Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space and let $W_t$ be standard Wiener process and

$$T_{a,b}=\inf\{t:W_t=a+bt\}$$ where $a$ and $b$ are costant.I want to get expectation of $T_{a,b}$ but I can not extract $f_{T_{a,b}}$.


  • $\begingroup$ Do you know how to solve this when $b=0$? $\endgroup$ – Math1000 Jul 23 '16 at 0:52
  • $\begingroup$ Yes. It is well-known. $\endgroup$ – math Jul 23 '16 at 2:05
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    $\begingroup$ See here: math.stackexchange.com/questions/133628/… $\endgroup$ – Math1000 Jul 23 '16 at 3:22
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    $\begingroup$ So thanks.It was so useful $\endgroup$ – math Jul 23 '16 at 6:53

Sasha's answer is so nice. I want to offer other way.

Other way

Let $\alpha\in \mathbb{R}^+$. It is well known that $X_t=\exp\left(\alpha W_t-\frac{1}{2}\alpha^2 t\right)$ is a martingale, therefore $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}[X_0]=1$$ so $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}\left[\exp\left(\alpha W_{T_{a,b}}-\frac{1}{2}\alpha^2 T_{a,b}\right)\right]=1$$ set $t=T_{a,b}$, we have $W_{T_{a,b}}=a+bT_{a,b}$ and $$\mathbb{E}\left[e^{\large{(\alpha b-\frac{1}{2}\alpha^2)T_{a,b}}}\right]=e^{-\alpha\,a}$$ define $\lambda=-\alpha b+\frac{1}{2}\alpha^2$, we have $\alpha=b+\sqrt{b^2+2\lambda}$. Hence $$\mathbb{E}\left[e^{\large{-\lambda T_{a,b}}}\right]=e^{-a\,(b+\sqrt{b^2+2\lambda}\,)}$$ By differentiating with respect $\lambda$, we have $$\mathbb{E}\left[T_{a,b}e^{\large{-\lambda T_{a,b}}}\right]=\frac{a}{\sqrt{b^2+2\lambda}}e^{-a\,(b+\sqrt{b^2+2\lambda}\,)}$$ Now let $\lambda=0$


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