---
tags: discussion
---
# Macro HW3
{%hackmd hackmd-dark-theme %}
## Q2
### Guess and Verify

Let $A$ be the resource the agent can use at each perios, wh have the following Bellman equation,

$$
v(A) = \max_c \{ \log c + \beta v(A')\},
$$

where $A' = A-c$. Let's guess $v(A) = E + F \log(RA + w)$, and the Bellman equation becomes,

$$
v(A) = \max_c \{ \log c + \beta[E + F \log (R(A-c) + w) ]\}.
$$

F.O.C,

$$
\frac{1}{c} = \frac{\beta F R}{R(A-c) + w} \implies \beta FRc = R(A-c) +w.
$$

After some algebra,

$$
c^* = \frac{RA + w}{R(1 + \beta F)},\quad A-c^* = \frac{R \beta F A - w}{R(1+ \beta F)},\quad R(A-c^*) + w=\frac{\beta F}{1 + \beta F} (RA + w).
$$

Plugging those results into the Bellman equation,

$$
v(A) = \log (RA + w) - \log (R(1 + \beta F)) + \beta F[\log (RA +w) + \frac{\beta F}{1 + \beta F}] = E + F\log(RA + w).
$$

Comparing the coefficients,

$$
F = \frac{1}{1-\beta}, \quad E = - \log(R(1+ \beta F)) + \frac{\beta}{1-\beta} \log \beta.
$$

Hence,

$$
v(A) = - \log(R(1+ \beta F)) + \frac{\beta}{1-\beta} \log \beta + \frac{1}{1-\beta} \log(RA + w).
$$

### Wrong solution
The Euler equation is

$$
\frac{c_{t+1}}{c_t} = \beta R
$$

Hence, given $c_0$, the utility based on the optimal consumption path is,

$$
v = \sum_{t=0}^\infty \beta^t \log (c_t)\\
= \sum_{t=0}^\infty \beta^t \log (c_0 (\beta R)^t)\\
= \sum_{t=0}^\infty \beta^t [ \log c_0 + t \log \beta R]\\
= \sum_{t=0}^\infty \beta^t \log c_0 + \sum_{t=0}^\infty \beta t \log \beta R\\
= \log c_0 \frac{1}{1-\beta} + \log \beta R \frac{\beta}{(1- \beta)^2}
$$

Clearly, higher $c_0$ implies high utility, but $c_0$ cannot be arbitrarily high. There are two constraints for $c_0$. First,

$$c_0 \in [0, a_0 +w].$$

Second, $c_0$ needs to ensure the optimal consumption path is feasible, that is, $c_1 = c_0 \beta R$. We can derive it based on the budget constraint of $c_1$:

$$
c_1 \le (A_0 - c_0) R + w\\
c_0 \beta R \le (A_0 - c_0) R + w\\
c_0 (\beta R +R) \le A_0 R + w\\
c_0 \le \frac{A_0 R + w}{R(1+ \beta)}.
$$

Let $A_0 = a_0 + w$

$$
c_0 = \min \{A_0, \frac{A_0 R + w}{R(1+ \beta)}\}
$$

We can observe that

$$
A_0 = \frac{A_0 R(1+ \beta)}{R(1+ \beta)} = \frac{A_0 R + A_0 R \beta}{R(1+ \beta)},
$$

and the minization depends on $$\min \{w, A_0 R \beta \}$$.