# Distribution of hitting time of Brownian motion

Let $$\tau = \inf\{t: B_t = 1\}$$ where $$B_t$$ is the standard brownian motion. How does one find the distribution of $$B_t$$ without strong markov property ?

• Can you use the reflection principle ? (despite of the fact that it's a consequence of the strong Markov property). If yes, then $\mathbb P\{\tau\leq T\}=2\mathbb P\{B_T>1\}$
– Surb
Apr 4, 2022 at 15:52
• Yes. I can use reflection principle for random walks. Is there any proof without using that ? Apr 4, 2022 at 15:59
• I assume you want the distribution of $\tau$ rather than the distribution of $B_t$? Apr 4, 2022 at 16:02

Note that finding the distribution of $$\tau$$ is equivalent to evaluating its Laplace transform $$\mathbb{E}[e^{-\mu \tau}]$$ for all $$\mu \ge 0$$.
Let $$M_t := e^{\lambda B_t - \frac 12 \lambda^2 t}$$ for some $$\lambda > 0$$ to be chosen later. Note that $$M$$ is a martingale, and by the optional stopping theorem so is $$M_{t \wedge \tau}$$. Since $$B_{t \wedge \tau} \le 1$$ for all $$t$$, we have $$M_{t \wedge \tau} = e^{\lambda B_{t \wedge \tau} - \frac 12 \lambda^2 (t \wedge \tau)} \le e^{\lambda}.$$ Since $$M$$ is bounded, sending $$t \rightarrow \infty$$ and appealing to the dominated convergence theorem implies $$\mathbb{E}[M_{\tau}] = M_0$$, i.e. \begin{align*} 1 &= M_0 \\ &= \mathbb{E}[M_\tau] \\ &= \mathbb{E}[e^{\lambda B_\tau - \frac 12 \lambda^2 \tau} ] \\ &= e^{\lambda} \mathbb{E}[e^{-\frac 12 \lambda^2 \tau}]. \end{align*} Rewriting, we've shown $$\mathbb{E}[e^{-\frac 12 \lambda^2 \tau}] = e^{-\lambda}$$. Letting $$\lambda := \sqrt{2\mu}$$, we therefore have $$\mathbb{E}[e^{-\mu \tau}] = e^{-\sqrt{2\mu}}$$, and hence we have identified the distribution of $$\tau$$ through its Laplace transform.
• I am not aware of Laplace transform. I will look it up. But can we start with the characteristic function of $M_t$ and get a similar thing? Apr 4, 2022 at 16:16
• I think it would be tricky to get a factor of $i$ to appear with $\tau$ in $M_t$, but certainly if you are able to that would work. The Laplace transform of $X$ is basically the same as the moment generating function $\mathbb{E}[e^{tX}]$, it's just only defined for $t \le 0$. Apr 4, 2022 at 17:07