Calculate $\mathbb{E}(T^2)$ and $\mathbb{E}(\int_0^T X_s \,d s)$ for exit time $T$ of Brownian motion $(X_t)_{t \geq 0}$

Question

Let $T$ be the exit time of from the interval $[-b,a]$ of a standard Brownian Motion $X_t$, then how would we go about calculating the following two expectations:

• $E[T^2]$ (and)
• $E[\int_0^T X_tds]$?

---

(Ideas which haven't been tied in:)

I want to use the optional stopping theorem but what martingale would I use? Also for the second I know (by Ito's formula) I can write: \begin{equation} \int_0^T X_tds = 6\int_0^T X_t^3 dX_t - 6\int_0^T X_t^2 ds \end{equation} but how can I put that to use? I'm thinking since $X_t -X_0$ is a normal random variable, I could put that to use in calculating the second term on the RHS of the above but I'm not certain how...

Answer 1

First of all, recall that it follows from Wald's identities that

$$\mathbb{P}(X_{T}=-b) = \frac{a}{a+b} \qquad \mathbb{P}(X_{T} = a) = \frac{b}{a+b} \qquad \mathbb{E}(T)=ab, \tag{1}$$

see e.g. this answer for a proof or Corollary 5.11 in Brownian Motion - An Introduction to Stochastic Processes (by René Schilling and Lothar Partzsch).

Part I: Calculate $\mathbb{E}(\int_0^T X_s \, ds)$:

It follows from Itô's formula that $$\int_0^t X_s \, ds = - \int_0^t X_s^2 \, dX_s + \frac{X_t^3}{3}.$$ Since the first term on the right-hand side is a martingale, the optional stopping theorem (applied to the bounded stopping time $T \wedge k$) yields

$$\mathbb{E} \left( \int_0^{T \wedge k} X_s \, ds \right) = \frac{1}{3} \mathbb{E}(X_{T \wedge k}^3).$$

As $|X_{s \wedge T}| \leq \max\{a,b\}$ for all $s \geq 0$, it now follows from the dominated convergence theorem that

$$\mathbb{E} \left( \int_0^T X_s \, ds \right) = \frac{1}{3} \mathbb{E}(X_T^3).$$

Finally, using $$X_T = a 1_{\{X_T=a\}} -b 1_{\{X_T=-b\}},$$ we obtain $$\mathbb{E} \left( \int_0^T X_s \, ds \right) = \frac{(-b)^3}{3} \mathbb{P}(X_T=-b) + \frac{a^3}{3} \mathbb{P}(X_T=a).$$

Plugging in the results from $(1)$, we are done.

Part II: Calculate $\mathbb{E}(T^2)$:

1. Show that $M_t := X_t^3 -3 t X_t$ is a martingale and that (with a similar argumentation as in part I) $$\begin{align*} \mathbb{E}(X_T^3) &= 3a \mathbb{E}(T 1_{\{X_T} = a\}) -3b \mathbb{E}(T 1_{\{X_T}=-b\}) \\ &= 3a \mathbb{E}(T) - 3 (a+b) \mathbb{E}(T 1_{\{X_T = -b\}}). \tag{2} \end{align*}$$
2. Use $(1)$ and Step 1 to calculate $\mathbb{E}(T 1_{\{X_T=-b\}})$.
3. Show that $N_t := X_t^4-6t X_t^2 + 3t^2$ is a martingale. Using the optional stopping theorem show that $$\mathbb{E}(N_T)=0. \tag{3}$$
4. Using the identity $$X_T = a 1_{\{X_T=a\}} -b 1_{\{X_T=-b\}},$$ step 2 and $(3)$ calculate $\mathbb{E}(T^2)$.