Compute moments of Brownian motion stopped at exit time of $[a,b]$ 
Given $B_t$ a standard brownian motion and $a < 0 < b$ Set $T = \inf\{ t : B_t = a \vee B_t = b\} $
For any $\alpha \in \mathbb{Z}^+$, find $EB_T^\alpha$.

I know I can use optional sampling here to get
$$ 0 = E B_T = a \frac{b}{b-a} + b\frac{a}{a-b}.$$
This question was partially solved (this above part is given) and this is just for studying purposes.
My questions: Do I need to use the above from optional sampling? I suppose it helps us know we can actually calculate the moments given that $T$ is a bounded stopping time.
Wouldn't the answer just be calculating the moments of $B_T$
Which comes down to $B_T \sim N(0,T)$ so I could use the MGF of this.
 A: It is true that $B_t$ is Gaussian with mean $0$ and variance $t$ for any deterministic time $t$. However, if $T$ is a stopping time, then $B_T$ is, in general, not Gaussian. In fact, there is some general theorem which says that for a large class of distributions $\mu$ there exists a stopping time $T$ such that $B_T$ has distribution $\mu$.
Consider, for instance, the stopping time in your question, i.e. $$T = \inf\{t; B_t = a \, \, \text{or} \, \, B_t = b\}$$ for $a<0<b$. Because of the continuity of the sample paths of Brownian motion, it follows that $B_T$ takes only the values $a$ and $b$. This already shows that $B_T$ cannot be Gaussian. 
To calculate the moments of $B_T$ you can use the fact that $$\mathbb{P}(B_T = a) = \frac{b}{b-a} \quad \text{and} \quad \mathbb{P}(B_T = b) = \frac{a}{a-b}$$ which is a consequence of the optional stopping theorem applied to the martingale $(B_t)_{t \geq 0}$. This means that we can write down explicitly the distribution of $B_T$: $$B_T \sim \frac{b}{b-a} \delta_a + \frac{a}{a-b} \delta_b$$ where $\delta_x$ denotes the Dirac measure. This allows us, in particular, to calculate arbitrary moments of $B_T$: $$\begin{align*} \mathbb{E}(B_T^k) = \mathbb{E}(B_T^k 1_{\{B_T=a\}}) + \mathbb{E}(B_T^K 1_{\{B_T=b\}}) &= a^k \mathbb{P}(B_T = a) + b^k \mathbb{P}(B_T=k) \\ &= a^k \frac{b}{b-a} + b^k \frac{a}{a-b}. \end{align*}$$
