Review_Questions_Solutions-1

Review_Questions_Solutions-1 - Economics W3213 Spring 2010...

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Economics W3213 Spring 2010 Bruce Preston Review Question Solutions Textbook Questions: Attached Long Answer Question 1. Consider the following constant returns to scale technology in the Solow model: Y t = AK α t N 1 α t for a positive constant A . Suppose the government purchases goods in the amount of g p erun i to fcap i ta l K t every year so that total government purchases are gK t .T h e government has a balanced budget so that its tax revenue in year t , T t , equals total government purchases. Total national saving, S t ,is S t = s ( Y t T t ) where 0 <s< 1 is the saving rate. Capital evolves according to the accumulation equation K t +1 =(1 δ ) K t + I t . The population grows according to N t +1 =(1+ n ) N t . (a) Write the production function in per worker terms. Then solve for an equation which determines the steady state capital stock per worker. Show your result graph- ically. What characteristics determine the steady state? Inperworkertermstheproduct ionfunct ionis y t = Ak α t = Af ( k t ) . Capital stock dynamics are therefore k t +1 (1 + n )= s ( Ak α t gk t )+(1 δ ) k t . Steady state requires constant capital leading to the condition sAk α t =( n + δ + sg ) k. Graph follows directly. Steady state capital k = sA

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(b) Suppose the government increases its purchases per unit of capital. What are the e f ects on the steady-state levels of capital per worker, output per worker and con- sumption per worker? Does the result imply the optimal level of government pur- chases is zero? Government purchases reduce steady capital and therefore output. This can be seen directly from the above expression for the steady state level of capital. (c) Consumption e f ects require a little more work. Consumption is just whatever is left out of disposable income after saving. Therefore c =(1 s )( y gk ) so that ∂c ∂g =( 1 s ) μ A ∂f ∂k ∂k ∂g k g ∂k ∂g =( 1 s ) μμ A ∂f ∂k g ∂k ∂g k To sign this note that the second term is negative and the f rst term in big brackets depends on the sign of μ A ∂f ∂k g . Note that ∂f ∂k = αk α 1 = α μ sA n + δ + sg 1 = α μ n + δ + sg sA . Therefore A ∂f ∂k g = α ( n + δ + sg ) s g = α ( n + δ + sg ) sg s = α ( n + δ ) (1 α ) sg s (1) Now ∂k ∂g = 1 1 α sA n + δ + sg ¸ 1 1 α 1 s 2 A ( n + δ + sg ) 2 = s (1 α )( n + δ + sg )
Hence ∂c ∂g =( 1 s ) μμ A ∂f ∂k g ∂k ∂g k =( 1 s ) μ ¡ g Aαk α 1 ¢ s (1 α )( n + δ + sg ) k k =( 1 s ) μμ (1 α ) sg α ( n + δ ) s s (1 α )( n + δ + sg ) 1 k =( 1 s ) μ (1 α ) sg α ( n + δ ) (1 α )( n + δ + sg ) 1 k =( 1 s ) μ (1 α ) sg α ( n + δ ) (1 α )( n + δ + sg ) (1 α )( n + δ +

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Review_Questions_Solutions-1 - Economics W3213 Spring 2010...

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