Distribution of stochastic integral to stopping time

Question

Let $B$ be a standard Brownian motion, and let $H$ be a continuous adapted process with $\int_0^\infty H_s^2ds=\infty$. For $\sigma>0$, let $T_\sigma=\inf\{t\geq0:\int_0^tH_s^2ds>\sigma^2\}$. Find the distribution of the random variable $$X_\sigma=\int_0^{T_\sigma}H_sdB_s.$$ I really don't know where to start with this one, any hints would be greatly appreciated, thanks! I am told that if I use Dubin-Schwarz I have to prove it (which implies to me that I don't have to use Dubin-Schwarz).

Answer 1

Given some real number $\lambda$, you may introduce the exponential local martingale defined by $$M_t := \exp\Big(i\int_0^t H_s \mathrm{d}B_s + \frac{1}{2} \int_0^t H_s^2 \mathrm{d}s\Big).$$ Ito calculus (or your lecture) shows that $M$ is actually a local martingale.

You have $M_t \in [0,e^{\sigma^2}]$ for $t \in [0,T_\sigma]$, so Doob optional stopping theorem applies to $0$ and $T_\sigma$. In particular, $E[M_{T_\sigma}] = E[M_0] = 1$.

You derive the characteristic function of $X_\sigma$.