Econ 510a (second half)Prof: Tony SmithTA: Theodore PapageorgiouFall 2004Yale UniversityDept. of EconomicsSolutions for Homework #1Question 1We know that Lct L and thus Lit1 − L. It is also true that kt kct kit. Moreoverwe have that:kctLctkitLitkctLkt− kct1 − L1 − kct kt− kct kct ktand:kit1 − ktThus we can write the capital-labor ratios above as:kctLctktLtktLtkitLitSince both Fand Ghave constant returns to scale we can write:Fkct,Lct LctfKctLct LfktLtGkit,Lit LitgKitLit 1 − LgktLtSince the total amount of labor is fixed, with out loss of generality we can normalize it sothat L  1:Fkct,1 fktGkit,11 − gktTotal capital evolves according to:kt11 − kt itkt11 − kt 1 − gktWe will assume the following: gktis strictly increasing in kt gktis strictly concave in kt g0 01 − g′0 k→lim1 − 1 − g′k 1Therefore based on all the above, as we did in class we see that if we graphkt11 − kt1 − gkt, it will start at zero and cross the 45∘degree line from aboveonly once at k∗,wherek∗is such that k∗1 − k∗1 − gk∗( k∗1 − gk∗).Thus for kt k∗it will be the case that kt1− kt 0 (since kt11 − kt1 − gktis above the 45∘degree line) and for kt k∗it will be the case that kt1− kt 0 (sincekt11 − kt1 − gktis below the 45∘degree line).Thus ktis monotone bounded sequence. Since it is also bounded, it has a limit which is k∗.Question 2Let’s define two new function hand lsuch that:hkt≡ fgktlkt≡ gfktObviously it will be the case that:kt2 hktif t is evenkt2 lktif t is oddWe now want to see whether there exists khsuch that hkh kh. We know that since fand gare strictly increasing and strictly concave, hhas to be strictly increasing andstrictly concave. Moreover we have that:f0 g0 0  h0f′0g′0 1  h′0 1and:k→lim f′gkg′k 1 k→lim h′k 1Therefore based on all the above, as we did in class we see that if we graph hk, it willstart at zero and cross the 45∘degree line from above only once at kh.Thus for kt khand t even, it will be the case that kt2− kt hkt− kt 0 and forkt khand t even, it will be the case that kt2− kt hkt− kt 0 (since his now belowthe 45∘degree line).Thus ktis monotone bounded sequence. Since it is also bounded, it has a limit which is kh.Similarly for lwe want to whether there exists klsuch that lkl kl. We know thatsince fand gare strictly increasing and strictly concave, lhas to be strictly increasingand strictly concave. Moreover we have that:f0 g0 0  l0f′0g′0 1  l′0 1and:k→lim g′fkf′k  1 k→lim l′k 1Therefore based on all the above, as we did in class we see that if we graph lk, it willstart at zero and cross the 45∘degree line from above only once at kl.Thus for kt kland t odd, it will be the case that kt2− kt lkt− kt 0 and for kt kland t odd, it will be the case that kt2− kt lkt− kt 0 (since lis now below the 45∘degree line).Thus ktis monotone bounded sequence. Since it is also bounded, it has a limit which is kl.In other words there is global convergence to a ”two cycle” in which ktoscillates betweenkhand kl.If we know assume that fkt aktand gkt bkt,wherea and b are positive constants,we will have that for t even:hkt fgkt abktand for t odd:lkt gfkt baktClearly if ab  1 capital grows indefinitely and if ab  1, capital will shrink and converge tozero. If ab  1 capital will stay at its initial level.Question 3The functional Euler equation is− u′fk− gk u′fgk− ggkf′gk 0Differentiate both sides with respect to k,wehave0  −u′′fk− gkf′k− g′k u′′fgk− ggkf′gkg′k−g′k2f′gk u′fgk− ggkf′′gkg′kEvaluate this equation at k  k∗, and notice that gk∗ k∗and fk∗ −1,wehave0  −u′′fk∗− k∗f′k∗− g′k∗ u′′fk∗− k∗f′k∗g′k∗−g′k∗2 u′fk∗− k∗f′′k∗g′k∗Plug in c∗ fk∗− k∗, after some manipulation, we getg′k∗2− 1 1u′c∗u′′c∗f′′k∗f′k∗g′k∗1 0Or:2− 1 1u′c∗u′′c∗f′′k∗f′k∗ 1 0where  is equal to g′k∗. We also know that the speed of convergence near the steadystate is inversely related to the slope of the decision rule at the steady state (i.e. g′k∗or ).Here we will see that curvature of production function will speed up convergence, whilecurvature of utility function will retard it. The economic intuition is as follows: (a) the higherthe curvature of production in the steady state, the sharper the change in the marginal return ofcapital when we are perturbed from steady state. Therefore, people want to invest more tomake use of this opportunity, which speeds up convergence. (b) the higher the curvature ofutility function in the steady state, the sharper the change in the marginal utility whenperturbed from the steady state. Since people wants to smooth their marginal utility, they willconsume more today to offset the change, which slows down the capital accumulation.Plug into the curvature of utility and