Continuous Time
Continuous Time Macro
Solving an HJB
The HJB usually takes the form
$$\rho V_t (Nt) = \max{C, a} \left{ u© + \mathcal{A} V_t(N_t)\right},$$
where $\mathcal{A} V_t(N_t)$ is the drift of $dV_T(N_t)$, $\rho$ is the discount rate, $u©$ is the flow payoff of $C$. To solve this equation follow these steps:
- Take FOC wrt controls (here $C, a$)
- Stare at it for a while and make a guess of the functional form of the solution to the PDE. In this case if $u© = \ln©$ we would have chosen $V(N) = v_0 + v_1 \ln(N)$. You really learn how to do this by practice.
- Plug the assumed functional form into the HJB (take necessary derivatives and replace all instances of $V(N)$)
- Plug in FOC from step 1
- Use method of undetermined coefficients to extract coefficients in assumed functional form. If you are unable to do this, try one more time because you might have made a dumb algebra mistake. After that start over at step 2 with a new guess
Note: If you aren’t able to provide an explicit functional form, but rather have a more general guess like $V(x, y) = f(x) + y g(x)$, then you should alter your approach slightly. Starting from step 5 you will need to do the following:
- You should be able to use the method of undetermined coefficients to come up with ODEs for each of the generic functions in your guess (in this case $f$, and $g$).
- Solve these ODEs however you know how
- If you can’t, you probably made a bad guess for the form of the solution, so start over at step 2.
Ito processes
GBM
A geometric Brownian motion solves the following SDE
$$dS_t = \mu S_t dt + \sigma S_t dW_t.$$
The solution is
$$S_t = S_0 \exp \left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t \right).$$
We often write GMB as
$$\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$$
or even in terms of $d \log S_t$. By Ito’s lemma we have that
$$d \log S_t = \frac{d S_t}{S_t} - \frac{1}{2} \sigma^2 dt.$$
Solving for $\frac{d S_t}{S_t}$ and matching coefficients we see that we must have
- drift of $d \log S_t = \mu - \frac{1}{2} \sigma^2$
- Volatility of $d \log S_t = \sigma$.