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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.

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might want to accept the answer below... – Lost1 Dec 17 '13 at 23:06
This is NOT my suggestion, please read better. – Did Dec 18 '13 at 10:48
@Did : Sorry for my typo. I replaced $dB_s$ for $ds$ for $U$ and the normalization coefficient of $\xi$. Please confirm if that is all or if I do misunderstood your suggestion. Thanks – Paul Dec 18 '13 at 11:45
The problem was indeed $dB_s$ instead of $ds$. – Did Dec 18 '13 at 12:12

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 $$ E[\mathrm e^{\mathrm ix\varepsilon_1+x^2U^2/2}]=1. $$

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I gave up this idea wich was my only one by the way since I prematurely guessed $\xi=\frac{1}{ \sqrt{\int_0^1\phi(t)^2\sigma_t^4 dt} } \int_0^1\phi(t)\sigma_t^2 dB_t$ would not be normal distributed. Thank you very much! – Paul Dec 17 '13 at 22:51
I feel I misunderstood something in the theory because I still don't see why it's normal distributed. Can you help me, please ? – Paul Dec 17 '13 at 23:10
Of course, this U is very far from being normal. What made you think it should be? – Did Dec 17 '13 at 23:19
not $U$ but the $\xi$ I defined in the first comment – Paul Dec 17 '13 at 23:36
The answer is not obvious. See Edit. – Did Dec 18 '13 at 12:29

An easy counter-example : $ 2 \int_0^1 B_s dB_s = B_1^2 -1$ which is not a symmetric random variable contrary to the product $U \xi$ with $\xi$ a centered gaussian random variable independent from $U$.

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This is not a counterexample since one would have to consider $\phi(t)\sigma^2_t=B_t$, which is impossible for sign reasons. – Did Nov 14 '14 at 11:08

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