2
$\begingroup$

Consider $B=(B_t)_{t\geq 0}$ real $\mathcal F_t$ - brownian motion starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider a new real $\mathcal F_t$ - brownian motion $\tilde{B}=(\tilde{B_0})_{t\geq t}$ independent of $B$ as weel as a process $H=(H_t)_{t\geq 0}$ given by

$$ H_t := \frac{1}{\int _0^t f^2(B_s) ~ds}\int _0^t f(B_s) ~d \tilde B_s \mathbf 1_{\{ \int _0^t f^2(B_s) ~ds>0\}}, \ t\geq 0,$$

where $f \in \mathcal C^0(\mathbb R, \mathbb R)$ and $f \not\equiv 0$.

What is the conditional distribution of $H_t$ knowing $B$?

I dont have any idea on how to start to approach it. Any advice will be strongly appreciate. Thank's in advance.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

Hi, If you know $B$ on the entire trajectory then $H_t$ is no more than a "scaled"-Wiener integral (as the integrand become deterministic).

As $\int _0^t f(B_s) ~d \tilde B_s \mathbf 1_{{ \int _0^t f^2(B_s) ~ds>0}}$ is then a normal random variable with null expectation and variance equal to $\int _0^t f(B_s)^2 \mathbf 1_{{ \int _0^t f^2(B_s) ~ds>0}}ds$, the "scaled"-Wiener integral is only a standard normal random variable.

Best regards

PS: By the way I think that the indicator function is not really necessary as $\int _0^t f(B_s)^2ds$ is almost surely strictly positive $\forall t>0$ unless mistaken.

$\endgroup$
2
  • $\begingroup$ Thank you very much! I believe you are right about the indicator function. $\endgroup$
    – Paul
    May 28, 2013 at 11:20
  • $\begingroup$ @Paul : By the way the process $H_t$ is in itself interesting as a canonical way to construct a process with the following property : being continuous (except at time 0) with the same law at every instant but different from the great disorder process which is nowhere continuous. Best regards. $\endgroup$
    – The Bridge
    May 28, 2013 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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