Vtyjkyuk

Let $(W_s)_s geq 0$ be a Wiener process and $tau$ be a random variable with an exponential distribution with parameter $lambda$. Suppose that $W$ and $tau$ are independent. In this question, we see that the distribution of the stopped Wiener process $W_tau$ corresponds to a Laplace distribution with scale parameter $fracsigmasqrt2lambda$ where $sigma$ is the instantaneous variance of the Wiener process.

Suppose now, that the variance $sigma^2$ also follows a stochastic process such that we have:
$$dS = sigma SdW_s \
dsigma^2 = alphasigma^2dt + xisigma^2dW_sigma$$
where $alpha$ and $xi$ are independent of $S$ and $dW_s$ and $dW_sigma$ are independent Wiener processes.

My aim is to derive the distribution of $log S_tau$.

First, if $barV_T$ denotes the mean variance over some time interval $[0,T]$ defined by
$$barV_T = frac1T intlimits_0^Tsigma^2(t)dt,$$
it is easy to show (Lemma of Îto) that
$$log S_T = log S_0 - barV_TT/2 + intlimits_0^TsqrtbarV_TdW_3 $$
where $W_3$ is a Wiener process.
Making use of the answer provided here I get for the characteristic function of the stopped Process $log S_tau$:

$$varphi_S_tau^b, sleft(uright) = Eleft( e^iu S_tau right) \
= Eleft( Eleft( e^iuS_tau midtau right)right) \
= Eleft( e^iubarV_tautau/2-tauleft(barV_tauright)u^2/2 right) \
=lambda int_0^infty e^-lambda + iubarV_tautau/2-tauleft(barV_tauright)u^2/2 , dtau.
$$
Here I am stuck as I do not see how to integrate $barV_tau$ over $tau$..is there any way to represent the distribution of the stopped Wiener Process $log S_tau$ as a function of $barV_tau$?