Integration of independent Brownian motions

Question

I am wondering if the following integral of stochastic Brownian motions has an analytical solution?

$$ \int_{0}^{t}e^{\nu \tilde{V}_{\tau} - \frac{1}{2}\nu^{2}\tau}d\tilde{W}_{\tau} $$

where $\tilde{W}_{\tau},\tilde{V}_{\tau}$ are two independent Brownian motions, $N(0,\tau)$. I did some search on the subject and found this reference based on Bougerol work (in french). However, I cannot find a satisfactory answer.

Thanks in advance for your replies.

Edit:

In the papers they give 2 results:

$$ \sinh(\tilde{V}_{t}) \stackrel{law}{=} \int_{0}^{t}e^{\tilde{V}_{s}}d\tilde{W}_{s} $$

or

$$ \sinh(\tilde{V}_{t}+\tilde{\varepsilon}t) \stackrel{law}{=} \int_{0}^{t}e^{\tilde{V}_{s}+s}d\tilde{W}_{s} $$

where $\tilde{\varepsilon}$ is a symmetric Bernoulli variable between {-1,1}.

I was looking for an answer like that but for the integral:

$$ \int_{0}^{t}e^{\nu \tilde{V}_{\tau} - \frac{1}{2}\nu^{2}\tau}d\tilde{W}_{\tau} \stackrel{law}{=} \dots $$

Answer 1

I am not sure how rigorous this is (it has been a while since I touched stochastic calculus) but here goes:

Let $S_t := \exp ( \nu V_t - 0.5 \nu^2 t)$. Trivially, $X_t = \int_{0}^{t} S_z \, dW_z$.

We will solve $X_t = X_0 + \int_{0}^{t} S_z \, dW_z$ for some non zero real number $X_0$.

In differential form we have $d X_t = S_t \, d W_t$. Assuming that $S_t$ is locally Lipschitz continuous (is it?), we have $$X_t = X_0 \, \exp \left( -0.5 \int_{0}^{t} S_z^2 \, dz + W_t \int_{0}^{t} S_z \, dz \right)$$ via Ito's formula.