Prove $ W_t=B_t^4 -6B_t^2t+3t^2$ is a martingale, and compute $E(T^2)$ where $T=\inf(t\ge0,B_t=-a, B_t=b)$ if $a=b$.
Ok, if $0\lt t\lt s$, $W_t$ is a martingale if $E(W_s|[B_r]_{r\le t})=W_t$
So $E(B_s^4-6B_s^2+3s^2|[B_r]_{r\le t})=E(B_s^4|[B_r]_{r\le t})-6tE(B_s^2|[B_r]_{r\le t})+3t^2$, Im not sure of the algebra necessary to prove how to prove $E(B_s^4|[B_r]_{r\le t})=B_t^4, E(B_s^2|[B_r]_{r\le t})=B_s^2$
My professor hasn't gone over any theorem to tackle these cases so any feedback would be appreciated.
Thanks
Update ok so I figured out $E(B_s^2|[B_r]_{r\le t})=B_t^2 +(s-t)$ but I'm stuck on the other case
The Brownian motion has independent increments, it's always a good idea to use this fact if you have to prove a process a martingale.
$$\begin{align*} \mathbb{E}(B_s^4 \mid [B_r]_{[r \leq t]}) &= \mathbb{E}((B_s-B_t+B_t)^4 \mid [B_r]_{r \leq t}) \\ &= \mathbb{E}((B_s-B_t)^4 \mid [B_r]_{[r \leq t]}) + 4 B_t \mathbb{E}((B_s-B_t)^3 \mid [B_r]_{[r \leq t]}) \\ &\quad+ 6 B_t^2 \mathbb{E}((B_s-B_t)^2 \mid [B_r]_{[r \leq t]}) + 4 B_t^3 \mathbb{E}(B_s-B_t \mid [B_r]_{[r \leq t]}) +B_t^4 \end{align*}$$
for $s \geq t$. Since $B_s-B_t$ is independent of $[B_r]_{r \leq t}$, we get
$$\begin{align*} \mathbb{E}(B_s^4 \mid [B_r]_{[r \leq t]})&= \mathbb{E}((B_s-B_t)^4) + 4 B_t \mathbb{E}((B_s-B_t)^3) \\ &\quad+ 6 B_t^2 \mathbb{E}((B_s-B_t)^2) + 4 B_t^3 \mathbb{E}(B_s-B_t) +B_t^4 \end{align*}$$
Now use $B_s-B_t \sim N(0,s-t)$ to calculate the remaining expectations...