Expectation of Stochastic Process Given First Hitting Time Information Let $V_t$ satisfy the SDE $dV_t = -\gamma V_t dt + \alpha dW_t$. Let $\tau$ be the first hitting time for 0, i.e., $\tau $ = min$(t | V_t = 0)$. Let $s =$ min$(\tau, 5)$. Let $\mathcal{F}_s$ be the $\sigma$-algebra generated by all $V_t$ for $t\leq s$. Calculate $G= E[V_5 ^2 | \mathcal{F}_s]$ by showing that it is given as a simple function of one random variable.
Found this on a practice final and really don't know how to start. I thought of a possible PDE  approach using say the Kolmogorov Backward Equation, but does anyone know a possible alternative to tackle this problem? Any hint would be helpful. Thanks in advance.
 A: Here is a partial answer as I didn't finished the calculation but I provide references that should allow to finish the work.
So we have for any stopping times $\sigma$ with enough regularity ( e.g. integrable finished a.s., etc...), from the result in the my comment above by conditionning by $\sigma$ and using Itô isometry (please check this calculation):
$$G_\sigma=E[V_\sigma^2]=\frac{\alpha^2}{2\gamma}-(V_0^2-\frac{\alpha^2}{2\gamma}).E[e^{-2\gamma.\sigma}]$$
So knowing the density function of $\sigma$ could allow to finish the work. 
The law of $\tau=inf\{t>0,V_t=0\}$ is explicitly known but rather tricky as it invlolves special functions (look here, here or here), and for $s=5\wedge \tau$ you can re-express the preceding expression by :
$$G_s=\frac{\alpha^2}{2\gamma}-(V_0^2-\frac{\alpha^2}{2\gamma}).[e^{-10.\gamma}.\mathbb{P}[\tau>5]+\mathbb{E}[e^{-2\gamma.\tau}.1_{\{\tau\ge 5\}}]]$$
A Girsanov transform should allow further simplification for the term $\mathbb{E}[e^{-2\gamma.\tau}.1_{\{\tau\ge 5\}}]$, to rewrite it as $E[e^{-2\gamma.\tau}].\mathbb{Q}[\tau\ge 5]$ but I don't know if this is really necessary.
Regards
A: If $\tau > 5 $ you know $V_5$ so on that set you are $V_5$.  On $\tau < 5 $, you know $\tau$ and you know $V_{\tau} = 0$ so on $\tau < 5 $, $V_5$ is an OU process started from $0$ and run for time $t-\tau$
