What is the distribution of this random variable?

Question

Find the distribution of this random variable: $$X_t=\exp\left(t \int_0^t sdW_s\right)$$
knowing that $W$ is a Brownian motion in the filtered space $(\Omega, \mathcal{F},P,(\mathcal{F}_t)_{t\geq0} )$.

Answer 1

Without using any techniques from Ito calculus, let's write:

$$\int_0^t s dW_s = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{kt}{n} \left ( W_{\frac{(k+1)t}{n}}-W_{\frac{kt}{n}} \right )$$

Let $X_k = W_{\frac{(k+1)t}{n}}-W_{\frac{kt}{n}}$, then the $X_k$ are iid and $N(0,t/n)$. So we have

$$\int_0^t s dW_s = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{kt}{n} X_k$$

Because $\frac{kt}{n} X_k$ are independent, we have

$$\text{Var} \left ( \sum_{k=0}^{n-1} \frac{kt}{n} X_k \right ) = \sum_{k=0}^{n-1} \frac{k^2 t^2}{n^2} \frac{t}{n}$$

This last sum converges to $\int_0^t x^2 dx = t^3/3$, so the integral is distributed as $N(0,t^3/3)$.

Incidentally, this shows that for any $f \in C([0,t])$, $\int_0^t f(s) dW_s$ is distributed as $N\left ( 0,\int_0^t f(s)^2 ds \right )$.

The remainder of the problem is to find the distribution of $\exp(X)$ when $X$ is $N(0,t^5/3)$. This is not a problem about stochastic processes, merely one about probability. Can you finish from here?