Hitting time of Brownian Motion with a drift

Let $X_t =x+bt+\sqrt{2}W_t$, where $W_t$ is a standard Brownian motion. Let $T=\inf\{t: |X_t|=1\}$. I am trying to find $\mathbb{E}[T]$ for the case $b\neq0$.

Firstly, I am going to apply Girsanov to change the measure and the drift: $$M_t = e^{-\frac{b}{\sqrt{2}}W_t-\frac{b^2}{4}t},$$ If $\frac{d\mathbb{P}}{d\mathbb{Q}}|\mathcal{F}_t=M_t$, then $\mathbb{E}[T|\mathcal{F_t}]=\mathbb{E}^\mathbb{Q}[TM_t|\mathcal{F_t}]$. And under $\mathbb{Q}$, $X_t$ is driftless BM.

So $\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{\sqrt{2}}W_t-\frac{b^2}{4}t}|\mathcal{F_t}]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{2}X_t-\frac{b^2}{4}t+\frac{b}{2}x}|\mathcal{F}_t]$

If I managed to show $T<\infty$ a.s., or otherwise, I could arrive at: $$\mathbb{E}[T]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b}{2}X_T-\frac{b^2}{4}T+\frac{b}{2}x}]=\mathbb{E}^\mathbb{Q}[Te^{-\frac{b^2}{4}T}]\mathbb{E}^\mathbb{Q}[e^{-\frac{b}{2}X_T}]e^{\frac{b}{2}x}$$ Now $$\mathbb{E}^\mathbb{Q}[e^{-\frac{b}{2}X_T}]=e^{-\frac{b}{2}}\mathbb{P}^\mathbb{Q}(X_T=1)+e^{\frac{b}{2}}\mathbb{P}^\mathbb{Q}(X_T=-1)=e^{-\frac{b}{2}}\frac{1-x}{2}+e^{\frac{b}{2}}\frac{x+1}{2}$$

I have two questions:

1) How to compute $\mathbb{E}^\mathbb{Q}[Te^{-\frac{b^2}{4}T}]$

2) How to get around the problem that $T$ might not be finite?

Edit: 2) it's clear that $T<\infty$ under $\mathbb{Q}$, and so $T$ is $\mathbb{P}$-a.s. finite.

Solution 1:

If I understand Tom's problem correctly, we are looking for the two-sided first passage time ${\bf T}$ in a Wiener process $\{X(t), t \geq 0\}$ starting at $X(0)=x, \, -1 < x < +1$, with drift $b$, variance $2$, and absorbing barriers at $-1$ and at $+1$. This is a standard problem treated in many sources such as Darling and Siegert (1953) who in their Theorem 6.1 and eq. (6.6) derive the basic differential equation for $f(x) = {\sf E}[{\bf T}|X(0)=x]$, which (in the present notation) is $f^{''}(x) + b \, f^{'}(x) \; = \; -1$, with $f(-1)=f(+1)=0$. The solution is easily seen to be $$ f(x) \; = \; {\sf E}[{\bf T}|X(0)=x] \; = \; \frac{1}{b} \, \left\{ 2 \, \frac{1 - \exp[-b(1+x)]}{1 - \exp(-2b)} \, - \, (1+x) \, \right\} $$