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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.

  • 0
    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.2012-05-18
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    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.2012-05-18
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    @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).2012-05-18

2 Answers 2