Econ 510a (second half)Yale UniversityFall 2006Prof. Tony SmithHOMEWORK #1This homework assignment is due at 5PM on Friday, November 3 in Marnix Amand’s mailbox.1. Consider a growth model with capital accumulation equation kt+1= f(kt) if t is evenand kt+1= g(kt) if t is odd. Assume that:(i) f(0) = g(0) = 0.(ii) f0(0)g0(0) > 1.(iii) limk→∞f0(g(k))g0(k) < 1 and limk→∞g0(f(k))f0(k) < 1.(iv) f and g are strictly increasing and strictly concave.Show that, from any initial condition k0> 0, there is global convergence to a “two-cycle” in which ktoscillates between two values. How are these values determined?Solution We have:• k2t= g(f (k2t−2))• k2t+1= f (g(k2t−1))Let h be defined as h(k) = g(f(k))−k for all k. Should the sequence k2tbe converging,the limit necessarily is a solution of h(k) = 0 (cf. your favorite math course onsequences). Using properties (i) and (ii), we have that h(0) = 0 and h is increasingat 0. By differentiating twice, h is strictly concave. Using property (iii), h0< 0 fork sufficiently large, and hence limk→∞h(k) = −∞ because, by concavity, h is “under”its tangents. It is thus proved that h(k) = 0 has one solution, and only one (usingconcavity again), for k > 0, calledˆk.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 k2tis increasing and bounded for k0<ˆk anddecreasing and bounded for k0>ˆk. In both cases, k2tis converging, necessarily towardˆk which is the only possible limit.Same story for k2t+1, which converges to the unique solution of f(g(k)) = k. This theglobal convergence to the “two-cycle”.2. Consider a neoclassical growth model similar to the one that we have discusse d inlecture, but in which the level of technology oscillates deterministically between twovalues AHand AL, where AH> AL. In particular, period-t output ytequals AHF (kt)if t is even and equals ALF (kt) if t is odd. The planner (“Robinson Crusoe”) seeks tomaximizeP∞t=0βtu(ct), given k0> 0, subject to the resource constraint that ct+kt+1=yt+ (1 − δ)ktand to the nonnegativity constraint kt+1≥ 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 inwhich the level of technology is low. Find a pair of Bellman equations that thesefunctions must satisfy.)(b) Let the felicity function u be logarithmic, let yt= Atkαt, and assume that capitaldepreciates fully in one period (i.e., set δ = 1). Use a guess-and-verify method tofind the two value functions in part (a). Describe fully the dynamic behavior ofthe capital stock.Solution (a) When A = AH, the planner’s problem is given by the following value function:VH(K) = maxK00{u (AHF (K) + (1 − δ) K − K0) + βVL(K0)}(1)when A = ALthe planner’s problem is going to be given by:VL(K) = maxK00{u (ALF (K) + (1 − δ) K − K0) + βVH(K0)}(2)(b) If u (c) = log c, δ = 1 and F (Kt) = Kat, then the above value functions become:VH(K) = maxK00{log (AHKa− K0) + βVL(K0)}VL(K) = maxK00{log (ALKa− K0) + βVH(K0)}We know guess that the value functions are of the form:VH= E + F log KVL= G + J log KIn that case, the equation above for VHbecomes:E + F log K = maxK00{log (AHKa− K0) + β (G + J log K0)}(3)The first-order conditions are:−1AHKa− K0+βJK0= 0 ⇔βJK0=1AHKa− K0⇔K0= AHβJKa− βJK0⇔gH(K) = K0=βJAHKa1 + βJPlugging the decision rule derived above back in the original equation:E + F log K = logAHKa−JAHβKa1 + βJ+ βG + J logJAHβKa1 + βJ= log AH+ a log K − log (1 + βJ) + βG + βJ log JAHβ− βJ log (1 + βJ) + βJa log KTherefore:E = log AH− (1 + βJ) log (1 + βJ) + βG + βJ log JAHβ(4)and:F = a + βaJ(5)Similarly the value function when A = AL, is given by:G + J log K = maxK00{log (ALKa− K0) + β (E + F log K0)}(6)The first-order condition gives the following decision rule:gL(K) = K0=βF ALKa1 + βF(7)Plugging the decision rule back into the equation above, we get:G + J log K = log(ALKa−βF ALKa1 + βF) + βE + F logβF ALKa1 + βF= log AL+ a log K − log (1 + βF ) + βE + βF log βF AL+ βF a log K − βF log (1 + βF )Therefore:G = log AL− (1 + βF ) log (1 + βF ) + βE + βF log βF AL(8)andJ = a + βaF(9)combining the expression above for J with the one we found for F, gives us:J = a + βa (a + βaF ) ⇔J =a + βa21 − β2a2=a (1 + βa)(1 + βa) (1 − βa)=a1 − βaand thus:F = a + βaa1 − βa=a1 − βa= J(10)Therefore the equation for E now becomes:E = log AH−1 +βa1 − βalog1 +βa1 − βa+ βG +βa1 − βalog AHβa1 − βa==11 − βalog AH+ βG +βa1 − βalogβa1 − βa−11 − βalog11 − βaLet M =βa1−βalogβa1−βa−11−βalog11−βa. Then E becomes:E =11 − βalog AH+ βG + M(11)And similarly plugging in for E, in the equation above we had for G we have:G =11 − βalog AL+ βE + M(12)Thus solving the above system of 2 equations and 2 unknowns, we get:E =1 − β2−1(1 − βa)−1(log AH+ β log AL) + M + βMG =1 − β2−1(1 − βa)−1(log AL+ β log AH) + M + βMTherefore the decision rule when A = AHis given by:gH(K) = βaAHKa(13)and when A = AL,is given by:gL(K) = βaALKa(14)We will now show that there is a ”global convergence” to a ”two-cycle” in whichKtoscillates between two values. These 2 values are:KH= βaAHKaH⇔KH= (βaAH)11−awhen A = AHand similarly:KL= (βaAL)11−a(15)when A = AL.We will show that the assumptions of Question 2 of Homework #1hold indeed in this case and thus there is a global convergence to a ”two-cycle”.We have:·gH(0) = gL(0) = 0· Moreover we have that: g0H(K) = βa2AHKa−1and g0L(K) = βa2ALKa−1andthus g0H(0) = ∞ and g0L(0) = ∞ since a − 1 < 0. Therefore:g0H(0) g0L(0) = ∞ > 1(16)· Furthermore:gH(gL(K)) = βa+1aa+1AHAaLKa2g0H(gL(K)) = βa+1aa+3AHAaLKa2−1and:g0H(gL(K)) g0L(K) = βa+1aa+3AHAaLKa2−1βa2ALKa−1== βa+2aa+5AHAa+1LKa2+a−2and since a2+ a − 2 < 0 (remember 0 < a < 1) we have that:limK→∞g0H(gL(K)) g0L(K) = 0(17)Similarly:g0L(gH(K)) g0H(K) = βa+2aa+5ALAa+1HKa2+a−2(18)and:limK→∞= g0L(gH(K)) g0H(K) = 0(19)· Finally gHand gLare strictly increasing and strictly concave3. Consider a neoclassical growth model in which the felicity function u has constantelasticity of intertemporal substitution σ−1:u(c) =c1−σ− 11 − σ,where σ > 0 and u(c) = log(c) if σ = 1.