My Questions
Consider the following problem. It is almost identical to the classic Merton portfolio choice problem. Here I'm solving it using the so-called Martingale method. I have provided my attempt at a derivation. I have three questions:
- Is this correct?
- Why does it seem like the consumption path is stochastic? I understand that we can perhaps interpret this problem as identical to the classic Merton problem where the agent has some starting wealth $W_0 > 0$. We can do this by saying that $W_0 = \int_0^\infty \pi_t w dt$. However, why would the agent simple not choose $C_t = w$? My suspicion is that this depends on the relative values of $\rho$ and the interest rate $r$.
- Under what conditions would $C_t = w$, if ever?
Problem setup
An agent has initial wealth $W_0 = 0$ but receives a constant stream of wages $w$. There is a riskless asset that pays the interest rate $r$ and a risky security that follows the dynamics $$ \frac{dS}{S} = \mu_S dt + \sigma_S dB_t, $$ where $B_t$ is a standard brownian motion.
Now, the agent has CRRA utility. Thus, his decision is modelled by the following program:
\begin{align*} \max_{\{C_t\}_{t=0}^\infty} \quad \mathbb E\left[\int_0^\infty e^{-\rho t} \left ( \frac{C_t^{1-\gamma}}{1 - \gamma} \right ) \, \mathrm d t \right ] \\ \text{ s.t. } \mathbb E \left [ \int_0^\infty \pi_t (c_t - w) dt \right ] \leq W_0, \end{align*} where $\pi_t$ is the stochastic discount factor, which can be written as $$ \frac{d \pi_t}{\pi_t} = - \mu_{\pi} dt - \sigma_{\pi} d B_t. $$
My solution attempt
Proceeding with the Martingale method, the first-order condition on the appropriate Lagrangian are $$ u_c(c_t, t) = e^{-\rho t} C_t^{-\gamma} = \lambda \pi_t, $$ where $\pi = e^{-r t} \xi_t$, $\lambda$ is the Lagrange multiplier, and the exponential martingale is $\xi_t = \exp\left (-\eta B_t - \frac{t}{2} \eta^2 \right )$. Note that this is based on the assumption of complete markets and is equivalent to $$ \frac{d \pi_t}{\pi_t} = - r dt - \eta d B_t, $$ where $\eta = \frac{\mu_s - r}{\sigma_s}$ is the market price of risk.
The first-order condition implies that $C^*_t = \left( \lambda \pi_t e^{\rho t} \right )^{-1/\gamma}$. We can then substitute this into the budget constraint and solve for $\lambda$: \begin{align*} W_0 &= \mathbb E \int_0^\infty \pi_t (C_t^* - w) \, \mathrm d t \\ 0 &= \mathbb E \int_0^\infty \pi_t^{\frac{\gamma - 1}{\gamma}} \lambda ^{\frac{-1}{\gamma}} \exp(-\rho t/\gamma) - \pi_t w \, \mathrm d t. \end{align*}
Now, in order to proceed, let us make the following intermediate calculations: \begin{align*} \mathbb E_0[\pi_t] &= \exp\{-r t\} \\ &\text{and} \\ \mathbb E_0 \left [\pi_t^{\frac{\gamma - 1}{\gamma}} \right ] &= \mathbb E_0 \exp \left \{ - \frac{\gamma - 1}{\gamma} \left (r + \frac 12 \eta^2 \right ) t - \frac{\gamma - 1}{\gamma} \eta B(t) \right \} \\ &= \exp \left \{ - \frac{\gamma - 1}{\gamma} (r + \frac 12 \eta^2 ) t + \frac 12 \frac{(\gamma - 1)^2}{\gamma^2} \eta^2 t \right \} \\ &= \exp \left \{ -t \frac{\gamma - 1}{\gamma} \left[ r + \frac 12 \eta^2 \frac{1}{\gamma} \right] \right \}. \end{align*} Because of the appropriate regularity conditions, we can exchange the order of integration to make the following calculations: \begin{align*} \mathbb E \int_0^\infty \pi_t w \, \mathrm d t &= w \int_0^\infty \mathbb E[\pi_t] \, \mathrm d t = w \int_0^\infty \exp(-r t) = \frac{w}{r} \\ \text{and} \\ \mathbb E \int_0^\infty \pi_t^{\frac{\gamma - 1}{\gamma}} \exp \left (\frac{-\rho t}{\gamma} \right ) \, \mathrm d t &= \int_0^\infty \exp(-a t) dt = a^{-1}, \end{align*} where $a = \frac{\rho}{\gamma} +\frac{\gamma - 1}{\gamma} \left[r + \frac 12 \eta^2 \frac{1}{\gamma} \right ] $. Then, continuing with the budget constraint, we can solve for $\lambda^{-1/\gamma}$, $$ \lambda^{-1/\gamma} = \frac{w a}{r}. $$ We can then substitute this back into our expression for the optimal path of consumption, $$ C^*_t = \frac{w a}{r} \pi_t^{-1/\gamma} \exp \left \{ -\frac{1}{\gamma} \rho t \right \}. $$