HOMEWORK #1 This homework assignment is due at 5PM on Friday, November 3 in Marnix Amand s mailbox.

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1 Econ 50a (second half) Yale University Fall 2006 Prof. Tony Smith HOMEWORK # This homework assignment is due at 5PM on Friday, November 3 in Marnix Amand s mailbox.. Consider a growth model with capital accumulation equation k t+ = f(k t ) if t is even and k t+ = g(k t ) if t is odd. Assume that: Solution We have: (i) f(0) = g(0) = 0. (ii) f (0)g (0) >. (iii) lim k f (g(k))g (k) < and lim k g (f(k))f (k) <. (iv) f and g are strictly increasing and strictly concave. Show that, from any initial condition k 0 > 0, there is global convergence to a twocycle in which k t oscillates between two values. How are these values determined? k 2t = g(f(k 2t 2 )) k 2t+ = f(g(k 2t )) Let h be defined as h(k) = g(f(k)) k for all k. Should the sequence k 2t be converging, the limit necessarily is a solution of h(k) = 0 (cf. your favorite math course on sequences). Using properties (i) and (ii), we have that h(0) = 0 and h is increasing at 0. By differentiating twice, h is strictly concave. Using property (iii), h < 0 for k sufficiently large, and hence lim k h(k) = because, by concavity, h is under its tangents. It is thus proved that h(k) = 0 has one solution, and only one (using concavity again), for k > 0, called ˆk. Furthermore, since g f is increasing, we have, for k < ˆk, g(f(k)) < g(f(ˆk)) = ˆk, and, for k > ˆk, g(f(ˆk)) < ˆk. So the sequence k 2t is increasing and bounded for k 0 < ˆk and decreasing and bounded for k 0 > ˆk. In both cases, k 2t is converging, necessarily toward ˆk which is the only possible limit. Same story for k 2t+, which converges to the unique solution of f(g(k)) = k. This the global convergence to the two-cycle.

2 2. Consider a neoclassical growth model similar to the one that we have discussed in lecture, but in which the level of technology oscillates deterministically between two values A H and A L, where A H > A L. In particular, period-t output y t equals A H F (k t ) if t is even and equals A L F (k t ) if t is odd. The planner ( Robinson Crusoe ) seeks to maximize t=0 βt u(c t ), given k 0 > 0, subject to the resource constraint that c t +k t+ = y t + ( δ)k t and to the nonnegativity constraint k t+ 0 for all t. (a) Formulate the planner s problem recursively. (Hint: Consider two value functions, one for periods in which the level of technology is high and one for periods in which the level of technology is low. Find a pair of Bellman equations that these functions must satisfy.) (b) Let the felicity function u be logarithmic, let y t = A t k α t, and assume that capital depreciates fully in one period (i.e., set δ = ). Use a guess-and-verify method to find the two value functions in part (a). Describe fully the dynamic behavior of the capital stock. Solution (a) When A = A H, the planner s problem is given by the following value function: () V H (K) = max K {u (A H F (K) + ( δ) K K ) + βv L (K )} when A = A L the planner s problem is going to be given by: (2) V L (K) = max K {u (A L F (K) + ( δ) K K ) + βv H (K )} (b) If u (c) = log c, δ = and F (K t ) = K a t, then the above value functions become: V H (K) = max{log (A H K a K ) + βv L (K )} K V L (K) = max{log (A L K a K ) + βv H (K )} K We know guess that the value functions are of the form: V H = E + F log K V L = G + J log K In that case, the equation above for V H becomes: (3) E + F log K = max K {log (A H K a K ) + β (G + J log K )}

3 The first-order conditions are: A H K a K + βj = 0 K βj = K A H K a K K = A H βjk a βjk g H (K) = K = βja HK a + βj Plugging the decision rule derived above back in the original equation: E + F log K = log (A H K a JA ) HβK a + β (G + J log JA ) HβK a + βj + βj = log A H + a log K log ( + βj) + βg + βj log JA H β Therefore: βj log ( + βj) + βja log K (4) E = log A H ( + βj) log ( + βj) + βg + βj log JA H β and: (5) F = a + βaj Similarly the value function when A = A L, is given by: (6) G + J log K = max K {log (A L K a K ) + β (E + F log K )} The first-order condition gives the following decision rule: (7) g L (K) = K = βf A LK a + βf Plugging the decision rule back into the equation above, we get: G + J log K = log(a L K a βf A LK (E a + βf ) + β + F log βf A ) LK a + βf = log A L + a log K log ( + βf ) + βe + βf log βf A L + βf a log K βf log ( + βf )

4 Therefore: (8) G = log A L ( + βf ) log ( + βf ) + βe + βf log βf A L and (9) J = a + βaf combining the expression above for J with the one we found for F, gives us: J = a + βa (a + βaf ) J = a + βa2 β 2 a 2 = a ( + βa) ( + βa) ( βa) = a βa and thus: (0) a F = a + βa βa = a βa = J Therefore the equation for E now becomes: E = ( log A H + βa ) ( log + βa ) + βg + βa βa βa βa log A H = βa log A H + βg + βa βa log βa βa βa log βa Let M = () βa βa log log. Then E becomes: βa βa βa βa E = βa log A H + βg + M βa βa = And similarly plugging in for E, in the equation above we had for G we have: (2) G = βa log A L + βe + M Thus solving the above system of 2 equations and 2 unknowns, we get: E = ( β 2) ( ( βa) (log A H + β log A L ) + M + βm ) G = ( β 2) ( ( βa) (log A L + β log A H ) + M + βm )

5 Therefore the decision rule when A = A H is given by: (3) g H (K) = βaa H K a and when A = A L, is given by: (4) g L (K) = βaa L K a We will now show that there is a global convergence to a two-cycle in which K t oscillates between two values. These 2 values are: when A = A H and similarly: K H = βaa H K a H K H = (βaa H ) a (5) K L = (βaa L ) a when A = A L. We will show that the assumptions of Question 2 of Homework # hold indeed in this case and thus there is a global convergence to a two-cycle. We have: g H (0) = g L (0) = 0 Moreover we have that: g H (K) = βa2 A H K a and g L (K) = βa2 A L K a and thus g H (0) = and g L (0) = since a < 0. Therefore: (6) g H (0) g L (0) = > Furthermore: g H (g L (K)) = β a+ a a+ A H A a LK a2 g H (g L (K)) = β a+ a a+3 A H A a LK a2 and: g H (g L (K)) g L (K) = β a+ a a+3 A H A a LK a2 βa 2 A L K a = = β a+2 a a+5 A H A a+ +a 2 L Ka2 and since a 2 + a 2 < 0 (remember 0 < a < ) we have that: (7) lim K g H (g L (K)) g L (K) = 0

6 Similarly: (8) g L (g H (K)) g H (K) = β a+2 a a+5 A L A a+ +a 2 H Ka2 and: (9) lim = K g L (g H (K)) g H (K) = 0 Finally g H and g L are strictly increasing and strictly concave 3. Consider a neoclassical growth model in which the felicity function u has constant elasticity of intertemporal substitution σ : u(c) = c σ σ, where σ > 0 and u(c) = log(c) if σ =. In addition, assume that f(k) = Ak α +( δ)k, where A > 0, α (0, ), and δ [0, ]. What happens to the speed of convergence to the steady state as u becomes linear (i.e., as σ approaches 0)? As f becomes linear (i.e., as α approaches )? Try to provide economic intuition for your findings. (Note: As discussed in Section 4.2 in Chapter 4 of the lecture notes, the speed of convergence near the steady state is inversely related to the slope of the decision rule at the steady state.) Solution Using the notes (section 4.2), especially equation 4.2 on p. 47 (simply use the algebraic expressions for u and f): for u becoming linear, we have λ 0, so convergence is immediate. The intuition is that with linear utility, there is hardly any intertemporal smoothing, so the agent immediately jumps to the equilibrium for α becoming linear, we have λ, so convergence is very slow (in fact, there is no convergence). The intuition is that there are no decreasing returns to scale for saving, so the agent doesn t keep investing until he reaches equilibrium. 4. Consider the planning problem for a basic finite-horizon neoclassical growth model: max {c t, k t+ } T t=0 T β t log(c t ), t=0 given k 0 = 0 and subject to the resource constraint that c t + k t+ = Akt α + ( δ)k t and to the nonnegativity constraint k t+ 0 for all t T. Set β = 0.95, δ = 0.,

7 and α = 0.4. Choose A so that the steady-state value of capital in the corresponding infinite-horizon model is 00. Solve the model numerically (say, in Matlab) using the shooting method described in lecture on October 23: start by guessing a value for k, solve for k 2 from the Euler equation at time 0, then solve for k 3 from the Euler equation at time, and so on, until k T + is found (T being the time horizon). Then vary k and repeat until the appropriate value of k T + (what is it?) is found. Find the lowest value for T such that the highest value of capital between periods 0 and T exceeds 90. Solution We first derive the Euler equation for this problem, then we solve the question numerically. We can solve for Euler equation either from nonlinear programming or dynamic programming method. Here we use dynamic programming. Define f (k t ) = Akt α + ( δ) k t. The recursive formulation for this problem is v t (k t ) = max k t+ ln (f (k t ) k t+ ) + βv t+ (k t+ ) Note that here value function depends on the time subscript t. F.O.C. for t T is From Envelope theorem, we have c t + βv t+ (k t+ ) = 0 v t (k t ) = c t f (k t ) Iterate forward and plug back time subscripts, we get the Euler equation β c t+ f (k t+ ) = c t Plug in c t = f (k t ) k t+ and f (k t ) = Ak α t + ( δ) k t, we have k t+2 = f (k t+ ) β (f (k t ) k t+ ) f (k t+ ) k t+2 = ( ) Akt+ α + ( δ) k t+ β (Ak α t + ( δ) k t k t+ ) ( Aαkt+ α + ( δ) ) k t+2 = A ( + αβ) kt+ α + ( δ) ( + β) k t+ β (Akt α + ( δ) k t ) ( Aαkt+ α + ( δ) ) We can see that it is a second-order nonlinear difference equation, which has boundary condition k 0 = 0, k T + = 0. Before we go on to the numerical step, we solve for A. The infinite horizon steady state k solves f (k ) = β A = β ( δ) α (k ) α =

8 Now we start to solve for it numerically by using shooting method. For an error bound k T + < 0.0, when time horizon T exceeds 37, the highest value of capital between periods 0 and T exceeds 90. The approximate period- capital stock for T = 37 is k = If you use grid point search, to get such a precision you have to define the step size of grid as fine as 0 7.