Instructor : Tony Smith November 13, 2007T.A. : Evrim Aydin SaherCourse : Econ 510a (Macroeconomics)Term : Fall 2007 (second half)Problem Set 2 : Suggested Solutions1. .(a) In this case, the capital accumulation equation becomes:KC;(t+1)+ KI;(t+1)= (1  )KC;t+ (1  )KI;t+ G(KI;t; (L  LC;t))So the dynamic programming problem is :V (KC;t; KI;t) = maxfCt;Lt;KC;(t+1);KI;(t+1)g1t=0u(Ct) + V (KC;(t+1); KI;(t+1))s:t:Ct= F (KC;t; LC;t)KC;(t+1)+ KI;(t+1)= (1  )KC;t+ (1  )KI;t+ G(KI;t; (L  LC;t))Substituting in the constraints:V (KC;t; KI;t) = maxfLt;KC;(t+1)g1t=0u(F (KC;t; LC;t))+V (KC;(t+1); (1  )KC;t+ (1  )KI;t+ G(KI;t; (L  LC;t))  KC;(t+1))Our state variables are: KC;t; KI;tOur choice (control) variables are: LC;t; IC;t(b) First order conditionsWith respect to LC;t:u0(Ct)F2(KC;t; LC;t)G2(KI;t; (L  LC))= V2(KC;(t+1); KI;(t+1))With respect to KC;(t+1):V1(KC;(t+1); KI;(t+1)) = V2(KC;(t+1); KI;(t+1))And hence:V1(KC;(t+1); KI;(t+1)) =u0(Ct)F2(KC;t; LC;t)G2(KI;t; (L  LC;t)):Envelope Theorem:V1(KC;t; KI;t) = u0(Ct)F1(KC;t; LC;t) + V1(KC;(t+1); KI;(t+1))(1  )V2(KC;t; KI;t) = V2(KC;(t+1); KI;(t+1))(1   + G1(KI;t; (L  LC;t; )))Update one period:V1(KC;(t+1); KI;(t+1)) = u0(Ct+1)F1(KC;(t+1); LC;(t+1)) + V2(KC;(t+2); KI;(t+2))(1  )V2(KC;(t+1); KI;(t+1)) = V2(KC;(t+2); KI;(t+2))(1   + G1(KI;(t+1); (L  LC;(t+1))))1(c) The set of equations to determine the steady state values of capital and labor in each sector (i.e.,KC, LC, KI, LI):The Euler equations are:1 = u0(Ct+1)u0(Ct)24F1(KC;(t+1); LC;(t+1))G2(KI;t; (L  LC;t))F2(KC;t; LC;t)+ (1  )F2(KC;(t+1);LC;(t+1))G2(KI;(t+1);(LLC;(t+1)))F2(KC;t;LC;t)G2(KI;t;(LLC;t))351 = u0(Ct+1)u0(Ct)F2(KC;(t+1);LC;(t+1))G2(KI;(t+1);(LLC;(t+1)))F2(KC;t;LC;t)G2(KI;t;(LLC;t))(1  ) + G1(KI;(t+1); (L  LC;(t+1))) At the steady state:1 = u0(C )u0(C )24F1(K C; L C)G2(K I; (L  L C))F2(K C; L C)+ (1  )F2(K C;L C)G2(K I;(LL C))F2(K C;L C)G2(K I;(LL C))35= F1(K C; L C)G2(K I; (L  L C))F2(K C; L C)+ (1  )1= F1(K C; L C)G2(K I; (L  L C))F2(K C; L C)+ (1  )And1 = u0(C )u0(C )F2(K C;L C)G2(K I;(LL C))F2(K C;L C)G2(K I;(LL C))[(1  ) + G1(K I; (L  L C))]=  [(1  ) + G1(K I; (L  L C))]1= (1  ) + G1(K I; (L  L C))Therefore we have:F1(K C; L C)G2(K I; (L  L C))F2(K C; L C)+ (1  ) = (1  ) + G1(K I; (L  L C))F1(K C; L C)G1(K I; (L  L C))=F2(K C; L C)G2(K I; (L  L C))Additionally we have:V2(K C; K I) = V2(K C; K I)(1   + G1(K I; (L  L C; )))G1(K I; (L  L C; )) = 1 (1  ):Finall we have the law of motion for capital:K C+ K I= (1  )K C+ (1  )K I+ G(K I; (L  L C))(K C+ K I) = G(K I; (L  L C)):2(d) We have F (KC;t; LC;t) = KC;tL(1)C;tand G(KI;t; LI;t) = KI;tL(1)I;t:And so:F1(K C; L C) = K CL C(1)F2(K C; L C) = (1  )K CL CG1(K I; (L  L C)) = K I(L  L C)(1)G2(K I; (L  L C)) = (1  )K I(L  L C)FromG1(K I; (L  L C; )) = 1 (1  )K I(L  L C)(1)= 1 (1  )K I(L  L C)= 1 (1  )!1(1):Plugging these back intoG2(K I; (L  L C))G1(K I; (L  L C))=F2(K C; L C)F1(K C; L C)(1  )K I(LL C)K I(LL C)(1)=(1  )K CL CK CL C(1)K I(L  L C)=(1  )(1  )K CL CK CL C=(1  )(1  ) 1 (1  )!1(1):3Finally, plugging these results into the budget constraint(K C+ K I) = G(K I; (L  L C))(K C+ K I) = (K I)(L  L C)(1)(K C+ K I)(L  L C)=K I(L  L C)K CL CL C(L  L C)=1K I(L  L C)K I(L  L C)L C(L  L C)=1K I(LL C)K I(LL C)K CL C=1(1(1))(1)(1(1))1(1)(1)(1)(1(1))1(1)=(1(1))1(1)1(1(1)) 1(1)(1)(1(1))1(1)=(1  )(1  )"1 1 (1  )! 1#SoL C=(1)(1)1(1(1)) 1(1)(1)1(1(1)) 1+ 1L:If we letA =(1  )(1  )"1 1 (1  )! 1#4We haveL C=A(A + 1)LK C=(1  )(1  ) 1 (1  )!1(1)L CL I= L  L C=1(A + 1)LK I= 1 (1  )!1(1)L I:2. (a) We start with the …rst perio d ’s budget constraint:c0+ qb1= b0+ !b1= q1(b0+ !  c0) :For t = 1 we have:c1+ qb2= b1+ !b2= q1(b1+ !  c1)= q1q1(b0+ !  c0) + !  c1= q2b0+ q2! + q1!  q2c0 q1c1:For t = 2 we have:c2+ qb3= b2+ !b3= q1(b2+ !  c2)= q1q2b0+ q2! + q1!  q2c0 q1c1+ !  c2= q3b0+ q3! + q2! + q1!  q3c0 q2c1 q1c2:And so on, until we have:b(t+1)= q(t+1)b0+ !(q(t+1)+ ::: + q1) q(t+1)c0+ qtc1+ ::: + q1ctq(t+1)b(t+1)= b0+ !(1 + ::: + qt) c0+ qc1+ ::: + qtct= b0+ !tPk=0qktPk=0qkck:Taking the limit gives us:limt!1q(t+1)b(t+1)= b0+ ! limt!1tPk=0qk limt!1tPk=0qkck0 = b0+!1  q1Pt=0qtct5where we used the no-Ponzi game condition on the LHS.So the consumer’s consolidated (or lifetime) budget constraint is1Pt=0qtct= b0+!1  q:(b) The transversality condition is:limt!1tu0(c t)b t= 0or,limt!1tu0(bt qb(t+1)+ !)bt= 0:The Euler equation isqu0(bt qb(t+1)+ !) = u0(b(t+1) qb(t+2)+ !):We want to showlimt!1qtbt= 0:Note that the Euler equation holds 8t :qu0(b0 qb1+ !) = u0(b1 qb2+ !)qu0(b1 qb2+ !) = u0(b2 qb3+ !)andq2u0(b0 qb1+ !) = 2u0(b2 qb3+ !):So we haveqtu0(b0 qb1+ !) = tu0(bt qb(t+1)+ !):Multiplying both sides by btand taking limits:limt!1qtu0(b0 qb1+ !)bt= limt!1tu0(bt qb(t+1)+ !)btu0(b0 qb1+ !) limt!1qtbt= 0 (using the transversality condition).And solimt!1qtbt= 0:(c) We want to prove that a sequence fb tg1t=0= 0 that satis…es the transversality condition and the E ulerequation maximizes the consumer’s objective function, subject to the se quenc e of budget constraintsand the nPg condition.Modi…ed