Compute $E(2B_T e^{-B_T}+ 4e^{-B_T})$ for $B = (B_t)_{t \geq 0}$ a standard Brownian motion and $T=\inf\{t \geq 0 : |B_t|=1\}$.

Question

Let $B = (B_t)_{t \geq 0}$ a standard Brownian motion, and $T=\inf\{t \geq 0 : |B_t|=1\}$. I then have to calculate the following expectation

$$E(B_T + \int_0^T B_s e^{-B_s} I(B_s \geq -1)ds)$$

where $I$ is the indicator function. By symmetry I can see that $E(B_T)=0$, thus I just have to calculate the expectation of the integral. Also between $0$ and $T$ the brownian motion clearly lies above $-1$ thus I can drop the indicator function. Through an application of Ito's lemma I found that

$$\int_0^t B_s e^{-B_s}ds = 2B_t e^{-B_t}+ 4e^{-B_t} -4 + 2 \int_0^t (B_se^{-B_s} + e^{-B_s}) dB_s$$

How would I continue from here? Should I try to show the integral is a martingale and potentially use the optional stopping theorem? If so I'd still need to calculate $E(2B_T e^{-B_T}+ 4e^{-B_T})$ which I'm not sure how to go about either? Or is there a completely different way to go about this problem?

Edit:

To show the integral is a martingale I believe it is sufficient to show that

$$\int_0^t E(B_s^2e^{-2B_s})ds \ \text{and} \int_0^tE(e^{-2B_s})ds$$

are both finite. How could I be sure of this? And then to use the optional stopping theorem I'd need to show that

$$\int_0^{t \wedge T} (B_se^{-B_s} + e^{-B_s}) dB_s$$

is bounded. How could I show this also?

Answer 1

From the definition of $T$ we know that $B_T = \pm 1$, and by the optional stopping theorem $\mathbb{E}[B_T] = 0$ so if we let $p := P(B_T = 1)$ then we know $p = \frac 12$. Hence \begin{align*} \mathbb{E}[2B_T e^{-B_T} + 4 e^{-B_T}] &= \frac 12 (2e^{-1}+4e^{-1}) + \frac 12 (-2e + 4e) \\ &= 3e^{-1} + e. \end{align*}

Now we will show $\int_0^{t \wedge T} (B_s e^{-B_s} + e^{-B_s})dB_s$ is a uniformly integrable martingale, so we can apply the optional stopping theorem to it. We have

\begin{align*} \mathbb{E}[\int_0^{t \wedge T} (B_s e^{-B_s} + e^{-B_s})^2ds] &= \mathbb{E}[\int_0^{t \wedge T} (B_s +1)^2e^{-2B_s}ds] \\ &\le \mathbb{E}[\int_0^{t \wedge T} (1 +1)^2e^{2}ds] \\ &= 4e^2 \mathbb{E}[t \wedge T] \\ &\le 4e^2 \mathbb{E}[T] < \infty \end{align*}

so the quadratic variation is bounded in $L^2$ and hence $\mathbb{E}[\int_0^{T} (B_s e^{-B_s} + e^{-B_s})dB_s]=0$.