ADVANCED MACROECONOMICS 2015
FINAL EXAMINATION FOR THE FIRST HALF OF SPRING SEMESTER
Hiroyuki Ozaki
Keio University, Faculty of Economics June 2, 2015
Important Remarks: You must write all your answers in English although I do not deduct it about the English mistake. More importantly, you need to justify your answers as much as possible. The mark depends on the correctness of your exposition.
Consider the problem of maximizing
U(0c) =U(c0, c1, c2, . . .) := lim
t→+∞c10−θ+
√ c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1 over the set of all consumption streams in the feasible set
{
0c∈R∞+ (∃1x∈R∞+
) c0+x1= 6x0 and (∀t≥1)ct+xt+1 = 6xt
}
given the parameters θ∈[0,1) and x0 >0.
Note that the utility function isrecursive with the aggregator functionW :R+×R→Rdefined by
W(c, u) =c1−θ+√
u+ 1−1.
Answer both of the following two problems. Important Hints: Notice that for anyc∈R+,
U(c, c, c, . . .) = (
1 +√
1 + 4c1−θ )2
4 −1.
(You may get some marks by proving this.) A good strategy to tackle these problems is to begin by finding a steady-state investment level, that is, an investment level which is optimal to hold forever if it is ever attained.
Problem 1. Assume that θ = 0. Derive the set of optimal consumption streams. Then, determine the value function.
Problem 2. Assume thatθ∈(0,1) and x0 = 61−θ1 /5. Derive the set of optimal consumption streams. Then, find the value of the utility function evaluated by these consumption streams.
ANSWER KEY Hiroyuki Ozaki
Keio University, Faculty of Economics June 3, 2015
In this paragraph, I prove thatU0 (the same utility function as in the exam) defined by
U0(0c) =U(c0, c1, c2, . . .) := lim
t→+∞c10−θ+
√ c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1 is recursive with the aggregator functionW :R+×R→R defined by
W(c, u) =c1−θ+√
u+ 1−1. (1)
That is, what I said right after the exam is wrong. To see this, note that U0(0c) = lim
t→+∞c10−θ+
√ c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1
= c10−θ+
√
t→lim+∞c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1
= c10−θ+
√
t→lim+∞c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1 + 1−1
= c10−θ+ vu ut lim
t→+∞
( c11−θ+
√
c12−θ+· · ·+
√
c1t−θ−1 )
+ 1−1
= c1−θ+√
U0(1c) + 1−1,
where the first line holds by the definition ofU0; the second line holds by the continuity of√
·; the third and the fourth lines are trivial; and the last line holds again by the definition of U0. This shows that U0 solves Koopmans’ equation and hence it is recursive with the aggregator function (1). Note that U0(0,0,0, . . .) =−1.
Next, I define the utility functionU1 by U1(0c) := lim
t→+∞c10−θ+
√ c11−θ+
√
c12−θ+· · ·+
√
c1t−θ+ 1−1
and I prove that U1 is recursive with the aggregator function (1). To this end, I do the same thing as above:
U1(0c) = lim
t→+∞c10−θ+
√ c11−θ+
√
c12−θ+· · ·+
√
c1t−θ+ 1−1
= c10−θ+
√
t→lim+∞c11−θ+
√
c12−θ+· · ·+
√
c1t−θ+ 1−1
= c10−θ+
√
t→lim+∞c11−θ+
√
c12−θ+· · ·+
√
c1t−θ+ 1−1 + 1−1
= c10−θ+ vu ut lim
t→+∞
( c11−θ+
√
c12−θ+· · ·+
√
c1t−θ+ 1−1 )
+ 1−1
= c1−θ+√
U1(1c) + 1−1. Note thatU1(0,0,0, . . .) = 0.
Clearly,U0 andU1are distinct from each other. Furthermore, both are recursive with the identical aggregator function (1). This means that the aggregator function dose not uniquely determine the utility function. This is something that I did not intend. I know some condition under which the aggregator function uniquely determines the utility function and I thought that this condition is now satisfied. I was wrong and I will talk more about this later.
Now I proceed in any case. To solve this maximization problem, we need to find the value of u := U(c, c, c, . . .). Since u satisfies Koopmans’ equation, we know that it must be the case that
u=c1−θ+√
u+ 1−1, from which it follows that u+ 1 = c1−θ+√
u+ 1. This is simply the quadratic equation of
√u+ 1 and it can be easily solved foru to get
u= (
1−√
1 + 4c1−θ )2
4 −1 (2)
and
u= (
1 +√
1 + 4c1−θ )2
4 −1. (3)
Both are increasing incand hence are the well-defined utility values over constant consumption streams. Furthermore, it is immediate that (2) corresponds toU0 and (3) corresponds toU1.
However, in what follows, I exploit the fact that there exists a steady-state consumption level which satisfies the Euler equation. And, to guarantee its existence, I require that the utility level corresponding to the steady-state consumption stream be given by (3). To sum up, at this very stage, the utility function need be defined by U1 instead of U0. While I do not stint my apology for this mistake, I conclude, after a moment’s reflection, that you are given enough information to solve this problem. That is, (1), (3) and the production function that are everything you need to solve this problem. So, I have decided not to give any special treatment in grading your examination papers.
Here is the answer. First, in order to derive the Euler equation, use Bellman’s equation along the line given in the lecture. This is justified because the utility function is recursive.
Bellman’s equation is
V(xt) = max{W (ct, V(xt+1))|ct+xt+1 =F(xt), ct, xt+1 ≥0}
= max{W (F(xt)−xt+1, V(xt+1))|F(xt)≥xt+1≥0} .
By assuming the interiority of the solution, the first-order necessary condition turns out to be W1(ct, ut+1) =W2(ct, ut+1)·V′(xt+1), (4) whereWi denotes the partial derivative ofW with respect to thei-th argument andut denotes U(ct, ct+1, ct+2, . . .). The value function V is known to be differentiable by the Benveniste- Scheinkman Theorem and the envelop theorem shows that its derivative V′ is given by
V′(xt) =W1(ct, ut+1)·F′(xt). (5) By combining (4) and (5), we finally reach the Euler equation for a one-sector growth model with a recursive-utility-maximizing agent:
W1(ct, ut+1) =W2(ct, ut+1)·W1(ct+1, ut+2)·F′(xt+1).
Specifying the aggregator function by (1) and the production function by F(x) = 6x leads to the Euler equation for the current problem:
c−tθ= 3 (ut+1+ 1)−1/2c−t+1θ . (6) I now find the steady-state consumption stream which satisfies the Euler equation (6).
Since (∀t)ct=c and ut+1 is given by (3), we have
1 = 3
(
1 +√
1 + 4c1−θ )2
4
−1/2
which can be easily solved for the steady-state consumption level ¯c:
¯
c= 61/(1−θ). Notice that I would get stuck here if I used (2).
Answer to Problem 1. Set θ= 0. Then, ¯c = 6 and the steady-state capital stock ¯x is given by ¯x= 6/5 from the feasibility constraint: ¯c+ ¯x= 6¯x. Define ˆt(x0) by
ˆt(x0) := min{
t≥0 6tx0>6/5} .
(I hereafter suppress (x0).) Consider the consumption stream0c∗ given by
0c∗= (0,0, . . . ,0,6ˆtx0−6/5,6,6, . . .), (7) where 6ˆtx0−6/5 is the ˆt-th period’s consumption, and the capital stream 1x∗ defined by
1x∗= (6x0,62x0, . . . ,6ˆt−1x0,6/5,6/5, . . .), where 6ˆt−1x0 is the (ˆt−1)-th period’s capital stock.
Notice that, in this particular problem, the Inada condition is not satisfied and hence a zero consumption can not be excluded. In fact, the first-order necessary condition (4) should be replaced by the inequality (recall the Kuhn-Tucker theory in the first lecture),
{ W1(ct, ut+1) =W2(ct, ut+1)·V′(xt+1) if ct>0 W1(ct, ut+1)≤W2(ct, ut+1)·V′(xt+1) if ct= 0 and hence, the Euler equation is now the Eulerinequality:
{ W1(ct, ut+1) =W2(ct, ut+1)·W1(ct+1, ut+2)·F′(xt+1) if ct>0 W1(ct, ut+1)≤W2(ct, ut+1)·W1(ct+1, ut+2)·F′(xt+1) if ct= 0. This means that the Euler equation (6) now turns out to be
{ 1 = 3 (ut+1+ 1)−1/2 if ct>0 1≤3 (ut+1+ 1)−1/2 if ct= 0.
(8)
Consider the consumption stream 0c∗ defined by (7). First, note that c∗ˆt = 6ˆtx0 −6/5 is positive by definition and hence that the first half of (8) is met. Second, note that for any t, U(tc∗) ≤U(6,6,6, . . .) because 6ˆtx0−6/5≤6. Therefore, the second half of (8) is also met. I thus conclude that (7) satisfies (8).
Next, I need to show that (7) is the unique consumption stream which satisfies (8). A lot of work might be necessary for this and I do not like to do that here. Instead, I simply assume it. (You do not lose any mark without doing this, off course.)
Finally, I show that the optimal consumption stream certainly exists by showing that the utility function (together with the given production function) is upper-convergent. (Here, the upper-convergence theorem in the lecture is invoked.) Since any consumption stream which violates the Euler equation can not be optimal and since we know that there exists an optimal consumption stream, (7) must be the unique optimal consumption stream I am seeking.
To this end, notice that the variable discount factor for the aggregator function (1) is given by (1/2) (u+ 1)−1/2. This shows that the discount factor is bounded above by a half. Fur- thermore, the higher the future utility is, the closer to zero the discount factor becomes. While the economy can grow without bound with the constant rate of growth (since the production
function is ofAk-type), the future is discounted nonlinearly. From analogy to L’Hospital’s rule, we see that far future will be less and less important for the consumer and the upper-convergence thus follows. This is only intuition behind the fact that the upper-convergence is actually satis- fied but I think that it is enough for the purpose of the exam. (I can give a formal proof that the utility function is upper-convergent given the aggregator function (1).) The value function J∗ is
J∗(x0) = U1(0,0, . . . ,0,6ˆtx0−6/5,6,6, . . .)
= W( c0, W(
c1,· · ·W(
cˆt−1, W(
ctˆ, ut+1ˆ ))
· · ·))
= W(0, W(0,· · ·W (0, W(ctˆ,8))· · ·))
= W
( 0, W
(
0,· · ·W (
0,(cˆt+ 3)1/2−1
)· · ·))
= W
( 0, W
(
0,· · ·(cˆt+ 3)1/4−1· · ·))
· · ·
= (ctˆ+ 3)(1/2)ˆt −1,
wherectˆ= 6ˆtx0−6/5. Since ˆtitself is the function ofx0,J∗ is the well-defined value function.
Answer to Problem 2. Sinceθ∈(0,1), the Inada condition requires that the consump- tion be strictly positive. Let ¯x= 61/(1−θ)/5. Then,
¯
c+ ¯x= 61/(1−θ)+ 61/(1−θ)/5 = 6·61/(1−θ)/5 = 6¯x= 6x0.
This shows that (¯c,¯c,c, . . .) and (¯¯ x,x,¯ x, . . .) are steady-state consumption and capital streams¯ which are feasible fromx0. Notice that here is no transition path and the economy stays in the steady state from the beginning. By the argument around the equation (not inequality) (6), this path clearly satisfies the Euler equation. The same logic shows that (¯c,c,¯ ¯c, . . .) is the unique optimal consumption stream from x0 and the value of the utility function is then given by
J∗ (
61/(1−θ)/5 )
=U1(¯c,c,¯ ¯c, . . .) = (
1 +√
1 + 4¯c1−θ )2
4 −1 = 8.
Important Announcement. As I mentioned above, I did not expect the aggregator function (1) generates multiple utility functions. As I said there, I though some condition which guarantees the uniqueness is satisfied but I missed something. What did I miss? If you figure out this problem completely, I will add some marks so as to upgrade your final score (for the first half of this semester), say, from “B” to “A.” Submit your result as a term paper via an email by June 20. You may like to see the book by Becker, R. E. and J. H. Boyd III, Capital Theory, Equilibrium Analysis and Recursive Utility, 1997, Blackwell.
