Hitting time of Brownian Motion combined with another Brownian Motion gives "Cauchyprocess"

Question

Let $W(t)_{t\ge 0}, B(t)_{t\geq0}$ be independent Brownian Motions on $\Bbb R$ an $\tau(a) := \inf \{ t \geq 0 : W(t) = a\}$ the first hitting time of $a\geq0$. Now I want the distribution of $$X(a) := B(\tau (a))$$

My attempt (EDITED):

I know the density of $\tau(a)$. It is $$\rho_{\tau(a)}(t)= \frac a {\sqrt{2 \pi t^3}} \exp\left({\frac {- a^2}{2t}}\right)$$ So further we can calculate for a suitable function $f$: $$\Bbb E[f(X(a))] = \Bbb E [f(B(\tau (a)))] =\int_0^\infty \frac a {\sqrt{2 \pi t^3}} \exp\left({\frac {- a^2}{2t}}\right) \int_\Bbb R \frac 1 {\sqrt{2\pi t}}\exp\left( \frac {- y^2}{2t} \right) f(y) \text{d}y \text{d}t \\ = \int_\Bbb R f(y) \int_0^\infty \frac{a}{2\pi t^2}\exp\left( \frac {-a^2 - y^2}{2t} \right) \text{d}t \text{d}y $$

But here I don't know how to get further. So far I know it should be $$\int_0^\infty \frac{a}{2\pi t^2}\exp\left( \frac {-a^2 - y^2}{2t} \right) \text{d}t \overset{!}{=} \frac {a}{\pi(a^2 +y^2)}$$ But I think that is not the way to do it, maybe someone has another idea for this.

EDIT: it is straight forward to show the (!)

Answer 1

You made a little mistake in your expression. You should have written $$ \begin{split} \Bbb E\left[f\big(B(\tau(a))\big)\right]&=\int_0^\infty\mathrm d t\frac{a}{\sqrt{2\pi t^3}}\mathrm e^{-a^2/2t}\int_{\Bbb R}f(y)\exp\left(-\frac{y^2}{2{\color{red} t}}\right)\frac{\mathrm dy}{\sqrt{2\pi t}}\\ &=\frac{a}{2\pi}\int_{\Bbb R}f(y)\mathrm dy\int_0^\infty\exp\left(-\frac{a^2+y^2}{2t}\right)\frac{\mathrm dt}{t^2}\\ &=\int_{\Bbb R}f(y)\underbrace{\frac{a}{\pi}\frac{1}{a^2+y^2}}_{\text{Cauchy law density}}\,\mathrm dy. \end{split} $$

Answer 2

Distributions are uniquely characterized by their characteristic function, and therefore it suffices to calculate the characteristic function. Using the independence of $(B_t)_{t \geq 0}$ and $\tau(a)$ we find

$$\mathbb{E}\exp(i \xi X(a)) = \int \mathbb{E}\exp(i \xi B_t) \, d\mathbb{P}_{\tau(a)}(t) = \int \exp \left(- \frac{t}{2} |\xi|^2 \right) \, d\mathbb{P}_{\tau(a)}(t).$$

(Here $\mathbb{P}_{\tau(a)}$ denotes the distribution of $\tau(a)$.) Plugging in the distribution of $\tau(a)$, we obtain

$$\begin{align*} \mathbb{E}\exp(i \xi X(a)) &= \frac{a}{\sqrt{2\pi}} \int_{(0,\infty)} \frac{1}{t^{3/2}} \exp \left(- \frac{t}{2} \xi^2- \frac{a^2}{2t} \right) \, dt \\ &= \frac{a}{\sqrt{2\pi}} e^{-|\xi| a}\int_{(0,\infty)} \frac{1}{t^{3/2}} \exp \left(- \left[\sqrt{\frac{t}{2}} |\xi|- \frac{a}{\sqrt{2t}} \right]^2 \right) \, dt. \tag{1} \end{align*}$$

If we change the variables according to $t=a^2/(|\xi|^2 s)$, i.e. $dt=-a^2/(\xi^2 s^2) ds$, we find

$$\begin{align*} \mathbb{E}\exp(i \xi X(a)) &= \frac{a}{\sqrt{2\pi}} e^{-|\xi| a} \frac{a^2}{\xi^2} \int_{(0,\infty)} \left( \frac{\xi^2 s}{a^2} \right)^{3/2} \exp \left(- \left[ \frac{a}{\sqrt{2s}} - |\xi| \sqrt{\frac{2}{s}} \right]^2 \right) \frac{ds}{s^2} \\ &= \frac{1}{\sqrt{2\pi}} e^{-|\xi| a} |\xi| \int_{(0,\infty)} \frac{1}{\sqrt{s}} \exp \left(- \left[ \frac{a}{\sqrt{2s}} - |\xi| \sqrt{\frac{s}{2}} \right]^2 \right)\,ds. \tag{2} \end{align*}$$

Writing

$$\mathbb{E}\exp(i \xi X(a)) = \frac{1}{2} \mathbb{E}\exp(i \xi X(a)) + \frac{1}{2} \mathbb{E}\exp(i \xi X(a))$$

we find from $(1)$ and $(2)$

$$\begin{align*} \mathbb{E}\exp(i \xi X(a)) &= \exp(-|\xi| a) \frac{1}{2\sqrt{2\pi}} \int_{(0,\infty)} \left( \frac{a}{t^{3/2}} + \frac{|\xi|}{\sqrt{t}} \right) \exp \left(- \left[\sqrt{\frac{t}{2}} |\xi|- \frac{a}{\sqrt{2t}} \right]^2 \right) \, dt. \end{align*}$$

Now we perform a further change of variables; we set $y := \sqrt{t/2} |\xi| - a/\sqrt{2t}$, i.e. $$dy = \frac{1}{2 \sqrt{2}} \left( \frac{|\xi|}{\sqrt{t}}+ \frac{a}{t^{3/2}} \right) \, dt,$$

and get

$$\begin{align*} \mathbb{E}\exp(i \xi X(a)) &= \exp(-|\xi| a) \frac{1}{\sqrt{\pi}}\int_{\mathbb{R}} \exp(-y^2) \, dy = e^{-|\xi|a}. \end{align*}$$

The right-hand side is the characteristic function of the Cauchy distribution (with parameter $a$), and this finishes the proof.