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.


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.

  • $\begingroup$ Thank you very much! I believe you are right about the indicator function. $\endgroup$ – Paul May 28 '13 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 '13 at 13:22

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