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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.

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1) That looks hard. 2) It will be finite (and have finite expectation), since there's some probability $\epsilon$ that it will hit in any time interval $[n,n+1)$ if it hasn't hit already. –  Ben Derrett May 18 '12 at 13:20
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I suspect applying OST to the stochastic exponentials of $\theta W_t$ and $-\theta W_t$ will help. –  Ben Derrett May 18 '12 at 13:21
    
Well OK it's easy to say $T<\infty$ $\mathbb{Q}$-a.s. by the properties of BM, hence it's a.s. finite under $\mathbb{P}$ (absolute continuity). So that is not a problem anymore. –  Tom Artiom Fiodorov May 18 '12 at 13:39
    
use an exponential martingale ($e^{\frac b 2 X_t}$ i think, but something like that) to find $\mathbb P (X_T = a )$ then apply Wald. –  mike May 18 '12 at 14:21
    
@mike, What is Wald? I have managed to use OST on the exponential martingale and then just differentiate the answer with respect to $b$ to compute the expression 1). –  Tom Artiom Fiodorov May 18 '12 at 14:38

2 Answers 2

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 $$ f(x) \; = \; {\sf E}[{\bf T}|X(0)=x] \; = \; \frac{1}{b} \, \left\{ 2 \, \frac{1 - \exp[-b(1+x)]}{1 - \exp(-2b)} \, - \, (1+x) \, \right\} $$

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I think $e^{-bX_t}$ is a martingale, something like that is right, anyway. Using it you have $e^{bx} = e^{-ba}\mathbb P(X_T = a) + e^{ba}\mathbb P(X_T = -a)$, and $ 1 = \mathbb P(X_T = a) +\mathbb P(X_T = -a)$, from which you determine $\mathbb P(X_T = a)$ and $\mathbb E(X_T)$. Then Wald says $\mathbb E (X_T) = b \mathbb E (T)$

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