Compute $E(M_σ)$ when $σ = \inf(t ≥ 0: |B_t| = 1)$ and $M_t = 4B^2_t +e^{4B_t−8t}−4t$

Question

Given $M_t = 4B^2_t +e^{4B_t−8t}−4t$ for $t ≥ 0$ and a Brownian motion $(B_t)_{t \geq 0}$. Compute $E(M_σ)$ when $σ = \inf(t ≥ 0: |B_t| = 1)$.

I have tried to show that $E|M_σ|\leq K$ to apply Doob's Optional Sampling Theorem, thus $E(M_σ)=E(M_0)=1$, but my I am unsure how to explain the reasoning.

So far I've got $$E|M_σ|=E|M_{t\wedgeσ}|=E|4B_{t\wedgeσ}^2+e^{4B_{t\wedgeσ}-8{t\wedgeσ}}-8t|\leq E|4B_{t\wedgeσ}^2+e^{4B_{t\wedgeσ}-8{t\wedgeσ}}|,$$ but don't know how to proceed. Any help would be greatly appreciated.

Answer 1

Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) \qquad Y_t := e^{4B_t-8t}.$$ In particular, $\mathbb{E}(M_{\sigma}) = \mathbb{E}(X_{\sigma})+\mathbb{E}(Y_{\sigma})$, and therefore it suffices to calculate $\mathbb{E}(X_{\sigma})$ and $\mathbb{E}(Y_{\sigma})$.

Calculation of $\mathbb{E}(X_{\sigma})$:

1. Check (or recall) that $(X_t)_{t \geq 0}$ is a martingale.
2. Apply the optional stopping theorem to show that $$\mathbb{E}(X_{t \wedge \sigma})=0,$$ i.e. $$\mathbb{E}(B_{t \wedge \sigma}^2) = \mathbb{E}(t \wedge \sigma).$$
3. Show that $|B_{t \wedge \sigma}| \leq 1$ for all $t \geq 0$.
4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$\mathbb{E}(B_{\sigma}^2) = \mathbb{E}(\sigma),$$ i.e. $$\mathbb{E}(X_{\sigma})=0.$$

Calculation of $\mathbb{E}(Y_{\sigma})$:

1. Check that $(Y_t)_{t \geq 0}$ is a martingale.
2. Apply the optional stopping theorem to show that $\mathbb{E}(Y_{t \wedge \sigma})=1$ for all $t \geq 0$.
3. Conclude from $|B_{t \wedge \sigma}| \leq 1$ that $|Y_{t \wedge \sigma}| \leq e^{4}$ for all $t \geq 0$.
4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $\mathbb{E}(Y_{\sigma})=1$.