3
$\begingroup$

(the paper link: 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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Mar 27, 2023 at 12:48
  • $\begingroup$ 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: drive.google.com/file/d/15pfPSzyrbik8QNMp9HTjimFyQBfVN8e2/… $\endgroup$ Mar 27, 2023 at 12:50

1 Answer 1

3
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.