Distribution of a stopped Brownian motion at the hitting time of an independent process

Question

I am trying to understand the distribution of $B^{(1)}_{\tau}$ where $(B^{(1)}_t, B^{(2)}_t)_t$ are independent standard Brownian motions and $\tau = \inf \{t \in \mathbb{R}\hspace{1mm} s.t. B^{(2)}_t = a \}$. I understand it should be a Cauchy distribution but I don't get where I can start, any help is welcome.

Answer 1

Hint

Reflexion principle gives you the distribution of $\tau$. Moreover, $$\mathbb P\{B^{(1)}_\tau\leq x\}=\int_0^\infty \mathbb P\{B_t^{(1)}\leq x\mid \tau=t\}\mathbb P\{\tau\in \,\mathrm d t\}=\int_0^\infty \mathbb P\{B_t^{(1)}\leq x\}\mathbb P\{\tau\in \,\mathrm d t\},$$

where the first equality comes from the total probability formula, and the second one from the fact that $B^{(1)}$ and $B^{(2)}$ are independents.

Answer 2

For anyone interested, here is what I got using Surb's Hint:

Thanks to the reflexion principle, we know the distribution of $\tau$:

\begin{align*} \mathbb{P}[\tau \leqslant t] &= \mathbb{P}[\sup_{0\leqslant s\leqslant t} B_s \geqslant a] &\text{by continuity of $B$ } \\ &= \mathbb{P}[ |B_t| \geqslant a] &\text{by the reflection principle}\\ &= \dfrac{1}{\sqrt{2\pi t}} \int_{a}^{\infty} e^{-\frac{x^2}{2t}}dx \\ &= \dfrac{1}{\sqrt{2\pi}} \int_{\frac{a}{\sqrt{t}}}^{\infty} e^{-\frac{x^2}{2}}dx \\ \end{align*}

Thus, $$\dfrac{d\mathbb{P}[\tau \leqslant t]}{dt}= \dfrac{a}{\sqrt{2\pi t^3}}e^{-\frac{a^2}{2t}}$$ Moreover, since $\tau$ is a deterministic function of $B^1$ and $B^1 \perp \!\!\! \perp B^2$, we have $\tau \perp \!\!\! \perp B^2$. This allows us to use the second inequality below for $x \in \mathbb{R}$:

\begin{align*} \mathbb{E}[e^{ixB_{\tau}^2}] &= \mathbb{E}_{\tau}[\mathbb{E}[e^{ixB_t^2}|\tau = t]]\\ &= \int_0^{\infty}\mathbb{E}[e^{ixB_t^2}] (\dfrac{a}{\sqrt{2\pi t^3}}e^{-\frac{a^2}{2t}})dt\\ &=\dfrac{a}{\sqrt{2\pi}} \int_0^{\infty} \dfrac{1}{t^{\frac{3}{2}}} e^{-(\frac{x^2t}{2}+ \frac{a^2}{2t})} dt \\ &=\dfrac{ae^{-a|x|}}{\sqrt{2\pi}} \int_0^{\infty} \dfrac{1}{t^{\frac{3}{2}}} e^{-(\frac{|x|\sqrt{t}}{\sqrt{2}}- \frac{a}{\sqrt{2t}})^2} dt\\ &=\dfrac{ae^{-a|x|}}{\sqrt{2\pi}} \int_0^{\infty} \dfrac{x}{a\sqrt{s}} e^{-(\frac{a}{\sqrt{2s}}-\frac{|x|\sqrt{s}}{\sqrt{2}})^2} ds &\text{by the change of variables } t = \dfrac{a^2}{x^2s} \end{align*}

And by taking the average of the two last expressions: \begin{align*} \mathbb{E}[e^{ixB_{\tau}^2}] &= \dfrac{1}{2}\mathbb{E}[e^{ixB_{\tau}^2}] + \dfrac{1}{2}\mathbb{E}[e^{ixB_{\tau}^2}] \\ &= \dfrac{ae^{-a|x|}}{2\sqrt{2\pi}} \int_0^{\infty} (\dfrac{1}{t^{\frac{3}{2}}} + \dfrac{x}{a\sqrt{t}}) e^{-(\frac{|x|\sqrt{t}}{\sqrt{2}}- \frac{a}{\sqrt{2t}})^2} dt \\ &= \dfrac{e^{-a|x|}}{\sqrt{\pi}} \int_0^{\infty} (\dfrac{a}{2\sqrt{2}t^{\frac{3}{2}}} + \dfrac{x}{2\sqrt{2}\sqrt{t}}) e^{-(\frac{|x|\sqrt{t}}{\sqrt{2}}- \frac{a}{\sqrt{2t}})^2} dt \\ &= \dfrac{e^{-a|x|}}{\sqrt{\pi}} \int_{-\infty}^{\infty}e^{-s^2}ds &\text{by the change of variables } s = \frac{|x|\sqrt{t}}{\sqrt{2}}- \frac{a}{\sqrt{2t}}\\ &= e^{-a|x|} \end{align*}

Thus, the characteristic function of $B_{\tau}^2$ is the one of the Cauchy distribution with parameter $a$ which implies that $B_{\tau}^2$ has this very distribution.