# 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.

-
Hi I think you should derive first the law ofthe Hitting time. Otherwise this is a Ornstein-Uhlenbeck dynamic and the solution is known explicitly, it should be something like $V_t=V_0.e^{-\gamma.t}+\alpha\int_0^Te^{-\gamma.(t-s)}dW_s$, from this you can get an closed form solution for $E[V_t^2]$. Hope this helps bestregards. –  TheBridge Dec 14 '11 at 12:48
Thanks for following up. I noticed this was an O-U process as well and $E(V_t ^2)$ itself is well known. However, conditional on the filtration $\mathcal{F}_{s}$, I figured it would be something different. Could you expand on what you meant by the law of the hitting time? –  David Dec 14 '11 at 19:19

## 2 Answers

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

-

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 $t-\tau$

-
that last should have been $5-\tau$ –  mike Mar 29 '12 at 22:16