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Unformatted text preview: Economics 1011b Problem Set 5 Professor Aleh Tsyvinski The problem set is due Tuesday, April 3, by 5 pm in your TFs mailbox in Littauer. Late problem sets will not be accepted. You can work in groups (discuss the solutions, etc). However, you must write the solutions by yourself. Please write down all derivations/explanations. For clarifications (not answers) please contact Leon Berkelmans (email@example.com). Exercise 1. Permanent Changes in Government Spending Assume that the government announces a permanent level of government spending, G that is funded by lump sum taxes. There is a representative house- hold whose lifetime utility is given by U = t =0 t u ( C t ), where = 1 1+ . As- sume that C has to be greater than zero. Capital ( K ) and government spending ( G ) are the only inputs to production, which is given by Y t = G K t . Suppose there is no depreciation. 1. For a given level of government spending, what is the steady state capital level in the economy? Solution: The households budget constraint is: C t = K t + G K t- K t +1- G Substituting this into the utility function of the representative agent gives: U = X t =0 t u ( K t + G K t- K t +1- G ) Taking first order conditions with respect to K t gives: 0 =- u ( K t- 1 + G K t- 1- Kt )+ (1+ G K - 1 t ) u ( K t + G K t- Kt + 1) In steady state K t- 1 = K t = K t +1 = K ss . Therefore the first order condition gives: 1 = (1 + G K - 1 ss ) K ss = (( 1 - 1) 1 G ) 1 - 1 = ( 1 G ) 1 - 1 2. What is the derivative of steady state consumption with respect to G ? 1 Solution: Steady state consumption in this model is: C ss = Y ss- G = G ( 1 G ) - 1- G = G (1- - 1 ) ( ) - 1- G = G 1- ( ) - 1- G Therefore: C ss G = 1- G 1- - 1 ( ) - 1- 1 3. What is the condition that assures that consumption has a finite maxi- mum? Solution: What I should have asked for here is the necessary and suffi- cient condition. It also worked out being a bit harder than first intended. My apologies. To have a finite maximum it is necessary that the functions second deriva- tive must be negative (it could also be zero, with a negative first derivative, assuming G cant be negative, but I am going to ignore this case)....
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