Distribution of $\int_0^t e^s dB(s)$

Question

Consider $$\left(\int_0^t e^s dB(s)\right)_{t\in [0,1]}$$ where $B$ is a Brownian motion.

What is the distribution of this process?

Since $f(s) = e^s$ is continuously differentiable, $B$ is process with uncorrelated increments and $\mathbb E [B(t)]\equiv 0$, one can use the fact that

$$\int_a^b f(s) dB(s) = f(b)B(b) - f(a)B(a) - \int_a^b B(s)f'(s) ds$$

This reduces in our specific case to

$$\int_0^t e^s dB(s) = e^t B(t) - \int_0^t B(s) s e^s ds$$

Now I am stuck to find the distribution of $\int_0^t B(s) s e^s ds$. Is there a convinient way to find its distribution or an easier way to find the distribution of the original process?

Answer 1

$$ M_t=\int_0^ta(s)\,\mathrm dB(s)\qquad (M_t)_{t\geqslant0}\stackrel{\text{dist.}}{=}(B(A_t))_{t\geqslant0}\qquad A_t=\int_0^ta(s)^2\,\mathrm ds$$

Answer 2

Call this process $M_t$. By the Dubins-Schwarz theorem (see page 54), your (a-priori local) martingale has the same distribution as a Brownian Motion evaluated at $[M]_t$ so it suffices to compute the quadratic variation of your process, which you can get from Ito's isometry.