Mid-term 2016 Solution

Macroeconomic Analysis ECON 6022 Fall 2016

Oct 22, 2016

INSTRUCTIONS: Points for each sub-question are indicated in the brackets. Explanations and formulas are expected. Therefore, answers without explanations will lose most of the points (the same for the homework).

1

The Solow Model [40 Points]

Consider a Solow model where the production function is Y = AK α N 1−α , where A is productivity, K and N are capital and the population in this economy, and α is output elasticity of capital. A constant fraction s of output is invested (or saved). And the depreciation rate for capital is δ, the population growth rate is ∆A = g. n, and the growth rate of TFP is constant, g > 0. Specifically, A 1. (6 points) Define the capital per effective labor as k¯ = mulation equation in terms of k¯ .

K . Please derive the capital accu1 A1 − α · N

2. (6 points) Calculate the steady state capital per effective worker, k¯∗ , output per effective worker level, y¯∗ and consumption per effective worker c¯∗ . Y , solve the steady state level of output per worker y ∗ . N ∗ 4. (6 points) Derive the golden rule steady state capital per effective worker k¯G and the steady state

3. (5 points) Define output per worker as y =

level of consumption per effective worker level c¯∗G . 5. Suppose the saving rate unexpectedly increases from s to s0 in Period t∗ . (Before the shock, the economy is at the steady state.) a. (5 points) Analyze the dynamics of the convergence to the new steady state with the Solow Diagram. b. (6 points) Plot the trajectory of ∆y/y over time, before and after Period t∗ . Justify your answer. d. (6 points) Plot the trajectory of ln(Yt ), before and after time t∗ . Justify your answer. Solution 1

K

1. (6 points) We assume k =

. Then

1

A 1−α N

∆K ∆k¯ 1 ∆A ∆N = − − ¯ K 1−α A N k sY − dK 1 = − g−n K 1−α   1 sA 1−α N y 1 = − d + g + n 1 ¯ 1−α A 1−α N k  1 ¯ ⇒ ∆k¯ = sk¯α − d + g + n k. 1−α 2. (6 points) We now set ∆k = 0. Then

k =

d+
¯∗ α

∗

1 ! 1−α

s

¯∗

y¯ = k

g 1−α

(1)

+n α ! 1−α

s

=

d+

∗

g 1−α

+n α ! 1−α

s

∗

c¯ = (1 − s) y¯ = (1 − s)

d+

g 1−α

.

+n

1

3. (5 points) For y = A 1−α y¯ follows ∗

y =A

1 1−α

∗

y¯ = A

α ! 1−α

s

1 1−α

d+

g 1−α

.

+n

4. (6 points) There holds c¯∗ = (1 − s) y¯∗ . Hence, in the steady state  1 g + n k¯∗ 1−α 
α
α c¯∗ = (1 − s) k¯∗ = k¯∗ − d +

sk¯∗ =

 d+

 1 g + n k¯∗ . 1−α

The first order condition with respect to k¯∗ is α k¯∗


α−1

∗ k¯G

=d+ =

1 g+n 1−α 1 ! 1−α

α d+

1 1−α g

.

+n

Therefore, we have c¯∗G

= (1 − s) k¯∗


α

= (1 − s)

α ! 1−α

α d+

1 1−α g

+n

5. (a) (5 points) The following figure shows the change from s to s0 .

2

.

(b) (6 points) yt = y¯t · A1/(1−α) Take log of both sides: log(yt ) = log(¯ yt ) +

1 log(A) 1−α

Then take time derivatives of both sides, the growth rate of output per capita can be written as 4yt 4¯ yt 1 ∆k¯ 1 = + g=α ¯ + g. yt y¯t 1−α 1−α k As it can be seen in the following picture, at time t∗ , the growth rate of y jumps and then decreases to the original level

1 1−α g.

(c) (6 points) We can see d log Y ∆Y ∆y ∆N ∆k¯ 1 = = + =α ¯ + g + n. dt Y y N 1−α k The following picture shows the behavior of log Y .

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2

Consumption and saving with borrowing constraint [30 points]

Suppose an agent lives for three periods without initial wealth. The agent earns labor income only in the second period. The agent’s net asset position at the end of period t is at+1 , and consumption in each period ct , where t = 1, 2, 3. The agent faces the following sequence of budget constraints during her life: c1 + a2 = 0 c2 + a3 = y + (1 + r)a2

(2)

c3 = (1 + r)a3 . where y is the labor income in Period 2, r is the interest rate. The agent maximizes his life-time utility: U = u(c1 ) + βu(c2 ) + β 2 u(c3 ) where u(c) = ln(c). Assume that in the first period the agent can borrow a certain fraction of his labor income, θ. That is, c1 ≤ θ

y . 1+r

1. (6 points) Derive the Euler equation between Period 2 and 3. 2. (6 points) Now further assume that β = 1. Solve for the agent’s optimal consumption for each period, when θ = 1. 3. (6 points) Derive the savings for each period in the above question. 4. (6 points) Solve for the agent’s optimal consumption for each period, when θ < 1/3. 5. (6 points) Does Euler equation hold between Period 1 and 2? Please elaborate. Solution

4

1. (6 points) From (2) we can derive the following inter-temporal budget constraint: c1 +

y c2 c3 = + . 2 1 + r (1 + r) 1+r

(3)

The Euler equations are: u0 (c1 ) = β 2 (1 + r)2 u0 (c3 )

(4)

u0 (c2 ) = β(1 + r)u0 (c3 ).

(5)

2. (6 points) With log utility, we further have c3 = β (1 + r) c2

⇒

c1 =

c2 = c3

β2

c3 . β (1 + r)

2.

(1 + r)

When θ = 1, the consumer is not borrowing constrainted. We have c1 ≤

y 1+r

Therefore, plug (6) and (7) into (3), we will have c3 c3 c3 y + + = . 2 2 2 + r) β(1 + r) (1 + r) 1+r

β 2 (1 So that

y , (1 + β + β 2 )(1 + r) βy , c2 = 1 + β + β2 (1 + r)β 2 y . c3 = 1 + β + β2 c1 =

Further impose the assumption that β = 1, we have y , 3(1 + r) y c2 = , 3 (1 + r)y . c3 = 3 c1 =

3. (6 points) The savings for each period are: y , 3(1 + r) 2y , s2 = y2 − c2 = 3 (1 + r)y s3 = −c3 = − . 3 s1 = −c1 = −

5

(6)

(7)

4. (6 points) With c1 = θ

y , the problem becomes a two-period decision. From (3), we have 1+r c3 = (1 − θ)(1 + r)y − (1 + r)c2 .

(8)

Combine (8) and (6), we can solve for the optimal consumption at Period 3: c3 = (1 − θ) (1 + r) y − c3 =

(1 − θ) (1 + r) y . 2

Therefore, c2 =

(1 − θ) y . 2

5. (6 points) From the answers above, we know u0 (c∗1 ) =

1+r θy

(1 + r)u0 (c∗2 ) =

(1 + r)2 (1 − θ)y

(9)

(10)

Since θ < 1/3, u0 (c∗1 ) > (1 + r)u0 (c∗2 ). Euler equation does not hold. That is because the consumer cannot borrow a sufficient amount to smooth consumption.

3

Precautionary Savings

In a two-period model, suppose the agent’s lifetime utility function is U (c1 , c2 ) = u(c1 ) + βu(c2 ), where u(·) is a concave function. The market interest rate is constant, r. The agent’s incomes are y1 and y2 in Period 1 and 2, respectively. The initial wealth endowment is w0 . 1. (6 points) Derive the inter-temporal budget constraint in this case. 2. (6 points) Further assume that β = 1 and r = 0, solve for the optimal consumption (c∗1 ,c∗2 ) in Period 1 and 2. 3. (6 points) Further assume that y2 is uncertain in Period 2. y2 can be y2 h and y2 l with equal probability. Specifically, y2 h = y1 + σ > 0 and y2 l = y1 − σ > 0. Moreover, the utility function takes the following form, u(c) = ln(c). What is the agent’s optimal consumption in Period 1 (i.e., c∗∗ 1 )? 4. (6 points) Do precautionary savings increase in w0 ? Please show it carefully. 5. (6 points) Do precautionary savings increase in σ 2 ? Please show it carefully. Solution: 1. (6 points) The agent’s inter-temporal budget constraint is

6

c1 +

c2 y2 = w0 + y1 + 1+r 1+r

2. (6 points) The Euler equation is u0 (c1 ) = (1 + r) · β · u0 (c2 ) If r = 0 and β = 1, the Euler equation would be reduced to u0 (c∗1 ) = u0 (c∗2 ) Then c∗1 = c∗2 =

w0 +y1 +y2 . 2

3. (6 points) The agent’s problem becomes max U (c1 , c2 ) = u(c1 ) + β[0.5u(c2 h ) + 0.5u(c2 l )] c1

subject to

c2 h = w0 + y1 + 1+r c2 l = w0 + y1 + c1 + 1+r

c1 +

y2 h 1+r y2 l 1+r

The first-order condition is u0 (c1 ) = (1 + r)β[0.5u0 (c2 h ) + 0.5u0 (c2 l )] If β = 1 and r = 0, this condition would be reduced to u0 (c1 ) = 0.5u0 (c2 h ) + 0.5u0 (c2 l ) If the utility function u(c) = ln(c), the FOC turns out to be 1 1 1 = h+ l c1 2c2 2c2 Combine with the budget constraints ch2 = w0 + y1 − c1 + y2h cl2 = w0 + y1 − c1 + y2l . the optimal consumption in Period 1 would satisfy 2c1 2 − 3c1 (w0 + 2y1 ) + (w0 + 2y1 )2 − σ 2 = 0 7

√ 3 3 The optimal consumption c∗∗ 1 = 4 w0 + 2 y1 −

(w0 +2y1 )2 +8σ 2 . 4

4. (6 points) The level of precautionary saving is

sp =

c∗1

− c∗∗ 1

w0 3 3 = + y1 − [ w0 + y1 − 2 4 2

p p (w0 + 2y1 )2 + 8σ 2 (w0 + 2y1 )2 + 8σ 2 1 1 ] = − w0 + y1 + 4 4 2 4

Then dsp 1 1 =− + dw0 4 4

s

(w0 + 2y1 )2 < 0. (w0 + 2y1 )2 + 8σ 2

Thus, precautionary saving decreases in the initial wealth w0 . 5. (6 points) Since

dsp dσ 2

=√

1 (w0 +2y1 )2 +8σ 2

> 0, then precautionary savings increase in σ 2 .

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