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.