Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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):

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.


share|cite|improve this answer

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 $t-\tau$

share|cite|improve this answer
that last should have been $5-\tau$ – mike Mar 29 '12 at 22:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.