14.462 Advanced Macroeconomics
Spring 2004
Problem Set 4 Solution
1. It is easy to check that pt = εt is the (bubbleless) REE equilibrium. Next consider the
learning process. Substituting the model equation into the learning equation yields
pet+1 = pte +
1
εt−1
(αpte + εt−1 − pet ) = ht pet +
.
gt
gt
By induction one obtain
pet+1 = p0e
t
�
t+1
t
�
εs−2 �
hs +
hτ
g
s−1
s=0
s=1
τ =s
Using this formula and the fact that the εt are iid, we can compute
t
t+1
t
t
�
�
�
�
�
2
1
� 2
e 2
h2s + σ 2
E
pet+1 − Et,REE pt+1 = (pe0 )2
h
=
(p
)
h2s + σ 2 m
t
τ
0
2
g
s=0
s=1 s−1 τ =s
s=0
2. Necessity is obvious. To show sufficiency notice that mt ≥
insures that the term in front of pe0 vanishes as t → ∞.
�t
τ =1
h2τ , so limt→∞ mt = 0
3. We have
h2t+1 mt +
=
h2t+1
1
2
gt+1
t+1
�
t
1 �
s=1
2
gs−1
τ =s
h2τ
+
1
2
gt+1
t+1
t+1
t+2
t+1
�
�
1
� 2
1
1
� 2
=
hτ + 2 =
hτ = mt+1
g2
gt+1
g2
s=1 s−1 τ =s
s=1 s−1 τ =s
4. First notice that mt ≥ g12 so limt→∞ mt = 0 clearly requires that limt→∞ gt = +∞.
t
�t
2
Also mt ≥ g21
h
τ =s τ for all s ∈ {1, . . . , t + 1}. It follows that limt→∞ mt = 0
s−1
requires
t
�
lim
hτ = 0
t→∞
τ =s
for all s ≥ 1, which is stronger than the requirement stated in the problem set since
it must hold for all s ≥ 1. Finally if α ≥ 1, then ht ≥ 1 for all t which is inconsistent
�
with limt→∞ tτ =s hτ = 0. Thus it must be the case that α < 1.
1
5. Suppose
lim
t
�
t→∞
for all s ≥ 1. The goal is to show that
hτ = 0
τ =s
�+∞
0
1
gt
= +∞.
for n ≥ 0. We want to
6. Pick T such that t ≥ T implies 1−α
< 12 and define an = g1−α
qt
T +n
�∞
�∞
show that n=0 (1 − an ) = 0 if and only if n=0 an = +∞. This follows from some
results on convergence of infinite products, see the book Mathematical Analysis, 2ed
by Apostol, pp. 206–209.
One direction is not so difficult: as 1 − an ≤ e−an , we have
�∞
�∞
− n=0 an
. At least to me it seems that the other direction is a bit
n=0 (1 − an ) ≤ e
more involved, but perhaps you found an easy proof. Otherwise take a look at the
book by Apostol.
�+∞ 1
1
= +∞
7. If gt is a constant, then limt→∞ gt = +∞ fails. If gt = (t+1)
2 , then
0
gt
fails.
8.
2