Homework II

Macroeconomics II - Department of EconomicsUniversidad Carlos III de Madrid

Richard Jaimes

February 26, 2015

1. Finite-horizon search with dependent draws. Consider the problem of a worker who lives two periodsand has linear utility: E[w1 +βw2]. In the first period, the worker draws a high wage offer w with probability0.5 and a low wage offer w < w with probability 0.5. If she accepts the offer, she gets the respective wagein both periods. If she rejects, in the second period she draws the same offer as in period 1 with probabilityp ≥ 0.5 and the other offer with probability 1−p. When rejecting an offer, the worker receives unemploymentassistance α < w in that period.

i. State this problem as a dynamic-programming problem: Say what state, feasible set, return function,value function and policy function for t = 1, 2.

Solution:

For t = 2: The state variable: t and w, where t is time and w is the wage drawn by the worker. If Adenotes accepting the wage offer and R rejecting it, the feasible set is defined as: {A,R}. Thus, the returnfunction will be:

F (C) =

w if C = A

α if C = R

The value function V2(w), on the other hand, is then the value of having drawn offer w in period 2:

V2(w) = maxC∈{A,R}

F (C) = max{w,α}

The policy function is:

g2(w) =

A if w ≥ α

R if w < α

Since w ∈ {w,w} with α < w < w, g2(w) = A for all w, and so V2(w) = w.

For t = 1: The state variable is w, where w is the wage drawn by the worker. The feasible set {A,R} andthe return function F (w) are the same as for t = 2. Thus, in order to define the value function for period1, we have to define the value contingent on the worker’s choice:

V1(w|A) = w + βw = (1 + β)w

V1(w|R) = α+ βE[V2(w′)|w]

We then define V1(w), the value of having drawn wage w in period 1, as

V1(w) = maxC∈{A,R}

V1(w|C)

1

Clearly, the worker will choose A if V1(w|A) ≥ V1(w|R). Hence, there will exist a value w∗ such that thepolicy function can be written as:

g1(w) =

A if w ≥ w∗

R if w < w∗

which implies that if w ≥ w∗, the optimal response is to accept the wage, and to reject otherwise.

ii. State the condition when the worker should reject the low wage offer in t = 1.

Solution:

The worker will reject the low wage offer if

V1(w|R) > V1(w|A)

which is true if and only if:

α+ βE[V2(w′)|w] > (1 + β)w

Computing E[V2(w′)|w], using the fact that α < w < w̄, we get that:

E[V2(w′)|w] = pV2(w) + (1− p)V2(w̄)

= pmax{α,w}+ (1− p) max{α, w̄}= pw + (1− p)w̄

Thus, the worker reject the low wage w if:

α+ βE[V2(w′)|w] > (1 + β)w

α+ β [pw + (1− p)w̄] > (1 + β)w

w [1 + β(1− p)] < α+ β(1− p)w̄

w <α+ β(1− p)w̄1 + β(1− p)

= w∗

iii. Now, assume that w1 is drawn from general cumulative distribution function F (w) and that we havedependence of the form w2 = w2 + ρw1, where w2 is another draw from F (w) (independent of the draww1) and where ρ > 0. Write out the integrals in the decision problem at t = 1 and show that the valueof rejecting at t = 1 is increasing in w1.

Solution:

Taking into account the new conditions, the value function will be:

V1(w1) = maxC∈A,R

{V1(w1|R), V1(w1|R)}

= maxC∈A,R

{α+ βE [V2(w2)|w1] , (1− β)w1}

where

E [V2(w2)|w1] =

∫ ∞0

[V2(w2)|w1] dF (w̄2) =

∫ ∞0

V2(w2 + pw1)dF (w̄2)

andV2(w2 + pw1) = max {w̄2 + pw1, α} = max {w̄2, α− pw1}+ pw1

Therefore it is possible to rewrite V1(w1) in the following way:

V1(w1) = maxC∈A,R

{α+ β

∫ ∞0

max {w̄2, α− pw1}+ pw1 dF (w̄2) , (1− β)w1

}

2

Finally, we proceed to show that the left hand part of the maximization equation above (V1(w1|R)) isincreasing in w1:

V1(w1|R) = α+ β

∫ ∞0

max {w̄2, α− pw1}+ pw1 dF (w̄2)

= α+ β

[∫ α−pw1

0

α− pw1 dF (w̄2) +

∫ ∞α−pw1

w̄2 dF (w̄2) +

∫ ∞0

pw1 dF (w̄2)

]= α+ β

[(α− pw1) F (α− pw1) +

∫ ∞α−pw1

w̄2 dF (w̄2) + pw1

∫ ∞0

dF (w̄2)

]= α+ β

[(α− pw1) F (α− pw1) +

∫ ∞α−pw1

w̄2 + (α− pw1)− (α− pw1) dF (w̄2) + pw1[1]

]= α+ β

[(α− pw1)F (α− pw1) +

∫ ∞α−pw1

(α− pw1)dF (w̄2) +

∫ ∞α−pw1

w̄2 − α+ pw1dF (w̄2) + pw1

]= α+ β

[(α− pw1)[F (α− pw1) + 1− F (α− pw1)] +

∫ ∞α−pw1

w̄2 − α+ pw1 dF (w̄2) + pw1

]= α+ β

[(α− pw1) +

∫ ∞α−pw1

w̄2 − α+ pw1 dF (w̄2) + pw1

]= α+ β

[α+

∫ ∞α−pw1

w̄2 − α+ pw1 dF (w̄2)

]Which is clearly increasing in w1 since p > 0 and it is at the lower limit of the integral with the sign ”−”and inside the integral with the sign ”+”.

2. Transversality condition in consumption-savings problem. Consider the deterministic consumption-savings problem without an endowment stream. The budget constraint is:

at+1 ≤ R(at − ct)

Given a0 > 0, the consumer wants to maximize

∞∑t=0

βtu(ct)

where 0 < β < 1, u′ > 0, u′′ < 0 and limc→0 u′(c) =∞ and the consumer cannot borrow, i.e. at+1 ≥ 0 ∀t.

(a) State the Euler equations and transversality condition for this problem (Will we always have interiorsolutions?).

Solution:

Life-cycle consumption-savings problem

max{ct,at+1}∞t=0

∞∑t=0

βtu(ct)

Subject to:

at − ct −at+1

R≥ 0 for all t = 0, 1, ...

at+1 ≥ 0 for all t = 0, 1, ...

3

ct ≥ 0 for all t = 0, 1, ...

a0 given

In order to characterize the solution to this problem we shall use an extension of the Kuhn-Tucker method.So, we maximize the following Lagrangian function with respect to ct and at+1:

L(ct, at+1, λt, µt, γt+1) =

∞∑t=0

βtu(ct) + λt

[at − ct −

at+1

R

]+ µtct + γt+1at+1

Solving, the first-order neccesary conditions are:

(ct) : =⇒ βtuc(ct, 1− nt)− λt + µt = 0

(at+1) : =⇒ −λtR

+ λt+1 + γt+1 = 0

λt

[at − ct −

at+1

R

]= 0

µtct = 0

γt+1at+1 = 0

Given that limc→0 u′(c) = ∞, the non-negativity constraint with respect to ct never binds. Moreover,

we know that if u is strictly increasing, the resource constraint holds with equality. Likewise, notice thatat+1 > 0 for all t. To see why, suppose by contradiction that at+1 = 0, then by the resource constraintswe get that ct = at and ct+2 + at+2

R = 0, that is ct+2 = −at+2

R , which implies that ct+1 = 0, which is a con-tradiction. Therefore, with the last arguments we can say that the problem has always an interior solution.

Thus, the necessary and sufficient conditions for an optimum are:

(ct) : =⇒ βtu′(ct) = λt for all t (1)

(at+1) : =⇒ λt+1

λt=

1

Rfor all t (2)

Resource constraint : ct +at+1

R= at for all t (3)

Transversality condition : limt→∞

λt at = limt→∞

βu′(ct)at = limt→∞

βu′(at −

at+1

R

)at = 0 (4)

Using the equations (1)-(2)-(3), we can write the Euler equation associated to this problem as:

βRu′(ct+1) = u′(ct) or βRu′(at+1 −

at+2

R

)= u′

(at −

at+1

R

)for all t (5)

(b) Repeat the proof from class for sufficiency of these conditions for this specific problem; state exactly atwhich point you use each assumption.

Solution:

Consider the following problem:

4

max{at+1}∞t=0

∞∑t=0

βtu(at, at+1)

Subject to

at+1 ≥ 0 for all t

a0 given

Claim: Let u(at, at+1) be continuously differentiable, concave in (at, at+1) and strictly increasing in at.If {a∗t+1}

∞t=0

satisfies

i. a∗t ≥ 0 for all t

ii. −u′(at − at+1

R

)+ βRu′

(at+1 − at+2

R

)= 0 for all t (Euler equation)

iii. limt→∞ βu′(at − at+1

R

)at = 0 (Transversality condition)

Then, {a∗t+1}∞t=0

maximizes the objetive function.

Proof : First, consider a finite T : {at}T+1t=0 with a0 = a∗0 and aT+1 = a∗T+1. Then the Euler equations

are sufficient for a∗1, a∗2, ..., a

∗T to maximize:

U(T )(a1, ..., aT ) =

∞∑t=0

βtu(at, at+1)

Because notice that:

∂U(T )

∂at+1= βtu′

(at −

at+1

R

)(− 1

R

)+ βt+1u′

(at+1 −

at+2

R

)= 0

=⇒ −u′(at −

at+1

R

)+ βRu′

(at+1 −

at+2

R

)= 0 for all t (Euler equations)

Since U(T ) is concave (it a sum of concave functions), we can say that:

U(T )(a1, ..., aT ) ≤ U(T )(a∗1, ..., a

∗T ) +

T−1∑t=0

∂U(T )

∂at+1(at+1 − a∗t+1)

If we re-arrange the last equation, it follows that:

U(T )(a∗1, ..., a

∗T )− U(T )(a1, ..., aT ) ≥

T∑t=0

βt[−u′

(at −

at+1

R

)+ βRu′

(at+1 −

at+2

R

)](a∗t+1 − at+1)

It is easy to check that by the Euler equation for t = 1, ..., T , we get: U(T )(a∗1, ..., a

∗T ) ≥ U(T )(a1, ..., aT ).

Now, consider {at+1}∞t=0 as a optimal control sequence and {at+1}∞t=0 as an alternative feasible se-quence with a∗0 = a0 given and for all t ≥ 0, at+1 > 0. In this case, we need to show that:

D = limT→∞

T∑t=0

βt[u

(a∗t −

a∗t+1

R

)− u

(a∗t −

a∗t+1

R

)]≥ 0

Since u is concave it is possible to apply the rule u(x)− u(y) ≥ Ou(x)′(x− y). Thus, it follows that:

u

(a∗t −

a∗t+1

R

)− u

(at −

at+1

R

)≥ u′

(a∗t −

a∗t+1

R

)(a∗t − at)−

1

Ru′(a∗t −

a∗t+1

R

)(a∗t+1 − at+1)

The last equation implies that:

5

D ≥ limT→∞

T∑t=0

βt[u′(a∗t −

a∗t+1

R

)(a∗t − at)−

1

Ru′(a∗t −

a∗t+1

R

)(a∗t+1 − at+1)

]

= u′(a∗0 −

a∗1R

)(a∗0 − a0) + lim

T→∞

T−1∑t=0

βt[− 1

Ru′(a∗t −

a∗t+1

R

)+ βu′

(a∗t+1 −

a∗t+2

R

)](a∗t+1 − at+1)

− limT→∞

[βT

1

Ru′(a∗T −

a∗T+1

R

)(a∗T+1 − aT+1)

]Given that a∗0 = a0, u′

(a∗0 −

a∗1R

)(a∗0 − a0) = 0. Hence,

D ≥ limT→∞

T−1∑t=0

βt[− 1

Ru′(a∗t −

a∗t+1

R

)+ βu′

(a∗t+1 −

a∗t+2

R

)](a∗t+1 − at+1)

− limT→∞

[βT

1

Ru′(a∗T −

a∗T+1

R

)(a∗T+1 − aT+1)

]The Euler equation implies that −u′

(a∗t −

a∗t+1

R

)+ βRu′

(a∗t+1 −

a∗t+2

R

)= 0 for all t ≥ 0, thus

limT→∞∑T−1t=0 βt

[− 1Ru′(a∗t −

a∗t+1

R

)+ βu′

(a∗t+1 −

a∗t+2

R

)](a∗t+1 − at+1) = 0. Therefore,

D ≥ − limT→∞

[βT

1

Ru′(a∗T −

a∗T+1

R

)(a∗T+1 − aT+1)

]= limT→∞

[βT+1u′

(a∗T+1 −

a∗T+2

R

)(a∗T+1 − aT+1)

]by Euler Equation

= limT→∞

[βTu′

(a∗T −

a∗T+1

R

)(a∗T − aT )

]by reindexization

≥ limT→∞

[βTu′

(a∗T −

a∗T+1

R

)a∗T

]since β, u′, and a > 0

Using the transversality condition in the previous equation, we get the following:

D ≥ limT→∞

[βTu′

(a∗T −

a∗T+1

R

)a∗T

]= 0

Thus, D ≥ 0, which implies that Euler equation and Transversality condition are sufficient conditionsito assure that the interior solution is the optimal solution. Q.E.D

(c) Now, assume the special case u(c) = c1−γ

1−γ

i. Show that a constant savings rate, i.e. setting ct = (1− s)at for all t, solves the Euler equations.

Solution:

Using u(c) = c1−γ

1−γ we get that u′(c) = c−γ ; thus the Euler Equation will be

c−γt = βRc−γt+1 for all t ≥ 0

{(1− s)at}−γ = βR{(1− s)at+1}−γ =⇒ at+1

at= (βR)

1/γfor all t ≥ 0

With ct = (1− s)at and ct = at − at+1

R , we can obtain the following:

at+1

at= sR for all t ≥ 0

6

Combining the last equation with the previous result we will have:

=⇒ s = β1/γR(1−γ)/γ

The last equation shows that there is a constant saving rate that solves the Euler equation.

ii. Under which conditions on R and β is the transversality condition fulfilled? At which rates do at, ct,u′(ct) and u(ct) grow?

Solution:

From the Euler Equation we can obtain that at = (βR)t/γa0. Then, using this condition together

with u′(c) = c−γ and ct = (1− s)at, it is easy to check that the transversality condition is fulfilled ifβ1/γR(1−γ)/γ < 1:

limt→∞

βtc−γat = 0

limt→∞

βt((1− s)at)−γat = 0

limt→∞

βt(1− s)−γa1−γt = 0

limt→∞

βt(1− s)−γ[(βR)

t/γa0

]1−γ= 0

limt→∞

(β1/γR(1−γ)/γ)t(1− s)−γa1−γ0 = 0

(1− s)−γa1−γ0 lim

t→∞(β1/γR(1−γ)/γ)t = 0

In order to define at which rates at, ct, u′(ct) and u(ct) grow, we know that the grow rate of at is:

at+1

at= (βR)

1/γfor all t ≥ 0

Therefore, using the fact that ct = (1− s)at and ct+1 = (1− s)at+1, it is true that:

ct+1

ct=

(1− s)at+1

(1− s)at=at+1

at

Thus ct grow at the same rate as at:

ct+1

ct= (βR)

1/γfor all t ≥ 0

We also know that u(c) = c1−γ

1−γ . Hence, u(ct+1)u(ct)

=c1−γt+11−γc1−γt1−γ

=(ct+1

ct

)1−γ. Using the previous result the

growth rate will be:

u(ct+1)

u(ct)= (βR)

(1−γ)/γfor all t ≥ 0

Finally, we have that u′(c) = c−γ , so u′(ct+1)u′(ct)

=c−γt+1

c−γt=(ct+1

ct

)−γ. Therefore, we get the following:

u′(ct+1)

u′(ct)=(β1/γR1/γ

)−γ= (βR)

−1for all t ≥ 0

iii. For the case where the transversality condition is not fulfilled, show that a (unique) maximizing se-quence does not exist.

Solution:

Suppose that transversality condition is not fullfilled, i.e. β1/γR(1−γ)/γ ≥ 1 (from previous part).Notice that it is possible to rewrite the objective sequence in the following way:

7

∞∑t=0

βtu(ct) =

∞∑t=0

βtc1−γt

1− γ

=

∞∑t=0

βt((1− s)at)1−γ

1− γ

=

∞∑t=0

βt((1− s)(β1/γR1/γ)ta0)1−γ

1− γ

=

∞∑t=0

βt((1− s)(β1/γR1/γ)ta0)1−γ

1− γ

=

∞∑t=0

βt((1− s)a0)1−γ

1− γ(β(1/γR1/γ)t(1−γ)

=

∞∑t=0

βt((1− s)a0)1−γ

1− γ(β(1−γ)/γR(1−γ)/γ)t

=

∞∑t=0

((1− s)a0)1−γ

1− γ(β1/γR(1−γ)/γ)t

=((1− s)a0)1−γ

1− γ

∞∑t=0

(β1/γR(1−γ)/γ)t

(6)

It is easy to see that (given a0 > 0) the previous sequence is strictly monotonic if β1/γR(1−γ)/γ > 1while it is constant when β1/γR(1−γ)/γ = 1. In both case does not exists a unique maximizingsequence. Therefore it is possible to conclude that when the transversality condition is not satisfiedthe consumer maximization problem has no a (unique) solution.

3. Lagrangian approach for neo-classical growth model with labor. Consider an economy with aggregateproduction function F (kt, nt), which is constant-returns-to-scale, continuously differentiable and concave in(k, n) and where Fk(k, n) > 0 and Fn(k, n) > 0 for all (k, n). We also impose the Inada conditions:

F (0, n) = 0 limk→0

Fk(k, 1) =∞ limk→∞

Fk(k, 1) = 0

F (k, 0) = 0 limn→0

Fn(1, n) =∞ limn→∞

Fn(1, n) =∞

The representative agent has an endowment of one unit of time per period, which may be divided betweenleisure lt and work nt. The agent orders sequences of consumption and leisure by:

∞∑t=0

βtu(ct, lt)

where 0 < β < 1, uc(c, l) > 0 and ul(c, l) > 0 for all (c, l). We impose the Inada conditions:

limc→0

uc(c, l) =∞ liml→0

ul(c, l) =∞

Investment is standard:

kt+1 + ct ≤ f(kt, nt) ≡ F (kt, nt) + (1− δ)kt

i. Write down the Lagrangian for this problem – be careful to argue exactly which constraints will bind andwhich will not. Find the first-order conditions, write them in terms of the allocation {kt+1, nt, ct} and in-terpret them. Which of these conditions are intratemporal (i.e. they only involve variables corresponding

8

to a single period t) and which are intertemporal (i.e. they involve variables corresponding to more thanone period)? Did the Euler equation for consumption change with respect to the case where leisure wasnot valued?

Solution:

The Consumer’s Problem is:

max{ct,nt,kt+1}∞t=0

∞∑t=0

βtu(ct, 1− nt)

Subject to

ct + kt+1 ≤ f(kt, nt) for all t

ct ≥ 0 for all t

kt+1 ≥ 0 for all t

1 ≥ nt ≥ 0 for all t

k0 given

In order to characterize the solution to this problem we shall use an extension of the Kuhn-Tucker method.So, we maximize the following Lagrangian function with respect to ct, nt and kt+1:

L(ct, nt, kt+1, λt) =

∞∑t=0

βtu(ct, 1− nt) + λt [f(kt, nt)− ct − kt+1] + µtct + γt+1kt+1

Solving, the first-order neccesary conditions are:

(ct) : =⇒ βtuc(ct, 1− nt)− λt + µt = 0

(nt) : =⇒ βtun(ct, 1− nt)− λtfn(kt, nt) = 0

(kt+1) : =⇒ −λt + λt+1fk(kt+1, nt+1) + γt+1 = 0

λt [f(kt, nt)− ct − kt+1] = 0

µtct = 0

γt+1kt+1 = 0

Given the Inada conditions defined as: limc→0 uc(c, l) = ∞, liml→0 ul(c, l) = ∞, limk→0 Fk(k, 1) = ∞,limn→0 Fn(1, n) =∞, F (k, 0) = 0 and F (0, n) = 0, the non-negativity constraints never bind. Moreover,we know that if u is strictly increasing in its arguments, the resource constraint holds with equality. Thus,the necessary and sufficient conditions for an optimum are:

(ct) : =⇒ βtuc(ct, 1− nt) = λt (7)

(nt) : =⇒ βtun(ct, 1− nt) = λtfn(kt, nt) (8)

9

(kt+1) : =⇒ λt = λt+1fk(kt+1, nt+1) (9)

Transversality condition : limt→∞

λt kt = 0 (10)

Feasibility : ct + kt+1 = f(kt, nt) (11)

Using the equations (6), (7) and (8), we can rewrite these conditions in the following way:

uc(ct+1, 1− nt+1)

uc(ct, 1− nt)=

1

βfk(kt+1, nt+1)(12)

un(ct, 1− nt) = uc(ct, 1− nt)fn(kt, nt) (13)

Interpretation of the above equations:

i. Equation (11): This is the intertemporal condition relating the trade-off between consumption andinvestment. It implies that the marginal cost of investing (decreasing consumption) today has to beequal to the discount marginal benefits of being able to consume more tomorrow times the amountof additional consumption tomorrow due to the increasing in fk(kt+1, nt+1). Notice that if R =fk(kt+1, nt+1) we obtain the same Euler equation calculated from the case where leisure is not valued.

ii. Equation (12): This is the intratemporal condition relating the trade-off between leisure and labor.It implies that the marginal disutility from labor supply has to be equal to the marginal benefit ofconsuming one unit more times the productivity of labor. In other words, the ratio between themarginal disutility from labor supply and the marginal utility of consumption has to be equal to thereal wage if we define w = fn(kt, nt).

ii. From these first-order conditions, find a minimal set of equations that characterize a steady state wherekt = k, ct = c and lt = l for all t.

Solution: To characterize the steady state we suppose that for all t:

kt = k (14)

ct = c (15)

nt = n (16)

Combining equations (10)-(15), we will have that:

fk(k, n) =1

β(17)

un(c, 1− n) = uc(c, 1− n)fn(k, n) (18)

c+ k = f(k, n) (19)

which is a system of three equations in three unknows (c, k, l). Notice that for all t, lt = 1 − nt, so insteady state it follows that n = 1− l.

iii. Now, assume that there is a final period T. State the problem as a dynamic-programming problem: Whatis the state, the return function, the feasible set for the choice variables (or controls), the value functionand the policy function? Hint: You will have to adapt the framework from class slightly here since theset of choice variables at t cannot be reduced to the state in t+ 1.

Solution:

10

(a) Period T :

• State variables: k

• Control variables: k′ and n

• The return function:

u(f(k, n)− k′, n) where k′ = f(k, n)− c (20)

• The feasible set: Γ(k) = {(k′, n) : f(k, n) ≥ k′ ≥ 0, 1 ≥ n ≥ 0}• The value function:

VT (k) = max(k′,n)∈Γ(k)

u(f(k, n)− k′, n) (21)

• The policy function:

gT (k) = 0 (22)

(b) Periods t = 0, 1, 2, ..., T − 2, T − 1:

• State variables: k

• Control variables: k′ and n

• The return function:

u(f(k, n)− k′, n) where k′ = f(k, n)− c (23)

• The feasible set: Γ(k) = {(k′, n) : f(k, n) ≥ k′ ≥ 0, 1 ≥ n ≥ 0}• The value function:

VT (k) = max(k′,n)∈Γ(k)

{u(f(k, n)− k′, n) + βVt+1(k′)} (24)

• The policy function:

gT (k) = arg max(k′,n)∈Γ(k)

{u(f(k, n)− k′, n) + βVt+1(k′)} (25)

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