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14.462 Advanced Macroeconomics
Spring 2004
Problem Set 6
(due May 7)
Problem 1
The economy is populated by a continuum of measure one of agents, indexed by i and
uniformly distributed over the [0, 1] interval. Agents are risk neutral with utility
1
ui = Aki − k
2
2
i
i
2
k2
iswhere ki is the individual investment of agent i, A is the return to investment, and
the cost of investment. Let
K =
1
0
kidi
denote the aggregate level of investment. The return to investment is given by
A = (1 − α)θ + αK
where α ∈
α > 0 there is a complementarity in that the return to individual investment is increasing
0,
1
2
The random variable θ parametrizes the fundamentals of the economy. If

in the aggregate level of investment, and the parameter α captures the degree of comple­ 

mentarity.

The fundamentals θ are not known at the time investment decisions are made. Further­ 

more, agents have heterogeneous beliefs about θ. The common prior is uniform over R.

Agent i has private information

xi = θ + σxξi
and there is public information
y = K + σyu.
The random variables ξi, i ∈ [0, 1] and u are standard normal and independent as well as
independent of θ. The precisions of the two sources of information are denoted as πx = σ−2
,
and πy = σ−2
, respectively. Let social welfare be given by a utilitarian aggregatory
� 1
w = uidi.
0
1

x
Submit Assignment For Help
Go To Answer Directly
info@economicshomeworkhelper.com
https://www.economicshomeworkhelper.com 1
1. Check that it is an equilibrium for investment to takes the form
ki = βxi + (1 − β)y.
Determine the coefficient β and describe how it varies with the degree of complemen­
tarity α and the two precisions πx and πy. Provide an intuitive explanation of your
findings.
2. How do heterogeneity Var(ki θ, y) and volatility Var(K θ) vary with the three param­| |
eters α, πx and πy. Provide intuition.
3. Show that social welfare conditional on fundamentals E[w θ] is a linear function of|
heterogeneity and volatility. Use your previous results to discuss how social welfare
varies with the parameters and provide intuition for your results.
4. Now suppose there is a second source of public information

z = θ + σzε

where ε is standard normal and independent of θ, u and the ξi, i ∈ [0, 1]. How are
the answers to parts 1.-3. affected?
Problem 2 (A simple Model of Savings)

Consider the problem of a consumer who wants to maximize the following program:

max
{ct}∞
t=0
E
∞�
t=0
βt c1−γ
t
1 − γ
(1)
s.t. (i) wt = et + Rtbt
wt = ct + bt+1 (2)
(ii) wo > 0
The endowment shock et and the interest rate Rt are i.i.d. Don’t worry about the non-
negativity constraint on consumption.
1. Rewrite the problem in recursive form.
2. Without solving for the value function, derive the first order condition (FOC) and
the envelope condition (EC). Combine the two to obtain the Euler Equation (EE).
3. Assume in this section that the endowment shock is 0 in all periods. Make a guess
for a value function. Using this guess, derive the consumption function. Using the
EE, solve for the constant. Then replace back into the Bellman equation, and verify
that you indeed found the value function.
2
4. Assume now that the endowment shock is stochastic, and that the interest rate is
deterministic and Rt = R. Use the EE to analyze consumption growth. What
happens if Rβ = 1? What is the R that makes expected consumption growth zero?
Discuss the implications of uninsurable risk.
5. Assume that Rt is stochastic, and that the consumer is the representative agent of
the economy. Assume that endowment is stochastic, and that the asset is in zero net
supply. Use the EE to price the asset when there is only aggregate risk. Discuss (but
do not solve) the case with only idiosyncratic risk.
6. Now assume that there is no uncertainty, that the interest rate is constant, and the
endowment shock 0. Solve for the value function, and the optimal consumption and
wealth path as a function of initial wealth. Using the optimal consumption path
that you derived from the recursive approach, derive the value function by replacing
consumption in expected utility. What condition in γ do you need to make sure that
the solution is indeed optimal? Discuss.
3
14.462 Advanced Macroeconomics
Spring 2004
Problem Set 6 Solution
Problem 1
1. The guess

ki = βxi + (1 − β)y.

implies

K = βθ + (1 − β)y.

Combining this with the relationship y = K + σyu yields

1

y = θ + σuu
β

Thus as a signal of θ the precision is β2
πy. Define

πx

φ =
πx + β2πy
Then we have
E[θ xi, y] = φxi + (1 − φ)y.|
Optimal investment is given by
ki = E[A xi, y] = (1 − α)E[θ xi, y] + αE[K xi, y]| | |
= [(1 − α) + αβ]E[θ xi, y] + α(1 − β)y|
Substituting for E[θ xi, z, y] we have|
ki = [(1 − α) + αβ]φxi + [[(1 − α) + αβ](1 − φ) + α(1 − β − γ)] y
Matching coefficients we get
β = [(1 − α) + αβ]φ
or equivalently
β
φ =
(1 − α) + αβ
1
Combining this with the equation definition φ gives the condition
β πx
=
(1 − α) + αβ πx + β2πy
The left hand side is increasing in β with a range [0, 1] while the right hand side is
πx
decreasing with a range
�
πx+πy
, 1
�
. Thus there is a unique solution for β. The left
hand side is increasing in α, so β is decreasing in α. The right hand side is increasing
in πx and decreasing in πy. It follows that β is increasing in πx and decreasing in πy,
that is
β = B(α, πx, πy)
with Bα < 0, Bπx > 0 and Bπy < 0. For future purposes it is useful to compute the
elasticities with respect to πx and πy explicitly. We get
Bπx (α, πx, πy)πx
=
Bπy (α, πx, πy)πy 1
<
1
= .
B(α, πx, πy)
−
B(α, πx, πy) (1−α) (πx+β2πy)2
+ 2 2
[(1−α)+αβ]2 βπxπy
What is the intuition for these results. If α increases, then complementarities are
stronger, and agents put more weight on the public signal since it helps predict what
others will do. Higher precision of the private signal induces agents to put more weight
on the private signal and higher precision of the signal about K induces agents to
put more weight on this public signal. However, notice one difference to the paper
by Angeletos and Pavan. If you increases α, this makes it more attractive to put
more weight on the public signal. But if agents put more weight on the public signal,
this makes the public signal less informative about θ, which makes it less attractive
to put weight on the public signal, partially offsetting the initial effect. Thus all the
effects on β are muted in comparison to Angeletos and Pavan. Why is the elasticity
with respect to πx and πy less than 1
in absolute value. Suppose we increase πx by2
one percent and β increases by more that 0.5 percent. Then relative precision of the
public signal y actually increases, in which case agents would not have wanted to put
more weight on the private signal in the first place. Similarly, suppose we increase πy
by one percent. If β decreases by more than 0.5 percent, than relative precision of
the public signal actually decreases, but in this case agents would not have wanted
to put more weight on the public signal in the first place.
It is also instructive to consider how φ depends on the parameters. We have
(1 − α)φ
β = (1)
1 − αφ
Substituting into the definition of φ gives the condition
πx
φ = .�
(1−α)φ
�2
πx + πy1−αφ
2
Again there is a unique solution
φ = Φ(α, πx, πy)
with Φα > 0, Φπx > 0 and Φπy < 0. Equation (1) implies
B(α, πx, πy) ≤ Φ(α, πx, πy)
with strict inequality if α > 0 and clearly the wedge is increasing in α.
2. We have
β2
Var(ki θ, y) = Var(βxi + (1 − β)y θ, y) =| |
πx
so heterogeneity as a function of parameters is given by
B(α, πx, πy)2
H(α, πx, πy) =
πx
It is decreasing in α and πy. Both higher α and higher πy induce agents to put less
weight on the private signal, and less weight on the private signal translates into less
heterogeneity. If πx increases, this directly reduces heterogeneity. But agents also
become more responsive to the private signal, which tends to increase heterogeneity.
But since the elasticity is less than 1
2
, we know that this does not overturn the direct
effect, and so heterogeneity falls. This differs from Angeletos and Pavan, where the
overall effect is ambiguous.
We have
1
Var(K θ) = Var((1 − β)y θ) = Var
�
1 − β
σyu
�
�
�
� θ
�
=
�
1 − β
�2
| |
β β πy
Thus volatility as a function of the parameters is given by
�
1 − B(α, πx, πy)
�2
1
V (α, πx, πy) =
B(α, πx, πy) πy
Clearly volatility is increasing in α and decreasing in πx. Higher α induces agents
to put more weight on the public signal, increasing volatility. Higher precision of
the private signal does the opposite. The effect of an increase in the precision πy is
more complicated. The direct effect is to reduce volatility. There are two indirect
effects, both related to the fact that agents become more responsive to the public and
thus less responsive to the private signal. Higher responsiveness to the public signal
increases volatility. This effect is also present in Angeletos and Pavan. In addition,
less responsiveness to the private signal reduces the precision of y as a signal about
θ, partially offsetting the increase in πy and thus increasing in volatility. Volatility
3
1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
π
y
V(α,1,π
y
)
α=0
α=0.25
α=0.5
Figure 1: Volatility as a function of πy
as a function of πy is analyzed in figure 1. Only the ratio of πx and πy matters for
the shape, so I restrict attention to the case πx = 1. Thus the figure shows volatility
as function of πy given πx = 1, and the graph is plotted for different values of α.
Of course one finds that higher α is associated with higher volatility. Volatility is
initially increasing in πy but eventually becomes decreasing.
3. By definition � 1
w = uidi.
0
Substituting the formula for ui = Aki − 1
k2
yields2 i
1� 1
1
� 1
w = A kidi − ki
2
di = AK −
1
�
ki
2
di
2 00 2 0
4
,
Since
� 1
k2
di =
�
0
1
(ki − K)2
di + K2
this can be written as0 i
1
�� 1 �
w = AK − (ki − K)2
di + K2
.
2 0
Substituting A = (1 − α)θ + αK yields
1
1
�� �
w = [(1 − α)θ + αK] (ki − K)2
di + K2
K −
2 0
1 1
� 1
= (1 − α)θK − (1 − 2α) K2
(ki − K)2
di.
2
−
2 0
Now notice that ki − K = β(xi − θ) and so
� 1
β2
(ki − K)2
di = β2
σ2
= .x
0 πx
Thus
1 1 β2
E[w θ] = (1 − α)θE[K θ] − (1 − 2α) E[K2
θ] .| |
2
| −
2 πx
Using the facts that E[K θ] = θ and E[K2
θ] = Var(K θ) + θ2
, this becomes| | |
1
E[w θ] = (1 − α)θ2
− (1 − 2α)
�
Var(K θ) + θ2
� 1 β2
|
2
| −
2 πx
1 1
�
β2
�
= (1 − 2α)Var(K θ) + .−
2
θ2
−
2
|
πx
Now recall from part 2. that Var(ki θ, y) = β2
. Using this factπx
|
1 1
E[w θ] = [(1 − 2α)Var(K θ) + Var(ki θ, y)] .| −
2
θ2
−
2
| |
So we can analyze welfare by looking at
Ω(α, πx, πy) = (1 − 2α)V (α, πx, πy) + H(α, πx, πy).
Since both volatility and heterogeneity are decreasing in πx, we immediately get
that Ω(α, πx, πy) is decreasing in πx. Thus making private information more precise
is unambiguously good for welfare. This is different from Angeletos and Pavan.
There more precise private information meant less uncertainty at the expense of lower
coordination, with ambiguous overall effects on welfare. But here precise private
5
information is also vital for the informativeness of the public signal and thus for
coordination. So it makes sense that here the effect is unambiguous.
β πx
=
(1 − α) + αβ πx + β2πy
β
�
πx + β2
πy
�
= πx [(1 − α) + αβ]⇐⇒
⇐⇒ ββ2
πy = πx(1 − α)(1 − β)
β2
= (1 − α)
(1 − β) 1
⇐⇒
πx β πy
Thus
�
(1 − β)
�2
1 β2
Ω(α, πx, πy) = (1 − 2α) +
β πy πx
(1 − 2α)
�
(1 − β)
�
β2
β2
=	 +
β πx πx
β2
1
�
−
(1
α
− 2α)(1 − β) + (1 − α)β
�
=
πx (1 − α)β
β
�
(1 − 2α) + αβ
�
=
πx (1 − α)
β
�
α
�
=
πx
1 − (1 − β)
1 − α
The condition α < 1
is sufficient for Ω(α, πx, πy) to be positive. Thus the last rela­2
tionship implies that Ω(α, πx, πy) is increasing in β for given πx. Since β is decreasing
in πy, it follows that making public information more precise also increases welfare.
Also notice that the right hand side is decreasing in α for given β and πx. Since
β is decreasing in α, it follows that Ω(α, πx, πy) is also decreasing in α. Making
complementarities stronger improves welfare.
4. For this part I will not try to sign derivatives analytically.	 Instead I derive the relevant
formulas and perform a limited numerical evaluation.
Now start with the guess
ki = βxi + γz + (1 − β − γ)y.
This implies

K = βθ + γz + (1 − β − γ)y.

6
do ,
Combining this with the relationship y = K + σyu yields
β γ 1
y = θ + z + σuu
β + γ β + γ β + γ
To obtain the information provided by y beyond what is provided by z define
(β + γ)y − γz 1
y˜ = = θ + σyu
β β
Thus we get an additional signal of precision β2
πy. Define
πz
δ =
πx + πz + β2πy
πx
φ =
πx + πz + β2πy
Then we have
E[θ xi, z, y] = φxi + δz + (1 − φ − δ)˜y|
= φxi + δz + (1 − φ − δ)
(β + γ)y − γz
β
�
γ
�
(β + γ)
= φxi + δ − (1 − φ − δ) z + (1 − φ − δ) y
β β
Optimal investment is given by
ki = E[A xi, z, y] = (1 − α)E[θ xi, z, y] + αE[K xi, z, y]| | |
= (1 − α)E[θ xi, z, y] + αE[K xi, z, y]| |
= [(1 − α) + αβ]E[θ xi, z, y] + αγz + α(1 − β − γ)y|
Substituting for E[θ xi, z, y] we have|
ki = [(1 − α) + αβ]φxi
� �
γ
� �
+ [(1 − α) + αβ] δ − (1 − φ − δ) + αγ z
β
� �
(β + γ)
� �
+ [(1 − α) + αβ] (1 − φ − δ) + α(1 − β − γ) y
β
Matching coefficients we get
β = [(1 − α) + αβ]φ
� �
γ
� �
γ = [(1 − α) + αβ] δ − (1 − φ − δ) + αγ
β
7
Eliminating φ, we now get the following equation for β

β πx
=
(1 − α) + αβ πx + πz + β2πy
Again there is a unique solution
β = B(α, πx, πz, πy)
with Bα < 0, Bπx > 0, Bπz < 0, Bπy < 0.
From the condition defining γ we get
�
β
�
γ
� �
γ = δ − (1 − φ − δ) + αγ
φ β
γ [φ(1 − α) + (1 − φ − δ)] = βδ
βπz
γ =
πx(1 − α) + β2πy
Thus

B(α, πx, πz, πy)πz

γ = C(α, πx, πy, πz) =
πx(1 − α) + B(α, πx, πz, πy)2πy
I will not try to sign the derivatives but instead do some limited numerical evaluation.
This is done in Figure 2, and the results contain nothing unexpected. An increase
in the degree of complementarity leads to an increase in the weight on z, as does an
increase in its own precision, while higher precision of the other signals reduces the
weight on z. Finally the coefficient on y is given by
1 − β − γ = D(α, πx, πy, πz) ≡= 1 − B(α, πx, πy, πz) − C(α, πx, πy, πz)
Figure 3 provides a limited numerical evaluation of the properties of D. Notice that
more complementarity does not necessarily lead to an increase in the weight on y.
This makes sense, since now there is an alternative public signal available. As α
increases, more weight is put on public information, but as the weight on private
information shrinks, y becomes less and less precise as a signal about θ, and thus at
high levels of α the signal z is the more attractive public signal.
Heterogeneity is once again
B(α, πx, πy, πz)2
H(α, πx, πy, πz) = ,
πx
but volatility is more complicated
Var(K θ) = Var(γz + (1 − β − γ)y θ)| |
8
Do
�
Substituting y yields
� �
β γ 1
��
�
�
�
�
Var(K θ) = Var γz + (1 − β − γ) θ + z + σuu θ|
β + γ β + γ β + γ
�
γ 1 − β − γ
�
�
�
�
�
= Var z + σu θ
β + γ β + γ
�
γ
�2
1
�
1 − β − γ
�2
1
= +
β + γ πz β + γ πy
Thus
�
C(α, πx, πy, πz)
�2
1
�
D(α, πx, πy, πz)
�2
1
V (α, πx, πy, πz) = +
1 − D(α, πx, πy, πz) πz 1 − D(α, πx, πy, πz) πy
Figure 4 provides a limited evaluation of volatility. Here it turns out that πy reduces

volatility.

Again we can analyze welfare by looking at

Ω(α, πx, πy, πz) = (1 − 2α)V (α, πx, πy, πz) + H(α, πx, πy, πz).
Figure 5 provides a limited numerical evaluation. Notice that an increase in πz can
reduce welfare.
Problem 2 (A simple Model of Savings)
1. The specification of the problem should of course include γ > 0 and γ = 1. Combining
the budget constraints, future wealth is

w�
= R�
(w − c) + e�

and so the Bellman equation is
�
c1−γ
�
V (w) = max
1 − γ
+ βE[V (R�
(w − c) + e�
)] .
c
2. The first order condition is

c−γ
= βE[R�
V �
(R�
(w − c) + e�
)]

and the envelope condition is

V �
(w) = βE[R�
V �
(R�
(w − c) + e�
)].

Thus V �
(w) = c−γ
and we obtain the Euler equation

� �
c�
�−γ
�
1 = βE R�
.
c
9
3. A guess that will work is
w1−γ
V (w) = a

1 − γ

for some a > 0.

The objective of the recursive problem then reduces to

c1−γ
R1−γ (w − c)1−γ
¯+ βa
1 − γ 1 − γ
1
¯where R = E[(R�
)1−γ
]1−γ is the certainty equivalent of R�
. The first order condition
becomes
¯c−γ
= βaR1−γ
(w − c)−γ

and so

1
c = w
¯1 1−γ
γ1 + (βa)γ R

which yields

1
γ R γ(βa) ¯
1−γ
w1
γ R γ
w − c =
1 + (βa) ¯
1−γ

Replacing into the objective yields the maximized value

1
γ
�
1
+ βaR1−γ [(βa)γ R¯
1−γ
]1−γ
�
w1−γ
¯
1 1
γ[1 + (βa)γ R γ ¯
1−γ
]1−γ 1 − γ¯
1−γ
]1−γ [1 + (βa)γ R
1
�
1 1 ¯
1−γ
]1−γ
�
w1−γ
¯
1−γ [(βa)γ R γ
γ=
γ R γ
+ [(βa)γ R ]γ
¯
1−γ
]1−γ 1 − γ1 1
γ[1 + (βa) ¯
1−γ
]1−γ [1 + (βa)γ R
1
¯
1−γ
�
w1−γγ
�
1 (βa)γ R
= +1 1
γ[1 + (βa)γ R γ ¯
1−γ
]1−γ 1 − γ¯
1−γ
]1−γ [1 + (βa)γ R
1
¯
1−γ
]γ w1−γ
γ= [1 + (βa)γ R

1 − γ

For our guess to be correct we need
1
γ R γγa = [1 + (βa) ¯
1−γ
] .
¯This equation has a unique positive solution if βR1−γ
< 1, and then it is given by
1 1−γ
�−γ
γ R¯ γa =
�
1 − β .
4. With Rβ = 1 the Euler equation becomes
c−γ
= Et[c−γ
t t+1]
10
�
and as marginal utility is strictly convex and endowment shocks are nondegenerate
Et[ct+1] > ct.
Thus Rβ < 1 is needed to obtain zero expected consumption growth.
5. The way the budget constraints are written the price of the asset is normalized to
one. Then it is more convenient to write wt = ct + ptbt+1 and wt = et + dtbt where
now dt ≥ 0 is a given i.i.d. dividend and the price pt has to adjust in equilbrium.
The Euler equation is then
e−γ
p = βE[d�
(e�
)−γ
]

and so the price is a function of the current endowment:

p(e) = eγ
βE[d�
(e�
)−γ
]
A high endowment today implies low expected consumption growth, which makes
transferring resources into the future attractive, so the price of the asset has to be
high for no trade to be an equilibrium.
If there is only idiosyncratic risk, then we have an endowment economy version of
Aiyagari. No trade is then of course not an equilibrium and the interest rate will
depend on the wealth distribution. In steady state we of course must have Rβ < 1.
6. Here we have a special case of part 3. If βR1−γ
< 1, then
1 ¯R γ
1−γ
)wt
1
γ
1
γ γ¯
1−γ
R
wt = (1 − βct =
1 + (βa)
and
1
γwt+1 = R(wt − ct) = (βR) wt,
so
wt =
�
(βR)
1
γ
�t
w0
and
)
�
(βR)
�t
γ¯
1−γ
R
1 1
γct = (1 − β γ w0.
Substituing into the utility function yields
1−γ
t
1−γ
0
∞ ∞� �
β(βR) γ
1−γ
�tc γ¯
1−γ
R )1−γ w
1 − γ
1
βt
(1 − β γ=
1 − γ
t=0 t=0
γ
γ
¯
1−γ
1−γ
R
1
γ
1−γ
0
¯R )1−γ
(1 − β w
= 1
γ 1 − γ(1 − β )
1−γ
0γ
1−γ
�
1 − β
γ
1−γ
w0
= a .
1 − γ
11
�1−γ w1
γ ¯R=
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.35
0.4
0.45
0.5
0.55
α
C(α,1,1,1)
1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
π
x
C(0.25,πx
,1,1)
1 2 3 4 5 6 7 8 9 10
0.1
0.2
0.3
0.4
0.5
0.6
πy
C(0.25,1,πy
,1)
1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
πz
C(0.25,1,1,πz
)
Figure 2: Properties of C(α, πx, πy, πz)
12
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.165
0.17
α
D(α,1,1,1)
1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
π
x
D(0.25,πx
,1,1)
1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
π
y
D(0.25,1,π
y
,1)
1 2 3 4 5 6 7 8 9 10
0
0.1
0.2
0.3
0.4
π
z
D(0.25,1,1,π
z
)
Figure 3: Properties of D(α, πx, πy, πz)
13
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.2
0.3
0.4
0.5
0.6
0.7
α
V(α,1,1,1)
1 2 3 4 5 6 7 8 9 10
0
10
20
30
40
π
x
V(0.25,π
x
,1,1)
1 2 3 4 5 6 7 8 9 10
0.3
0.4
0.5
0.6
0.7
π
y
V(0.25,1,πy
,1)
1 2 3 4 5 6 7 8 9 10
0
10
20
30
40
π
z
V(0.25,1,1,π
z
)
Figure 4: Properties of V (α, πx, πy, πz)
14
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
0.1
0.2
0.3
0.4
0.5
α
Ω(α,1,1,1)
1 2 3 4 5 6 7 8 9 10
0
5
10
15
20
π
x
Ω(0.25,π
x
,1,1)
1 2 3 4 5 6 7 8 9 10
0.2
0.25
0.3
0.35
0.4
0.45
0.5
πy
Ω(0.25,1,πy
,1)
1 2 3 4 5 6 7 8 9 10
0
5
10
15
20
πz
Ω(0.25,1,1,π
z
)
Figure 5: Properties of Ω(α, πx, πy, πz)
15

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Economics Assignment Sample Problems Set 6 with Solutions

  • 1. � � � . 14.462 Advanced Macroeconomics Spring 2004 Problem Set 6 (due May 7) Problem 1 The economy is populated by a continuum of measure one of agents, indexed by i and uniformly distributed over the [0, 1] interval. Agents are risk neutral with utility 1 ui = Aki − k 2 2 i i 2 k2 iswhere ki is the individual investment of agent i, A is the return to investment, and the cost of investment. Let K = 1 0 kidi denote the aggregate level of investment. The return to investment is given by A = (1 − α)θ + αK where α ∈ α > 0 there is a complementarity in that the return to individual investment is increasing 0, 1 2 The random variable θ parametrizes the fundamentals of the economy. If in the aggregate level of investment, and the parameter α captures the degree of comple­ mentarity. The fundamentals θ are not known at the time investment decisions are made. Further­ more, agents have heterogeneous beliefs about θ. The common prior is uniform over R. Agent i has private information xi = θ + σxξi and there is public information y = K + σyu. The random variables ξi, i ∈ [0, 1] and u are standard normal and independent as well as independent of θ. The precisions of the two sources of information are denoted as πx = σ−2 , and πy = σ−2 , respectively. Let social welfare be given by a utilitarian aggregatory � 1 w = uidi. 0 1 x Submit Assignment For Help Go To Answer Directly info@economicshomeworkhelper.com https://www.economicshomeworkhelper.com 1
  • 2. 1. Check that it is an equilibrium for investment to takes the form ki = βxi + (1 − β)y. Determine the coefficient β and describe how it varies with the degree of complemen­ tarity α and the two precisions πx and πy. Provide an intuitive explanation of your findings. 2. How do heterogeneity Var(ki θ, y) and volatility Var(K θ) vary with the three param­| | eters α, πx and πy. Provide intuition. 3. Show that social welfare conditional on fundamentals E[w θ] is a linear function of| heterogeneity and volatility. Use your previous results to discuss how social welfare varies with the parameters and provide intuition for your results. 4. Now suppose there is a second source of public information z = θ + σzε where ε is standard normal and independent of θ, u and the ξi, i ∈ [0, 1]. How are the answers to parts 1.-3. affected? Problem 2 (A simple Model of Savings) Consider the problem of a consumer who wants to maximize the following program: max {ct}∞ t=0 E ∞� t=0 βt c1−γ t 1 − γ (1) s.t. (i) wt = et + Rtbt wt = ct + bt+1 (2) (ii) wo > 0 The endowment shock et and the interest rate Rt are i.i.d. Don’t worry about the non- negativity constraint on consumption. 1. Rewrite the problem in recursive form. 2. Without solving for the value function, derive the first order condition (FOC) and the envelope condition (EC). Combine the two to obtain the Euler Equation (EE). 3. Assume in this section that the endowment shock is 0 in all periods. Make a guess for a value function. Using this guess, derive the consumption function. Using the EE, solve for the constant. Then replace back into the Bellman equation, and verify that you indeed found the value function. 2
  • 3. 4. Assume now that the endowment shock is stochastic, and that the interest rate is deterministic and Rt = R. Use the EE to analyze consumption growth. What happens if Rβ = 1? What is the R that makes expected consumption growth zero? Discuss the implications of uninsurable risk. 5. Assume that Rt is stochastic, and that the consumer is the representative agent of the economy. Assume that endowment is stochastic, and that the asset is in zero net supply. Use the EE to price the asset when there is only aggregate risk. Discuss (but do not solve) the case with only idiosyncratic risk. 6. Now assume that there is no uncertainty, that the interest rate is constant, and the endowment shock 0. Solve for the value function, and the optimal consumption and wealth path as a function of initial wealth. Using the optimal consumption path that you derived from the recursive approach, derive the value function by replacing consumption in expected utility. What condition in γ do you need to make sure that the solution is indeed optimal? Discuss. 3
  • 4. 14.462 Advanced Macroeconomics Spring 2004 Problem Set 6 Solution Problem 1 1. The guess ki = βxi + (1 − β)y. implies K = βθ + (1 − β)y. Combining this with the relationship y = K + σyu yields 1 y = θ + σuu β Thus as a signal of θ the precision is β2 πy. Define πx φ = πx + β2πy Then we have E[θ xi, y] = φxi + (1 − φ)y.| Optimal investment is given by ki = E[A xi, y] = (1 − α)E[θ xi, y] + αE[K xi, y]| | | = [(1 − α) + αβ]E[θ xi, y] + α(1 − β)y| Substituting for E[θ xi, z, y] we have| ki = [(1 − α) + αβ]φxi + [[(1 − α) + αβ](1 − φ) + α(1 − β − γ)] y Matching coefficients we get β = [(1 − α) + αβ]φ or equivalently β φ = (1 − α) + αβ 1
  • 5. Combining this with the equation definition φ gives the condition β πx = (1 − α) + αβ πx + β2πy The left hand side is increasing in β with a range [0, 1] while the right hand side is πx decreasing with a range � πx+πy , 1 � . Thus there is a unique solution for β. The left hand side is increasing in α, so β is decreasing in α. The right hand side is increasing in πx and decreasing in πy. It follows that β is increasing in πx and decreasing in πy, that is β = B(α, πx, πy) with Bα < 0, Bπx > 0 and Bπy < 0. For future purposes it is useful to compute the elasticities with respect to πx and πy explicitly. We get Bπx (α, πx, πy)πx = Bπy (α, πx, πy)πy 1 < 1 = . B(α, πx, πy) − B(α, πx, πy) (1−α) (πx+β2πy)2 + 2 2 [(1−α)+αβ]2 βπxπy What is the intuition for these results. If α increases, then complementarities are stronger, and agents put more weight on the public signal since it helps predict what others will do. Higher precision of the private signal induces agents to put more weight on the private signal and higher precision of the signal about K induces agents to put more weight on this public signal. However, notice one difference to the paper by Angeletos and Pavan. If you increases α, this makes it more attractive to put more weight on the public signal. But if agents put more weight on the public signal, this makes the public signal less informative about θ, which makes it less attractive to put weight on the public signal, partially offsetting the initial effect. Thus all the effects on β are muted in comparison to Angeletos and Pavan. Why is the elasticity with respect to πx and πy less than 1 in absolute value. Suppose we increase πx by2 one percent and β increases by more that 0.5 percent. Then relative precision of the public signal y actually increases, in which case agents would not have wanted to put more weight on the private signal in the first place. Similarly, suppose we increase πy by one percent. If β decreases by more than 0.5 percent, than relative precision of the public signal actually decreases, but in this case agents would not have wanted to put more weight on the public signal in the first place. It is also instructive to consider how φ depends on the parameters. We have (1 − α)φ β = (1) 1 − αφ Substituting into the definition of φ gives the condition πx φ = .� (1−α)φ �2 πx + πy1−αφ 2
  • 6. Again there is a unique solution φ = Φ(α, πx, πy) with Φα > 0, Φπx > 0 and Φπy < 0. Equation (1) implies B(α, πx, πy) ≤ Φ(α, πx, πy) with strict inequality if α > 0 and clearly the wedge is increasing in α. 2. We have β2 Var(ki θ, y) = Var(βxi + (1 − β)y θ, y) =| | πx so heterogeneity as a function of parameters is given by B(α, πx, πy)2 H(α, πx, πy) = πx It is decreasing in α and πy. Both higher α and higher πy induce agents to put less weight on the private signal, and less weight on the private signal translates into less heterogeneity. If πx increases, this directly reduces heterogeneity. But agents also become more responsive to the private signal, which tends to increase heterogeneity. But since the elasticity is less than 1 2 , we know that this does not overturn the direct effect, and so heterogeneity falls. This differs from Angeletos and Pavan, where the overall effect is ambiguous. We have 1 Var(K θ) = Var((1 − β)y θ) = Var � 1 − β σyu � � � � θ � = � 1 − β �2 | | β β πy Thus volatility as a function of the parameters is given by � 1 − B(α, πx, πy) �2 1 V (α, πx, πy) = B(α, πx, πy) πy Clearly volatility is increasing in α and decreasing in πx. Higher α induces agents to put more weight on the public signal, increasing volatility. Higher precision of the private signal does the opposite. The effect of an increase in the precision πy is more complicated. The direct effect is to reduce volatility. There are two indirect effects, both related to the fact that agents become more responsive to the public and thus less responsive to the private signal. Higher responsiveness to the public signal increases volatility. This effect is also present in Angeletos and Pavan. In addition, less responsiveness to the private signal reduces the precision of y as a signal about θ, partially offsetting the increase in πy and thus increasing in volatility. Volatility 3
  • 7. 1 2 3 4 5 6 7 8 9 10 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 π y V(α,1,π y ) α=0 α=0.25 α=0.5 Figure 1: Volatility as a function of πy as a function of πy is analyzed in figure 1. Only the ratio of πx and πy matters for the shape, so I restrict attention to the case πx = 1. Thus the figure shows volatility as function of πy given πx = 1, and the graph is plotted for different values of α. Of course one finds that higher α is associated with higher volatility. Volatility is initially increasing in πy but eventually becomes decreasing. 3. By definition � 1 w = uidi. 0 Substituting the formula for ui = Aki − 1 k2 yields2 i 1� 1 1 � 1 w = A kidi − ki 2 di = AK − 1 � ki 2 di 2 00 2 0 4 ,
  • 8. Since � 1 k2 di = � 0 1 (ki − K)2 di + K2 this can be written as0 i 1 �� 1 � w = AK − (ki − K)2 di + K2 . 2 0 Substituting A = (1 − α)θ + αK yields 1 1 �� � w = [(1 − α)θ + αK] (ki − K)2 di + K2 K − 2 0 1 1 � 1 = (1 − α)θK − (1 − 2α) K2 (ki − K)2 di. 2 − 2 0 Now notice that ki − K = β(xi − θ) and so � 1 β2 (ki − K)2 di = β2 σ2 = .x 0 πx Thus 1 1 β2 E[w θ] = (1 − α)θE[K θ] − (1 − 2α) E[K2 θ] .| | 2 | − 2 πx Using the facts that E[K θ] = θ and E[K2 θ] = Var(K θ) + θ2 , this becomes| | | 1 E[w θ] = (1 − α)θ2 − (1 − 2α) � Var(K θ) + θ2 � 1 β2 | 2 | − 2 πx 1 1 � β2 � = (1 − 2α)Var(K θ) + .− 2 θ2 − 2 | πx Now recall from part 2. that Var(ki θ, y) = β2 . Using this factπx | 1 1 E[w θ] = [(1 − 2α)Var(K θ) + Var(ki θ, y)] .| − 2 θ2 − 2 | | So we can analyze welfare by looking at Ω(α, πx, πy) = (1 − 2α)V (α, πx, πy) + H(α, πx, πy). Since both volatility and heterogeneity are decreasing in πx, we immediately get that Ω(α, πx, πy) is decreasing in πx. Thus making private information more precise is unambiguously good for welfare. This is different from Angeletos and Pavan. There more precise private information meant less uncertainty at the expense of lower coordination, with ambiguous overall effects on welfare. But here precise private 5
  • 9. information is also vital for the informativeness of the public signal and thus for coordination. So it makes sense that here the effect is unambiguous. β πx = (1 − α) + αβ πx + β2πy β � πx + β2 πy � = πx [(1 − α) + αβ]⇐⇒ ⇐⇒ ββ2 πy = πx(1 − α)(1 − β) β2 = (1 − α) (1 − β) 1 ⇐⇒ πx β πy Thus � (1 − β) �2 1 β2 Ω(α, πx, πy) = (1 − 2α) + β πy πx (1 − 2α) � (1 − β) � β2 β2 = + β πx πx β2 1 � − (1 α − 2α)(1 − β) + (1 − α)β � = πx (1 − α)β β � (1 − 2α) + αβ � = πx (1 − α) β � α � = πx 1 − (1 − β) 1 − α The condition α < 1 is sufficient for Ω(α, πx, πy) to be positive. Thus the last rela­2 tionship implies that Ω(α, πx, πy) is increasing in β for given πx. Since β is decreasing in πy, it follows that making public information more precise also increases welfare. Also notice that the right hand side is decreasing in α for given β and πx. Since β is decreasing in α, it follows that Ω(α, πx, πy) is also decreasing in α. Making complementarities stronger improves welfare. 4. For this part I will not try to sign derivatives analytically. Instead I derive the relevant formulas and perform a limited numerical evaluation. Now start with the guess ki = βxi + γz + (1 − β − γ)y. This implies K = βθ + γz + (1 − β − γ)y. 6 do ,
  • 10. Combining this with the relationship y = K + σyu yields β γ 1 y = θ + z + σuu β + γ β + γ β + γ To obtain the information provided by y beyond what is provided by z define (β + γ)y − γz 1 y˜ = = θ + σyu β β Thus we get an additional signal of precision β2 πy. Define πz δ = πx + πz + β2πy πx φ = πx + πz + β2πy Then we have E[θ xi, z, y] = φxi + δz + (1 − φ − δ)˜y| = φxi + δz + (1 − φ − δ) (β + γ)y − γz β � γ � (β + γ) = φxi + δ − (1 − φ − δ) z + (1 − φ − δ) y β β Optimal investment is given by ki = E[A xi, z, y] = (1 − α)E[θ xi, z, y] + αE[K xi, z, y]| | | = (1 − α)E[θ xi, z, y] + αE[K xi, z, y]| | = [(1 − α) + αβ]E[θ xi, z, y] + αγz + α(1 − β − γ)y| Substituting for E[θ xi, z, y] we have| ki = [(1 − α) + αβ]φxi � � γ � � + [(1 − α) + αβ] δ − (1 − φ − δ) + αγ z β � � (β + γ) � � + [(1 − α) + αβ] (1 − φ − δ) + α(1 − β − γ) y β Matching coefficients we get β = [(1 − α) + αβ]φ � � γ � � γ = [(1 − α) + αβ] δ − (1 − φ − δ) + αγ β 7
  • 11. Eliminating φ, we now get the following equation for β β πx = (1 − α) + αβ πx + πz + β2πy Again there is a unique solution β = B(α, πx, πz, πy) with Bα < 0, Bπx > 0, Bπz < 0, Bπy < 0. From the condition defining γ we get � β � γ � � γ = δ − (1 − φ − δ) + αγ φ β γ [φ(1 − α) + (1 − φ − δ)] = βδ βπz γ = πx(1 − α) + β2πy Thus B(α, πx, πz, πy)πz γ = C(α, πx, πy, πz) = πx(1 − α) + B(α, πx, πz, πy)2πy I will not try to sign the derivatives but instead do some limited numerical evaluation. This is done in Figure 2, and the results contain nothing unexpected. An increase in the degree of complementarity leads to an increase in the weight on z, as does an increase in its own precision, while higher precision of the other signals reduces the weight on z. Finally the coefficient on y is given by 1 − β − γ = D(α, πx, πy, πz) ≡= 1 − B(α, πx, πy, πz) − C(α, πx, πy, πz) Figure 3 provides a limited numerical evaluation of the properties of D. Notice that more complementarity does not necessarily lead to an increase in the weight on y. This makes sense, since now there is an alternative public signal available. As α increases, more weight is put on public information, but as the weight on private information shrinks, y becomes less and less precise as a signal about θ, and thus at high levels of α the signal z is the more attractive public signal. Heterogeneity is once again B(α, πx, πy, πz)2 H(α, πx, πy, πz) = , πx but volatility is more complicated Var(K θ) = Var(γz + (1 − β − γ)y θ)| | 8 Do
  • 12. � Substituting y yields � � β γ 1 �� � � � � Var(K θ) = Var γz + (1 − β − γ) θ + z + σuu θ| β + γ β + γ β + γ � γ 1 − β − γ � � � � � = Var z + σu θ β + γ β + γ � γ �2 1 � 1 − β − γ �2 1 = + β + γ πz β + γ πy Thus � C(α, πx, πy, πz) �2 1 � D(α, πx, πy, πz) �2 1 V (α, πx, πy, πz) = + 1 − D(α, πx, πy, πz) πz 1 − D(α, πx, πy, πz) πy Figure 4 provides a limited evaluation of volatility. Here it turns out that πy reduces volatility. Again we can analyze welfare by looking at Ω(α, πx, πy, πz) = (1 − 2α)V (α, πx, πy, πz) + H(α, πx, πy, πz). Figure 5 provides a limited numerical evaluation. Notice that an increase in πz can reduce welfare. Problem 2 (A simple Model of Savings) 1. The specification of the problem should of course include γ > 0 and γ = 1. Combining the budget constraints, future wealth is w� = R� (w − c) + e� and so the Bellman equation is � c1−γ � V (w) = max 1 − γ + βE[V (R� (w − c) + e� )] . c 2. The first order condition is c−γ = βE[R� V � (R� (w − c) + e� )] and the envelope condition is V � (w) = βE[R� V � (R� (w − c) + e� )]. Thus V � (w) = c−γ and we obtain the Euler equation � � c� �−γ � 1 = βE R� . c 9
  • 13. 3. A guess that will work is w1−γ V (w) = a 1 − γ for some a > 0. The objective of the recursive problem then reduces to c1−γ R1−γ (w − c)1−γ ¯+ βa 1 − γ 1 − γ 1 ¯where R = E[(R� )1−γ ]1−γ is the certainty equivalent of R� . The first order condition becomes ¯c−γ = βaR1−γ (w − c)−γ and so 1 c = w ¯1 1−γ γ1 + (βa)γ R which yields 1 γ R γ(βa) ¯ 1−γ w1 γ R γ w − c = 1 + (βa) ¯ 1−γ Replacing into the objective yields the maximized value 1 γ � 1 + βaR1−γ [(βa)γ R¯ 1−γ ]1−γ � w1−γ ¯ 1 1 γ[1 + (βa)γ R γ ¯ 1−γ ]1−γ 1 − γ¯ 1−γ ]1−γ [1 + (βa)γ R 1 � 1 1 ¯ 1−γ ]1−γ � w1−γ ¯ 1−γ [(βa)γ R γ γ= γ R γ + [(βa)γ R ]γ ¯ 1−γ ]1−γ 1 − γ1 1 γ[1 + (βa) ¯ 1−γ ]1−γ [1 + (βa)γ R 1 ¯ 1−γ � w1−γγ � 1 (βa)γ R = +1 1 γ[1 + (βa)γ R γ ¯ 1−γ ]1−γ 1 − γ¯ 1−γ ]1−γ [1 + (βa)γ R 1 ¯ 1−γ ]γ w1−γ γ= [1 + (βa)γ R 1 − γ For our guess to be correct we need 1 γ R γγa = [1 + (βa) ¯ 1−γ ] . ¯This equation has a unique positive solution if βR1−γ < 1, and then it is given by 1 1−γ �−γ γ R¯ γa = � 1 − β . 4. With Rβ = 1 the Euler equation becomes c−γ = Et[c−γ t t+1] 10
  • 14. � and as marginal utility is strictly convex and endowment shocks are nondegenerate Et[ct+1] > ct. Thus Rβ < 1 is needed to obtain zero expected consumption growth. 5. The way the budget constraints are written the price of the asset is normalized to one. Then it is more convenient to write wt = ct + ptbt+1 and wt = et + dtbt where now dt ≥ 0 is a given i.i.d. dividend and the price pt has to adjust in equilbrium. The Euler equation is then e−γ p = βE[d� (e� )−γ ] and so the price is a function of the current endowment: p(e) = eγ βE[d� (e� )−γ ] A high endowment today implies low expected consumption growth, which makes transferring resources into the future attractive, so the price of the asset has to be high for no trade to be an equilibrium. If there is only idiosyncratic risk, then we have an endowment economy version of Aiyagari. No trade is then of course not an equilibrium and the interest rate will depend on the wealth distribution. In steady state we of course must have Rβ < 1. 6. Here we have a special case of part 3. If βR1−γ < 1, then 1 ¯R γ 1−γ )wt 1 γ 1 γ γ¯ 1−γ R wt = (1 − βct = 1 + (βa) and 1 γwt+1 = R(wt − ct) = (βR) wt, so wt = � (βR) 1 γ �t w0 and ) � (βR) �t γ¯ 1−γ R 1 1 γct = (1 − β γ w0. Substituing into the utility function yields 1−γ t 1−γ 0 ∞ ∞� � β(βR) γ 1−γ �tc γ¯ 1−γ R )1−γ w 1 − γ 1 βt (1 − β γ= 1 − γ t=0 t=0 γ γ ¯ 1−γ 1−γ R 1 γ 1−γ 0 ¯R )1−γ (1 − β w = 1 γ 1 − γ(1 − β ) 1−γ 0γ 1−γ � 1 − β γ 1−γ w0 = a . 1 − γ 11 �1−γ w1 γ ¯R=
  • 15. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.35 0.4 0.45 0.5 0.55 α C(α,1,1,1) 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 π x C(0.25,πx ,1,1) 1 2 3 4 5 6 7 8 9 10 0.1 0.2 0.3 0.4 0.5 0.6 πy C(0.25,1,πy ,1) 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 πz C(0.25,1,1,πz ) Figure 2: Properties of C(α, πx, πy, πz) 12
  • 16. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.165 0.17 α D(α,1,1,1) 1 2 3 4 5 6 7 8 9 10 0 0.05 0.1 0.15 0.2 π x D(0.25,πx ,1,1) 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 π y D(0.25,1,π y ,1) 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 π z D(0.25,1,1,π z ) Figure 3: Properties of D(α, πx, πy, πz) 13
  • 17. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.2 0.3 0.4 0.5 0.6 0.7 α V(α,1,1,1) 1 2 3 4 5 6 7 8 9 10 0 10 20 30 40 π x V(0.25,π x ,1,1) 1 2 3 4 5 6 7 8 9 10 0.3 0.4 0.5 0.6 0.7 π y V(0.25,1,πy ,1) 1 2 3 4 5 6 7 8 9 10 0 10 20 30 40 π z V(0.25,1,1,π z ) Figure 4: Properties of V (α, πx, πy, πz) 14
  • 18. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0 0.1 0.2 0.3 0.4 0.5 α Ω(α,1,1,1) 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 π x Ω(0.25,π x ,1,1) 1 2 3 4 5 6 7 8 9 10 0.2 0.25 0.3 0.35 0.4 0.45 0.5 πy Ω(0.25,1,πy ,1) 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 πz Ω(0.25,1,1,π z ) Figure 5: Properties of Ω(α, πx, πy, πz) 15