<p>Okay, I know how the authors got their solution. I am just not sure it is justified. I'll present it anyway.</p>
<p>Judging by the appendix in the paper, the authors solve a deterministic problem and then slap on an Ito term and an expectation.</p>
<p>So, take the original problem, but without the expectation:</p>
<p><span class="math-container">$$
\max_{\{C_t, B_t\}} \int_0^\infty e^{-\rho t}\left\{\frac{C_t^{1- \vartheta}}{1 -\vartheta} + y_t^{1- \vartheta} v\left(\frac{B_t}{y_t}\right)\right\}\mathrm{d}t,
$$</span>
subject to
<span class="math-container">$$
C_t + \dot{B}_t \leq (R_t - G)B_t + (1-\mu)w_tn_t - T_t.
$$</span></p>
<p>This is a standard deterministic optimal control problem with FOC's (where <span class="math-container">$\lambda_t$</span> is the co-state on <span class="math-container">$B_t$</span>)</p>
<p><span class="math-container">\begin{align}
\lambda_t &amp;= C_t^{-\vartheta} = y_t^{-\vartheta}c_t^{-\vartheta},\\
\mathrm{d}\lambda_t &amp;= \lambda_t(\rho + G - R_t)\mathrm{d}t - y_t^{\vartheta}v'\left(\frac{B_t}{y_t}\right)\mathrm{d}t.
\end{align}</span>
From the first FOC, we also have by Ito's Lemma,
<span class="math-container">$$
\mathrm{d}\lambda_t = \frac{1}{2}(\vartheta \sigma)^2\lambda_t\mathrm{d}t - \vartheta\lambda_t \frac{\mathrm{d}c_t}{c_t} - \vartheta\sigma \lambda_t \mathrm{d}Z_t.
$$</span>
Plugging that into the second FOC and re-arranging, we get
<span class="math-container">$$
\vartheta \lambda_t \frac{\mathrm{d}c_t}{c_t} = \lambda_t \left(R_t - G - \rho + \frac{1}{2}\vartheta^2 \sigma^2\right)\mathrm{d}t + \lambda_t c_t^\vartheta v'\left(\frac{B_t}{y_t}\right)\mathrm{d}t - \vartheta \sigma \lambda_t \mathrm{d}Z_t,
$$</span>
or,
<span class="math-container">$$
\vartheta \mathbb{E}_t\left( \frac{\mathrm{d}c_t}{c_t}\right) = \left(R_t - G - \rho + \frac{1}{2}\vartheta^2 \sigma^2\right)\mathrm{d}t + c_t^\vartheta v'\left(b_t\right)\mathrm{d}t,
$$</span>
which is the expression you are looking for, given the correct interpretation.</p>
<p>Now, I am not fully certain this approach is justified. Since the state <span class="math-container">$B_t$</span> evolves deterministically, it might be OK to use the solution method above, since the variance in the co-state induced by the endowment fluctuations enters into the dynamics of the co-state implicitly. I don't have a formal way to justify this, though.</p>
<p>Hope this helps anyway!</p>

CC BY-SA 4.0
2023-03-27T16:29:07.240Z
2023-03-27T21:42:18.403Z
2023-03-27T21:42:18.403Z
Wittgenstein's Poker
<blockquote>
<p>(<a href="https://scholar.harvard.edu/sites/scholar.harvard.edu/files/straub/files/goldilocks.pdf" rel="nofollow noreferrer">the paper link</a>: p.32, Equation (26), (27))The output process: <span class="math-container">$d\log y_t = \sigma dZ_t$</span>. The problem is:
<span class="math-container">$$\max_{C_t, B_t} \mathbb E_0 \int_0^\infty e^{-\rho t}  \left(\frac{C_t^{1-v}}{1-v} + (y_t)^{1-v} v(\frac{B_t}{y_t}) \right) dt.$$</span> The budget constraint is
<span class="math-container">$$C_t + \dot{B_t} \le (R_t - G) B_t + (1-\mu) w_t n_t - T_t.$$</span> Let's say <span class="math-container">$b_t = B_t/y_t, c_t = C_t/y_t, w_tn_t/y_t = 1$</span>, and <span class="math-container">$T_t/y_t = \tau_t$</span>. Then, the authors say that
<span class="math-container">$$v \frac{\dot c_t}{c_t} = R_t - G - \rho + \frac12 v^2 \sigma^2 + c_t^v v'(b_t).$$</span></p>
</blockquote>
<p>I am having trouble deriving the last consumption process. My attempt is as follows. Using Ito's lemma, <span class="math-container">$dy_t = \frac{y_t}2 \sigma^2 dt + y_t \sigma dZ_t$</span>. Using Ito's lemma once again, we can get <span class="math-container">$db_t = [(R_t - G + \frac{\sigma^2}2)b_t + (1-\mu) - \tau_t - c_t] dt - b_t\sigma dZ_t$</span>. This makes <span class="math-container">$b_t$</span> a state variable (is that correct?). Then, we can write down HJB:
<span class="math-container">$$\rho V(b) = \max_{c} \left(\frac{c_t^{1-v}}{1-v} + v(b) + V'(b)[(R - G + \frac{\sigma^2}{2})b + (1-\mu) - \tau - c] + \frac{V''(b)}{2}b^2 \sigma^2 \right).$$</span>
We can have two equations from this HJB: 1) <span class="math-container">$c^{-v} = V'(b)$</span>; 2)<span class="math-container">$\rho V'(b) = v'(b) + V''(b)\mu_b(b) + V'(b)(R - G + \frac{\sigma^2}{2}) + \frac{V'''(b)}{2} b^2 \sigma^2 + V''(b) b\sigma^2$</span>, where <span class="math-container">$\mu_b(b)$</span> is the drift term of the process <span class="math-container">$b$</span>. I am stuck here. I was thinking of a guess-verify approach (say, <span class="math-container">$V(b) = K \log b$</span> for some <span class="math-container">$K$</span>), but still don't know how to handle <span class="math-container">$c^{-v} = V'(b)$</span> because this FOC says the process of <span class="math-container">$c$</span> should have the volatility term. I would appreciate if you give some help.</p>

CC BY-SA 4.0
2023-03-27T12:48:57.997Z
I don't think that solution is correct. It cannot be that consumption itself follows a deterministic process. Also, I don't even think expressions like $\dot{c}_t$ make sense in a stochastic setting in continuous time,  since $dZ_t/dt$ doesn't make sense. That being said, I cannot get to the exact expression they have, even if I assume $\dot{c}_t/c_t$ is just a stand in for the drift of the process $dc_t/c_t$. I think emailing the authors might be useful here.
Wittgenstein's Poker
CC BY-SA 4.0
2023-03-27T12:50:11.377Z
Btw, an HJB approach would be difficult here. Guessing $K \log(a)$ won't work, as clearly the objective function does not have that form. I took the Stochastic Maximum Principle approach, as briefly outlined in these notes by Brunnermeier and others: https://drive.google.com/file/d/15pfPSzyrbik8QNMp9HTjimFyQBfVN8e2/view?pli=1
Wittgenstein's Poker
<macroeconomics><mathematical-economics><optimal-control>
stochastic optimal control/derive the consumption process
shk910
# stochastic optimal control/derive the consumption process

(Asked by: username_0 on 2023-03-23)

> ([the paper link](https://scholar.harvard.edu/sites/scholar.harvard.edu/files/straub/files/goldilocks.pdf): p.32, Equation (26), (27))The output process: $d\log y\_t = \sigma dZ\_t$. The problem is:
> $$\max\_{C\_t, B\_t} \mathbb E\_0 \int\_0^\infty e^{-\rho t} \left(\frac{C\_t^{1-v}}{1-v} + (y\_t)^{1-v} v(\frac{B\_t}{y\_t}) \right) dt.$$ The budget constraint is
> $$C\_t + \dot{B\_t} \le (R\_t - G) B\_t + (1-\mu) w\_t n\_t - T\_t.$$ Let's say $b\_t = B\_t/y\_t, c\_t = C\_t/y\_t, w\_tn\_t/y\_t = 1$, and $T\_t/y\_t = \tau\_t$. Then, the authors say that
> $$v \frac{\dot c\_t}{c\_t} = R\_t - G - \rho + \frac12 v^2 \sigma^2 + c\_t^v v'(b\_t).$$

I am having trouble deriving the last consumption process. My attempt is as follows. Using Ito's lemma, $dy\_t = \frac{y\_t}2 \sigma^2 dt + y\_t \sigma dZ\_t$. Using Ito's lemma once again, we can get $db\_t = [(R\_t - G + \frac{\sigma^2}2)b\_t + (1-\mu) - \tau\_t - c\_t] dt - b\_t\sigma dZ\_t$. This makes $b\_t$ a state variable (is that correct?). Then, we can write down HJB:
$$\rho V(b) = \max\_{c} \left(\frac{c\_t^{1-v}}{1-v} + v(b) + V'(b)[(R - G + \frac{\sigma^2}{2})b + (1-\mu) - \tau - c] + \frac{V''(b)}{2}b^2 \sigma^2 \right).$$
We can have two equations from this HJB: 1) $c^{-v} = V'(b)$; 2)$\rho V'(b) = v'(b) + V''(b)\mu\_b(b) + V'(b)(R - G + \frac{\sigma^2}{2}) + \frac{V'''(b)}{2} b^2 \sigma^2 + V''(b) b\sigma^2$, where $\mu\_b(b)$ is the drift term of the process $b$. I am stuck here. I was thinking of a guess-verify approach (say, $V(b) = K \log b$ for some $K$), but still don't know how to handle $c^{-v} = V'(b)$ because this FOC says the process of $c$ should have the volatility term. I would appreciate if you give some help.

**Tags:** macroeconomics, mathematical-economics, optimal-control
> **Comments:**
> **username_1**: I don't think that solution is correct. It cannot be that consumption itself follows a deterministic process. Also, I don't even think expressions like $\dot{c}_t$ make sense in a stochastic setting in continuous time,  since $dZ_t/dt$ doesn't make sense. That being said, I cannot get to the exact expression they have, even if I assume $\dot{c}_t/c_t$ is just a stand in for the drift of the process $dc_t/c_t$. I think emailing the authors might be useful here.  
> **username_1**: Btw, an HJB approach would be difficult here. Guessing $K \log(a)$ won't work, as clearly the objective function does not have that form. I took the Stochastic Maximum Principle approach, as briefly outlined in these notes by Brunnermeier and others: https://drive.google.com/file/d/15pfPSzyrbik8QNMp9HTjimFyQBfVN8e2/view?pli=1  

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## 1 answers

### Answer by username_1 on 2023-03-27. Score: 3

Okay, I know how the authors got their solution. I am just not sure it is justified. I'll present it anyway.

Judging by the appendix in the paper, the authors solve a deterministic problem and then slap on an Ito term and an expectation.

So, take the original problem, but without the expectation:

$$
\max\_{\{C\_t, B\_t\}} \int\_0^\infty e^{-\rho t}\left\{\frac{C\_t^{1- \vartheta}}{1 -\vartheta} + y\_t^{1- \vartheta} v\left(\frac{B\_t}{y\_t}\right)\right\}\mathrm{d}t,
$$
subject to
$$
C\_t + \dot{B}\_t \leq (R\_t - G)B\_t + (1-\mu)w\_tn\_t - T\_t.
$$

This is a standard deterministic optimal control problem with FOC's (where $\lambda\_t$ is the co-state on $B\_t$)

\begin{align}
\lambda\_t &= C\_t^{-\vartheta} = y\_t^{-\vartheta}c\_t^{-\vartheta},\\
\mathrm{d}\lambda\_t &= \lambda\_t(\rho + G - R\_t)\mathrm{d}t - y\_t^{\vartheta}v'\left(\frac{B\_t}{y\_t}\right)\mathrm{d}t.
\end{align}
From the first FOC, we also have by Ito's Lemma,
$$
\mathrm{d}\lambda\_t = \frac{1}{2}(\vartheta \sigma)^2\lambda\_t\mathrm{d}t - \vartheta\lambda\_t \frac{\mathrm{d}c\_t}{c\_t} - \vartheta\sigma \lambda\_t \mathrm{d}Z\_t.
$$
Plugging that into the second FOC and re-arranging, we get
$$
\vartheta \lambda\_t \frac{\mathrm{d}c\_t}{c\_t} = \lambda\_t \left(R\_t - G - \rho + \frac{1}{2}\vartheta^2 \sigma^2\right)\mathrm{d}t + \lambda\_t c\_t^\vartheta v'\left(\frac{B\_t}{y\_t}\right)\mathrm{d}t - \vartheta \sigma \lambda\_t \mathrm{d}Z\_t,
$$
or,
$$
\vartheta \mathbb{E}\_t\left( \frac{\mathrm{d}c\_t}{c\_t}\right) = \left(R\_t - G - \rho + \frac{1}{2}\vartheta^2 \sigma^2\right)\mathrm{d}t + c\_t^\vartheta v'\left(b\_t\right)\mathrm{d}t,
$$
which is the expression you are looking for, given the correct interpretation.

Now, I am not fully certain this approach is justified. Since the state $B\_t$ evolves deterministically, it might be OK to use the solution method above, since the variance in the co-state induced by the endowment fluctuations enters into the dynamics of the co-state implicitly. I don't have a formal way to justify this, though.

Hope this helps anyway!


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