Distribution of hitting time of Brownian motion

Question

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 ?

Answer 1

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.