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.



Conditional law of an Ito's stochastic integral
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.


