Problem Set #2Mphil Macroeconomics, M200

Michaelmas, 2011.

Question 1

(i) Using a variational argument, one can show that

c−θt =1 + r

1 + ρEt[c

−θt+1]

(ii) If ln ct+1 is normally distributed, N (Et[ct+1], σ2), then −θ ln ct+1 is normally dis-

tributed with N (−θEt[ct+1], θ2σ2). As a consequence, we can write the Euler

equation as

c−θt =1 + r

1 + ρe−θEt[ln ct+1]+ θ2σ2

2

Taking logs yields

−θ ln ct = ln(1 + r)− ln(1 + ρ)− θ[Et[ln ct+1] +θ2σ2

2]

(iii) Rearranging the expression above yields

Et[ln ct+1] =ln(1 + r)− ln(1 + ρ)

θ+θσ2

2+ ln ct

⇒ ln ct+1 =ln(1 + r)− ln(1 + ρ)

θ+θσ2

2+ ln ct + ut+1

(iv) Rearranging the expression for Et[ln ct+1] gives

Et[ln ct+1]− ln ct =ln(1 + r)− ln(1 + ρ)

θ+θσ2

2

An increase in r increases consumption growth. An increase in σ2 also increases

consumption growth. The reason is that the third derivative of the momentary

utility function is positive.

Question 2

(i) The third derivative is given by γ3e−γc, which is positive.

(ii) Consumption in period two is given by C2 = Y1 + Y2 − C1. As a consequence,

conditional on C1, C2 is normally distributed with N (Y1 + Y2 − C1, σ2).

Expected lifetime utility is given by

−e−γC1 − E[e−γ(Y1 + Y2 − C1)]

2

The term logarithm of e−γ(Y1 + Y2 − C1) is N (−γ(Y1 + Y2 − C1), γ2σ2). Thus,

using the hint from the previous question reveals that

E[e−γ(Y1 + Y2 − C1)] = e−γ(Y1+Y2−C1)+ γ2σ2

2

As a consequence, lifetime utility is given by

−e−γC1 − e−γ(Y1+Y2−C1)+ γ2σ2

2

(iii) The first order condition of the above expression is

γe−γC1 = γe−γ(Y1+Y2−C1)+ γ2σ2

2

Taking logs and simplifying yields

C1 =Y1 + Y2

2− γσ2

4

As a consequence, an increase in σ2, decreases C1. This is due to precautionary

savings.

Question 3

(i) The Lagrangian is

L = max{it,kt+1}∞t=0

∞∑t=0

e−rt{π(Kt)kt − it − C(it)− qt(kt+1 − (1− δ)kt − it)}

(ii) The first order conditions are

qt = 1 + C ′(it) (it)

qt = e−r(π(Kt+1) + qt+1(1− δ)) (kt+1)

(iii) The problem is now

L = max{it,kt+∆}∞t=0

∞∑t=0

e−rt{∆(π(Kt)kt − it − C(it))− qt(kt+∆ − (1−∆δ)kt −∆it)}

with first order conditions

qt = 1 + C ′(it) (it)

qt = e−∆r(∆π(Kt+∆) + qt+1(1−∆δ)) (kt+∆)

(iv) The first order conditions in continuous time is

qt = 1 + C ′(it)

(r + δ)qt = π(Kt) + qt

If δ = 0, these are exactly the same as those in the textbook

3

(v) A positive δ has the same effect as a permanent increase in the interest rate