(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.

  • $\begingroup$ 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. $\endgroup$ Commented 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/… $\endgroup$ Commented Mar 27, 2023 at 12:50

1 Answer 1


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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