k0 (c + k 0 ) if Rkn + w < c + k 0 1 (c + k 0 ) if Rkn + w c + k 0 Rkn + w: Thus, (cn ; kn0 ) 2 (kn ) 8n 2 N; …nishing the proof. 4. Prove part (i) of Berge’s Maximum Theorem, which is stated as follows: If u(c) + V (k 0 ) is continuous and is continuous, then (ii) for (k) = arg maxu(c) + V (k 0 ) 8k 2 A: is upper hemicontinuous c;k0 2 (k) Solution: Suppose that is not uhc. Then for some sequence kn 2 A; (cn ; kn0 ) 2 (kn ) 8n 2 N such that lim kn = k 2 A; it is not true that lim (cn ; kn0 ) = (c; k 0 ) 2 (k): As n!1 n!1 is uhc, there is a subsequence, wlog the original sequence, such that lim (cn ; kn0 ) = (c; k 0 ) 2 n!1 (k): Then, it must be that (c; k 0 ) 2 = (k); even though (cn ; kn0 ) 2 (kn ) 8n 2 N: As is lhc, c) + then for every c^; k^0 2 (k); in particular c^; k^0 2 (k) such that u(c) + V (k 0 ) < u(^ V (k^0 ); there exists a sequence c^n ; k^n0 2 (kn ) for n 2 N such that lim c^n ; k^n0 = c^; k^0 : n!1 Since u and V are continuous, lim u(cn ) + V (kn0 ) = u(c) + V (k 0 ) < u(^ c) + V (k^0 ) = lim u(^ cn ) + V (k^n0 ): n!1 n!1 As x c^n ; k^n0 2 (kn ) for n 2 N; then this contradicts that (cn ; kn0 ) 2 (kn ) for n 2 N: 5. Prove part (ii) of Berge’s Maximum Theorem, which is stated as follows: If u(c) + V (k 0 ) is continuous and T V (k) = max u(c) + V (k 0 ) 8k 2 A: is continuous, then (ii) T V is continuous for c;k0 2 (k) Solution: To prove part (ii), we use part (i): is upper hemicontinuous. Recall that if is upper hemicontinuous, then 8k 2 A; the set (k) is nonempty and compact. Recall the de…nition T V (k) = max u(c) + V (k 0 ) 8k 2 A: c;k0 2 (k) Consider any sequence fkn g such that kn 2 A 8n 2 N and fkn g ! k: Consider a corresponding sequence fcn ; kn0 g such that (cn ; kn0 ) 2 (kn ) 8n 2 N: There exists such a sequence due to the Extreme Value Theorem (compact (kn ) and continuous u(c) + V (k 0 )). Since is upper hemicontinuous, then there is a subsequence (without loss of generality the original sequence) such that fcn ; kn0 g ! (c; k 0 ) 2 (k): Since the objective function u(c) + V (k 0 ) is continuous, then the sequence fT V (kn )g = u(cn ) + V (kn0 ) ! u(c) + V (k 0 ) = T V (k): Thus, T V is continuous. 3 6. Recall that CB(A) is the set of continuous, bounded, and concave functions with compact domain A (a complete metric space). Show that T : CB(A) ! CB(A) ; where the mapping T is de…ned as before: max u(c) + V (k 0 ) (T V )(k) = : (1) c; k 0 c + k 0 Rk + w Solution: We begin by assuming that the value function V is continuous, bounded, and concave. From Lecture 10, we know that the new value function T V is continuous and bounded (citing Berge’s Theorem). We will now show that T V is also concave. Consider any two capital levels k1 and k2 : If k1 is the initial capital, de…ne c1 and k1 0 as the optimal solution of the optimization problem (1). By de…nition, (T V )(k1 ) = u(c1 ) + V (k1 0 ): Similarly, if k2 is the initial capital, de…ne c2 and k2 0 as the optimal solution (c; k 0 ) of the optimization problem (1). By de…nition, (T V )(k2 ) = u(c2 ) + V (k2 0 ): For any 2 [0; 1] ; the convex combination of initial capital is k = k1 + (1 )k2 : Given the initial capital k ; the 0 0 0 choice (c ; k ) = (c1 ; k1 ) + (1 ) (c2 ; k2 ) satis…es all constraints (c 0; 0 k 0 k; and 0 0 c +k Rk + w). Thus, the choice (c ; k ) may be chosen by the consumer, so any optimal solution (c ; k 0 ) given the initial capital k must satisfy: (T V )(k ) = u(c ) + V (k 0 ) u(c ) + V (k 0 ): Given that both u and V are concave, then u(c ) + V (k 0 ) [ u(c1 ) + (1 u(c1 ) + V (k1 0 ) + (1 = = V (k1 0 ) + (1 ) u(c2 )] + (T V )(k1 ) + (1 )V (k2 0 ) ) u(c2 ) + V (k2 0 ) )(T V )(k2 ): In sum, (T V )(k ) (T V )(k1 ) + (1 )(T V )(k2 ); …nishing the argument that T V is a concave function. 7. Assume that u is strictly concave, and prove that there exists a unique solution to max u(c) + V (k 0 ) : c; k 0 c + k 0 Rk + w Solution: From the previous question, we know that V is concave. Suppose, in order to obtain a contradiction, that there exists two solutions to the optimization problem: (c1 ; k10 ) and (c2 ; k20 ) : If both are optimal solutions, then u(c1 ) + V (k10 ) = u(c2 ) + V (k20 ): Consider the convex combination (c ; k 0 ) = (c1 ; k10 ) + (1 ) (c2 ; k20 ) for any 2 (0; 1): This choice 0 satis…es all constraints (c 0; 0 k k; and c + k 0 Rk + w) and provides a strictly higher payo¤ as: u (c ) > V k 0 u (c1 ) + (1 k10 V + (1 ) u (c2 ) ) V k20 : The strict inequality u(c ) + V (k 0 ) > u(c1 ) + V (k10 ) = u(c2 ) + V (k20 ) contradicts our original supposition that (c1 ; k10 ) and (c2 ; k20 ) are optimal solutions. 4 8. Write down the …rst order conditions of the Bellman equation. Use them to argue that the value function V is di¤erentiably strictly increasing. Solution: The …rst order condition with respect to consumption is: Du(c) = 0; where 0 is the Lagrange multiplier of the budget constraint. Since u is di¤erentiably strictly increasing, then Du(c) > 0 for any c and > 0: The …rst order condition with respect to k 0 is: + DV (k 0 ) = 0: We know that V is di¤erentiable from the class notes. Combining the …rst order conditions yields: Du(c) = DV (k 0 ): Since u is di¤erentiably strictly increasing, then Du(c) > 0 for any c: Thus, DV (k 0 ) > 0; meaning that V is di¤erentiably strictly increasing. 9. Recall that the policy function is de…ned as g : A ! A such that k 0 ) + V (k 0 ): g(k) = arg max u(Rk + w k0 2A Assume that u is strictly concave. Then g is a function. From Berge’s Theorem, g is continuous. Prove that g satis…es the following properties: (a) g is continuously di¤erentiable (C 1 ) (b) g is strictly increasing (Dg(k) > 0 8k 2 A) Hint: Use the Implicit Function Theorem. You may assume that V is twice continuously di¤erentiable (C 2 ), a fact shown by Santos, Econometrica (1991). Solution: Consider the maximization problem in terms of two variables (k; k 0 ) : k 0 ) + V (k 0 ): max u(Rk + w 0 k 2A The …rst order condition is necessary and su¢ cient for an optimal solution k problem: Du(Rk + w k 0 ) + DV (k 0 ) = 0: De…ne the function F : A A ! R by F k; k 0 = Du(Rk + w 0 to the above k 0 ) + DV (k 0 ); where the optimal solution k 0 satis…es F (k; k 0 ) = 0: F is C 1 (as both u and V are C 2 ). The variable k 0 is an implicit function of k: Claim: D2 F (k; k 0 ) 6= 0: D2 F (k; k 0 ) = D2 u(Rk + w k 0 ) + D2 V (k 0 ): As u is strictly concave and V is concave, then D2 u(Rk + w k 0 ) < 0 and D2 V (k 0 ) 0; implying that D2 F (k; k 0 ) < 0: Applying the Implicit Function Theorem, there exists open sets around k and k 0 (call them O(k) and O(k 0 ); respectively) such that the implicit function g : O(k) ! O(k 0 ) de…ned by g(k) = k 0 is C 1 and Dg(k) = D2 F k; k From above, D2 F (k; k 0 ) < 0: D1 F (k; k 0 ) = (+) ( ) > 0: 5 0 1 D1 F k; k 0 R D2 u(Rk + w : k 0 ) > 0: Thus, Dg(k) =
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