Let $B = (B_t)_{t \geq 0}$ be a standard Brownian motion and set $M_t = 4B_t^2 + e^{4B_t -8t} -4t$. Find $\mathbb{E}(M_\sigma)$ for $\sigma = \inf\{t \geq 0 : |B_t| = 1\}$.
I have shown that $M_t$ is a martingale and am looking to use the optional stopping theorem. I can see that
$|M_{t \wedge \sigma}| = |4B_{t \wedge \sigma}^2 + e^{4B_{t \wedge \sigma} -8({t \wedge \sigma})} -4({t \wedge \sigma})|$
$\hspace{1.25cm}$ $\leq 4|B_{t \wedge \sigma}^2| + |e^{4B_{t \wedge \sigma}}| +4|{t \wedge \sigma}|$
$\hspace{1.25cm}$ $\leq 4 + e^4 +4|{t \wedge \sigma}|$.
I'm not entirely sure how to deal with the $|{t \wedge \sigma}|$ part. If I can find a bound for this part then I can use the optional stopping theorem to conclude that
$\mathbb{E}(M_\sigma) = \mathbb{E}(M_0) = 1$.
I feel there might be something obvious that I'm missing but I'm not too sure, thanks in advance.
You can show that $B_t^2-t$ is a martingale, so $0 = \mathbb{E}[B_{t \wedge \sigma}^2 - t\wedge \sigma]$ and hence $\mathbb{E}[t \wedge \sigma] = \mathbb{E}[B_{t \wedge \sigma}^2] \le 1$. By the monotone convergence theorem, we have $\mathbb{E}[\sigma] = \lim_{t \rightarrow \infty} \mathbb{E}[t \wedge \sigma] \le 1$ and therefore $$|M_{t \wedge \sigma}| \le 4 + e^4 + 4(t \wedge \sigma) \le 4 + e^4 + 4 \sigma.$$ Now we can use the dominated convergence theorem to conclude $$\mathbb{E}[M_\sigma] = \lim_{t \rightarrow \infty} \mathbb{E}[M_{t \wedge \sigma}] = \lim_{t \rightarrow \infty} \mathbb{E}[M_0] = 1.$$