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Homework I

Macroeconomics II - Department of EconomicsUniversidad Carlos III de Madrid

Richard Jaimes

February 19, 2015

1. Euler Approach: Choose trajectory at. Consider the life-cycle consumption-savings problem from theclass. Derive the Euler equation in a different way now:

i. Write ct as a function of at and at+1 (argue that the budget constraint has to hold with equality) andwrite the agent’s objective as a function of (a1,..., aT ). Recall that a0 is given and fix the optimal choiceaT+1 = 0.

Solution: The consumer’s problem in a general way:

max{ct,at+1}Tt=0

T∑t=0

βtu(ct)

Subject to

at + wt − ct −at+1

R≥ 0 for all t = 0, 1, 2, ..., T

aT+1 ≥ 0

ct ≥ 0 for all t = 0, 1, 2, ..., T

a0 given

Thus, in order to write ct as a function of at and at+1 and the agent’s objective as a function of (a1,...,aT ), we need to argue that the budget constraint has to hold with equality.

Claim: If the utility function is strictly increasing, then the budget constraint is binding.

Proof : Assume the opposite is true. That is, let ct be a solution and

ct +at+1

R< at + wt for some t

Then, there exists ∆ > 0 such that:

(ct + ∆) +at+1

R= at + wt for some t

and

u(ct + ∆) > u(ct)

It follows that ct cannot be a solution. A contradiction. QED

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Then, the consumer’s problem can be written in the following manner:

max{at+1}Tt=0

T∑t=0

βtu(at + wt −

at+1

R

)Subject to

aT+1 ≥ 0

a0 given

Notice that at the optimum aT+1 = 0 and the objetive function depends only on at for all t = 1, 2, ..., T

ii. Take the first-order conditions for at, show that they are equivalent to the Euler equations that we derivedin class and interpret them.

Solution:

FOC:

(at) : =⇒ βtu′(at + wt −

at+1

R

)− βt−1

Ru′(at−1 + wt−1 −

atR

)= 0 for all t = 1, 2, ..., T

Using the budget constraint of part i: ct =(at + wt − at+1

R

), it is easy to see that:

βtu′(ct) =βt−1

Ru′(ct−1) for all t = 1, 2, ..., T

Hence,

u′(ct) = βRu′(ct+1) for all t = 0, 1, 2, ..., T

which is the same equation that we derived in class. The interpretation is straightforward: At theoptimum the marginal cost of saving AC1 at period t has to be equal to the discounted marginal benefitsof consuming ACR at t+ 1.

2. Hall’s random-walk hypothesis. Consider a consumer who lives for two periods t = 1, 2 and faces stochasticincome draws y1, y2 (which may be dependent random variables). The consumer seeks to maximize:

u(c1) + βE1 [u(c2)]

where β ∈ (0, 1), u′(c) > 0, u′′(c) < 0 for all c and limc→0u′(c) = 1. The consumer has access to saving and

borrowing at interest rate R > 1; she has to leave period 2 with non-negative assets.

i. Find the state and the value function for t = 2.

Solution:

The state for period t = 2 is given by (a, y), where a and y are the amount of assets and income level fortoday, respectively. If we define a′ as the amount of assets for tomorrow. Then, given that the consumerhas to ”leave” period 2 with a non-negative assets, a′ = 0. Hence, the value function for t = 2 is

V2(a, y) = u(a+ y)

ii. Find the state for t = 1. Write the value function V1(.) at t = 1 recursively. i.e. using V2(.).

Solution:

The state for period t = 1 is given by (a, y1), where a is the given initial level of assets for today andy1 the given initial income for today. Thus, the optimal saving rule must satisfy the budget constrainta′ = R(a+ y1 − c).

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Since c ≥ 0, the feasible set is defined as: a′ ≤ R(a+ y1). Therefore, the value function at t = 1 is givenby:

V (a, y1) = maxa′≤R(a+y1);a′∈Γ(a,y1)

u

(a+ y1 −

a′

R

)+ βE1 [V2(a′, y2)]

iii. Find the Euler equation for the consumer.

Solution:

Given that limc→0 u′(c) = ∞, we know that ct > 0. It implies that a′ < R(a + y) is an interior

solution. So, in this case we can apply the envelope theorem. Thus, by the envelope theorem we get that:δV2(a′+y2)

δa′ = u′(a′ + y2). Therefore, from ii. we will have that:

V (a, y1) = maxa′≤R(a+y1);a′∈Γ(a,y1)

u

(a+ y1 −

a′

R

)+ βE1 [V2(a′, y2)]

The first order condition with respect to a′ will be:

− 1

Ru′(a+ y1 −

a′

R

)+ βE1

[δV2(a′ + y2)

δa′

]= 0

u′(a+ y1 −

a′

R

)= RβE1 [u′(a′ + y2)]

u′ (c1) = RβE1 [u′(c2)]

The last equation is the Euler Equation in terms of consumption levels (”today and tomorrow”) where

c1 = a+ y1 − a′

R and c2 = a′ + y2.

iv. Assume that Rβ = 1. Simplify the Euler equation and interpret it.

Solution:

Assuming Rβ = 1, we get that:u′ (c1) = E1 [u′(c2)]

Its interpretation is straightforward: the cost of saving 1 unit of consumption in period 1 has to be equalto the expected benefits of consuming 1 unit more in period 2.

v. Now, assume additionally that u(c) = − 12 (c̄ − c)2; the resulting equation is called ”Hall’s random-walk

hypothesis”. Why?

Solution:

Using the fact that u(ci) = − 12 (c̄ − ci)2, we will have that u′(ci) = (c̄ − ci). Therefore, replacing this in

the Euler equation derived in the last apart:

u′ (c1) = E1 [u′(c2)]

c̄− c1 = E1 [c̄− c2]

c̄− c1 = c̄− E1 [c2]

c1 = E1 [c2]

Thus, consumption tomorrow depends only on consumption today, which implies that the consumptionstream is a random walk.

The fact that c1 = E1 [c2] means that we can write c1 = c2 + ε, where ε is an error term with zero-mean.Thus, the consumption tomorrow is equal to the consumption today plus a random term.

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vi. Hall showed that the random-walk hypothesis holds for models with any number of periods (not onlyT = 2). To test it, he ran the following regressions on data:

ct = β0 + β1ct−1 + β2ct−2 + β3ct−3 + β4ct−4 + εt

Which coefficients do we expect under the random-walk hypothesis? Which test could be used to rejectthe random-walk hypothesis?

Solution:

Bearing in mind the regression ct = β0 + β1ct−1 + β2ct−2 + β3ct−3 + β4ct−4 + εt and the random walkhypothesis, we get the following:

E(ct|ct−1, ..., ct−4) = β0 + β1ct−1 + β2ct−2 + β3ct−3 + β4ct−4

Noting that the random walk hypothesis imply that consumption tomorrow depends only on consumptiontoday (and not on previous periods), we expect β2 = β3 = β4 = 0 and β1 = 1 (in order to get the sameequation as before E(ct|ct−1) = β0 + ct−1).

To test the random walk hypothesis we need to test these last two statements. Then, it is possible toapply an F-test to the test jointly H0: β2 = 0, β3 = 0, and β4 = 0; and a t-test to test H0: β1 = 1.Rejecting one of the previous null hypothesis will imply the rejection of the random walk hypothesis.

vii. Another test Hall ran isct = β0 + β1ct−1 + β2yt−1 + εt

Which coefficients should we expect in this regression under the random-walk hypothesis and how couldwe test it?

Solution:

Applying the same reasoning as before we will have that:

E(ct|ct−1, yt−1) = β0 + β1ct−1 + β2yt−1

Using the random walk hypothesis again, we expect β2 = 0 and β1 = 1 (in order to get E(ct|ct−1) =β0 + ct−1).

To test the random walk hypothesis it is possible to apply a t-test to test H0: β2 = 0; and a t-test to testH0: β1 = 1. Rejecting one of the previous null hypothesis will imply the rejection of the random walkhypothesis.

viii. Hall could not reject the random-walk hypothesis in 1978. The later literature, however, could find pre-dictability in ∆ct = ct+1ct: Predictable changes in income were found to lead to predictable changes inconsumption, in which borrowing constraints play a crucial role. Explain how in the two-period modelabove the presence of a borrowing constraint (a2 ≥ 0) can lead to a violation of the Euler equation andhence the random-walk hypothesis.

Solution:

After the borrowing constraint a2 ≥ 0 is introduced, the Euler equation becomes:

u′ (c1) ≥ RβE1 [u′(c2)]

which holds with equality only if a2 > 0 (as in the previous case). Thus, if u′ is convex, agents willmake some precautionary savings by reducing consumption (because of the uncertainty). If we assume

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Rβ = 1 and u(ci) = − 12 (c̄− ci)2, the optimal consumption decision implies that u′ (c1) = E1 [u′(c2)] and

E1 [c2] = c1. However, if u’ is convex it follows that

u′ (c1) = u′ (E1 [(c2)])

< E1 [u′(c2)]

which contradicts the Euler equation (u′ (c1) ≥ E1 [u′(c2)]).

3. Equivalence of dynamic-programming and sequence approach in finite horizon.Consider a finite-horizon dynamic-programming problem in general form (as we did in class). Prove that the set of policiesprescribed by the dynamic-programming approach is identical to the set of policies that attain the optimumin the original sequence problem.

Solution:

Let X be an unrestricted state space and x ∈ X any element of this set. Likewise, let Ft : X ×X → R be asequence of return functions with t ∈ {0, 1, ..., T}, Γt : X → X be a a sequence of feasibility correspondences,and β > 0.

Firstly, consider the following Bellman equation for the value functions Vt : X × R with VT+1(x) = 0 in adynamic-programming approach:

Vt(x) = supx′∈Γt(x)

{Ft(x, x′) + βVt+1(x′)} for all t = 0, 1, ..., T (1)

In this case, the policy correspondence gt : X → X is given by1:

gt(x) = arg maxx′∈Γt(x)

{Ft(x, x′) + βVt+1(x′)} for all t = 0, 1, ..., T (2)

Secondly, consider the following sequence problem. Let Xs(xs) be the set of feasible sequences at times ∈ {0, 1, ..., T}. Thus,

Xs(xs) = {{xt+1}Tt=s : xt+1 ∈ Γ(xt) for all t = s, ..., T}

Therefore, the sequence problem given xs at time t = s, s = 0, ...T in order to find out an optimal plan{xt+1}Tt=s is defined as2:

V ∗s (xs) = sup{xt+1}Tt=s∈Xs(xs)

T∑t=s

βt−sFt(xt, xt+1) (3)

Proposition 1: For any given x0 ∈ X, the sequence {x∗t+1}T

t=0attains the supremum V ∗0 (xs) in (3) if and

only if x∗t+1 ∈ gt(x∗t ) in (2) for all t = 0, 1, ..., T given x∗0 = x0.

Proof :

⇐= Any sequence satisfying (2) solves the sequence problem defined by (3). Thus, given that x∗1 ∈ g0(x0),we get:

V0(x0) = F0(x0, x∗1) + βV1(x∗1)

Notice that since also x∗t+1 ∈ gt(x∗t ) for t = 1, 2, ..., T , we can write the following:

1Notice also that gt may be empty for some x if the supremum of the equation (1) is not reached.2Notice that we are only interested in the sequence problem for s = 0.

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V0(x0) = F0(x0, x∗1) + βF1(x∗1, x

∗2) + β2V2(x∗2) = ... =

T∑t=0

βtFt(x∗t , x∗t+1) (4)

Now, we need to show that there is no other feasible sequence {xt+1}Tt=1 which can reach a higher value thanV0(x0) in (3). Therefore, note that from (1) and for t = 0, we know that:

V0(x0) ≥ F0(x0, x1) + βV1(x1) for any x1 ∈ Γ0(x0)

It is easy to check that from (1) and now for t = 1, we get

V0(x0) ≥ F0(x0, x1) + βV1(x1) ≥ F0(x0) + β [F1(x1, x2), βV2(x2)]

For any {x1, x2} such that x1 ∈ Γ0(x0) and x2 ∈ Γ1(x1). So, we can conclude that:

V0(x0) ≥T∑t=0

Ft(xt, xt+1) for any {xt+1}Tt=1 ∈ X0(x0) (5)

Finally, using (5) and (4) we can argue that these two equations imply that {x∗t+1}T

t=1weakly dominates any

other feasible allocation {xt+1}Tt=1 and that {x∗t+1}T

t=1attains the supremum in (3).

=⇒ Any solution to (3) satisfies (2) for all t. Assume the opposite is true: That is, suppose that a solution

{x∗t+1}T

t=1to (3) does not satisty x∗t+1 ∈ gt(x∗t ) for some s ∈ {0, ..., T}. Hence, for some s ∈ {0, ..., T} there is

a deviation xs+1 6= x∗s+1 which is strictly better than x∗t+1:

F (x∗s, xs+1) + βVs+1(xs+1)− δ = F (x∗s, x∗s+1) + βVs+1(x∗s+1) ≥

T∑t=s

βt−sFt(x∗t , x∗t+1) with δ > 0 (6)

The last inequality follows from using (1) recursively, applying the same procedure as before in the ”onedirection” proof.

Now, we have to find a sequence {xt+1}Tt=s which comes less than δ-close to the value of F (x∗s, xs+1) +βVs+1(xs+1) in (6). Thus, we have established a contradiction since it is a sequence that dominates the

sequence {x∗t+1}T

t=s. From (1), we can find xs+2 ∈ Γs+1(xs+1) such that, for arbitrary εs+1 > 0, we will have

that:

Fs+1(xs, xs+1) + βVs+2(xs+2) > Vs+1(xs+1)− εs+1

We can keep doing this and construct, for an arbitrary sequence {εt}Tt=s+1, a feasible policy {xt+1}Tt=s ∈ Xs(x∗s)such that:

T∑t=s+1

βt−sF (xt, xt+1) > Vs+1(xs+1)−T∑

t=s+1

βt−sεt

Since εt can be arbitrarily small, we have found a sequence that gets more than δ-close to F (x∗s, xs+1) +

βVs+1(xs+1) in (6). Clearly, this implies that the sequence {x∗1, ..., x∗s, xs+1, ..., xT+1 dominates {x∗t+1}T

t=0in

the sequence problem at time t = 0, which contradicts the fact that {x∗t+1}T

t=0is a optimal sequence. QED

4. Matlab computation of finite-horizon growth model. Consider the neo-classical growth model in finitehorizon: The production function is kα and capital depreciates at the rate δ, so the law of motion for thecapital stock is:

kt+1 = (1− δ)kt + kαt − ct

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Use the parameterization α = 0.3 and δ = 0.1. The representative agent orders consumption streams by:

T∑t=1

βt ln ct

where β = 0.96 and T = 50. In order to know what are reasonable quantities of capital in this economy, wewill take the steady-state capital stock of the corresponding infinite-horizon economy:

k∗ =

(α

1β − (1− δ)

) 11−α

as a reference point.

Solution:

Find Matlab code attached and its respective graphs.

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Homework_1_Exercise_4.m%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐%% MACROECONOMICS II ‐ HOMEWORK I ‐ Ex 4. %% RICHARD JAIMES %% Finite horizon iteration growth model %%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐%clear all

%% (a)alpha = 0.3;beta = 0.96;delta = 0.1;k = (alpha/((1/beta)‐(1‐delta)))^(1/(1‐alpha));N=100;x=linspace(2*k/N,2*k)';

%% (b)f = x.^alpha+(1‐delta).*x;

%% (c) % c(i,j) = consumption when k=x(i) and k'=x(j) c=zeros(N,N);for i=1:N c(i,:)=(f(i)*ones(1,N))‐x';end

%% (d) % Let set the utility = ‐inf for consumption <= 0cneg = (c<=0);u = log(c);u(cneg) = ‐Inf;

%% (e)T=50; %% First change T arbitrary V = zeros(N,T);P = zeros(N,T);

% in period T: optimality is consuming everythingv=log(f); % f is everything availableV(:,T)=v;

%% (f)for i=T‐1:‐1:1 A=zeros(N,N); % A(a,b) is the value in period t when k=a and k'=b is chosen for j=1:N A(j,:)=u(j,:)+beta*V(:,i+1)'; %Stepback induction. end [v,p]=max(A,[],2); V(:,i)=v; % V(k,i)=max(A(k,:)) P(:,i)=p;end

%% (g)figure(1);surf(V)title('Value b=0.96 d=0.1') print GRAPHS/figure(1);

figure(2);surf(P)title('Policies b=0.96 d=0.1')print GRAPHS/figure(2);

%% (h) % See next figures (beta=0.96 / 0.98 / 0.94) (alpha=0.1 / 0.05 / 0.15).

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