Distribution of $\int^T_t \sigma (T-u)dW_u$ where $W_t$ is a Brownian motion

Question

I am trying to find the distribution of $\int^T_t \sigma (T-u)dW_u$ where $W_t$ is a Brownian motion.

One (very hand-wavey) way is to assume a priori that it is Normally distributed. Then one can see that it has mean $0$ due to being a local Martingale and then variance $$E[X_t^2]=E[\int^T_t \sigma^2 (T-u)^2du]=\int^T_t \sigma^2 (T-u)^2du$$ using Ito isometry, so it is distributed with $\mathcal N(0, \int^T_t \sigma^2 (T-u)^2du)$.

Is there a better way to do this, in particular not assuming it is Normally distributed?

Answer 1

It can be shown that the integral is a time-changed Brownian Motion and from there we can deduce the distribution from looking at this BM.

Define $f(u) = \sigma(T-u)$ and $X_t = \int^t_0 f(u) dW_u$. Define the stopping time $$ \tau_t = \inf\lbrace u \geq0 : [X]_t > t \rbrace $$ and look at the process $B_t := X_{\tau_t}$. We see that by definition of the stopping time, $$ X_t = X_{\tau_{[X]_t}} = B_{[X]_t}. $$ We can show that $B_t$ is an $\lbrace \mathcal F_{\tau_t} \rbrace$-Brownian Motion by Levy's Theorem for BM:

1. Clearly, $B_t$ is adapted to $\lbrace \mathcal F_{\tau_t} \rbrace$
2. $B_t$ is a local martingale: Define a sequence of stopping times $$ S_n=\inf\lbrace t\geq0: |X_t|>n \rbrace $$ The stopped process $B_t^{S_n}$ for martingale at each $S_n$ since $$ \mathbb E[B_t^{S_n} \mid \mathcal F (\tau_s) ] \equiv \mathbb E[X^{S_n}(\tau_t) \mid \mathcal F (\tau_s) ] $$ Since any bounded local martingale is uniformly integrable (by definition of the stopping time $S_n$, $X^{S_n}_t$ is bounded), it follows from Optional Stopping Theorem that we have $$ \mathbb E[X^{S_n}(\tau_t) \mid \mathcal F (\tau_s) ] = X^{S_n}(\tau_s) \equiv B_s^{S_n} $$
3. Using the definition of $\tau_t$ again, the quadratic variation of $B_t$ is $$ [B]_t = [X]_{\tau_t} = t $$

Hence $B_t$ is a Brownian Motion, and therefore the random variable $$ X_t \equiv B_{[X]_t} \sim \mathcal N (0, [X]_t) $$ so $$ \int^T_t f(u) dW_u = X_T-X_t \sim \mathcal N (0, [X]_T-[X]_t) $$ and we have our result.

Answer 2

Of course it is normal. The integral (if taken as the Ito integral) is the limit of a linear combination of brownian motions which are normally distributed (To see this, discretize the integral). Therefore, the integral is normally distributed. You're second claim is just an application of the fundamental theorem of stochastic calculus. So, you're all good.