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

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?

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.