Consider a probability space $(\Omega,\mathcal F, (\mathcal F_t)_{t\geq0},\mathbb P)$ where $\mathbb F=(\mathcal F_t)_{t\geq0}$ is generated by $B=(B_t)_ { t \geq 0}$ a standard brownian motion starting at zero.
Also, consider a process $\varepsilon=(\varepsilon_t)_{t \in [0,1]}$ given by
$$ \varepsilon_t,=\sqrt{2}\int_0^t\phi(s) \sigma_s^2 dB_s \ , \quad\forall t \in [0,1] $$ where $\phi \in \mathcal C^1([0,1])$ deterministic and $ (\sigma_t)$ a progressively measurable process.
I am interested in representing this process at time $t=1$ as a product between two random variables $$ \varepsilon_1 =U\xi $$ where $U$ is a $\mathcal F_1$- measurable r.v. whose the law is to be determined and $\xi$ is a standard gaussian variable independent of $U$.
I have no idea how to start to show that, if it's possible in this case. Since $\sigma$ is a process whose his law is totally unknown , it seems strange for a priori be able to determine U ( I insist, if it's possible).
Edit: Following the suggestion given by Did in his answer. IF we define $$ \xi =\frac{1}{\sqrt{\int_0^1\phi(s)^2 \sigma_s^4 ds}} \int_0^t\phi(s) \sigma_s^2 dB_s $$
and $$U= \sqrt{2\int_0^1\phi(s)^2 \sigma_s^4 ds}$$
Why should $\xi$ be normal distributed? The fact is totally clear whenever $\sigma$ is deterministic however in this case $\sigma$ is not.
All advices are appreciated.
Thank's in advance.
When $\sigma$ and $B$ are independent, a solution is $$ U=\sqrt{2\cdot\int_0^1\phi(t)^2\sigma_t^4\,\mathrm dt}.$$ When $\sigma$ is progressively measurable with respect to the filtration of $B$, the same solution might apply in the sense that $U\xi$ could be distributed like $\varepsilon_1$. As first steps in this direction, note that, for every real $x$, by independence of $(U,\xi)$, $$ E[\mathrm e^{\mathrm ixU\xi}\mid U]=\mathrm e^{-x^2U^2/2}, $$ hence $$ E[\mathrm e^{\mathrm ixU\xi}]=E[\mathrm e^{-x^2U^2/2}]. $$ On the other hand, $U^2=\langle\varepsilon\rangle_1$ hence $$ E[\mathrm e^{\mathrm ix\varepsilon_1+x^2U^2/2}]=1. $$