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Let $W(t)$ be a Brownian motion on a filtered probability space $(\Omega,\mathcal{F}, \mathcal{F}_t,\mathrm{P})$ and let $\tau$ be a stopping time with all moments finite. How do I show that

$\int_0^{\tau}e^{-s}dW(s)$

has a smooth density?

My first guess was to use the martingale $M_t=e^{iu \int_0^te^{-s}dW(s)+\frac{u^2}{2}\int_0^te^{-2s}ds}$ and the optional stopping theorem to write

$1=\mathrm{E}M_{0}=\mathrm{E} M_{\tau}=\mathrm{E}e^{iu\int_0^{\tau}e^{-s}dW(s)+\frac{u^2}{2}\int_0^{\tau}e^{-2s}ds}$

If $\tau$ is non-random then I can bring the second integral on the left hand side to get the bound on the characteristic function

$\phi(u)=\mathrm{E}e^{iu\int_0^{\tau}e^{-s}dW(s)}=e^{-\frac{u^2}{2}\int_0^{\tau}e^{-2s}ds}$

and therefore $|u|^k\phi(u)$ is integrable in $u$ for every $k$ meaning the density of $\int_0^{\tau}e^{-s}dW(s)$ is smooth by Fourier inversion. However, this clearly does not work when $\tau$ is random. Help!

While at it, it would be great to go one step further and have the same conclusion for

$\int_0^{\tau}e^{-s}f(s)dW(s)$

for some predictable process $f(s)$ with $c_1\leq f^2(s)\leq c_2$. (Perhaps a simple time change after the previous case has been established)

Thanks in advance!

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  • $\begingroup$ What if we choose $\tau := 0$? Then the stopped integral does clearly not have a density. $\endgroup$ – saz Apr 6 '17 at 5:21

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