# Hitting time by the minimum of two Brownian motions

Consider the hitting time, $$\tau_0$$, of $$0$$ by a Brownian motion $$B_t$$, started from $$B_0=1$$. It is well known that $$\mathbb{E}\tau_0 = \infty$$.

However we also know that the minimum of two random variables which each have infinite expectation is not necessarily infinite. Can we say anything at all about $$\mathbb{E}[\tau_0\wedge \tau_0']$$, the first hitting time of $$0$$ by two independent Brownian motions started from $$B_0=1$$, $$B'_0=1$$?

Denote $$W_t = -1+ B_t$$, it's easy to verify that $$W_t$$ is a standard Brownian motion started at $$0$$ (i.e. $$W_0 = 0$$).

We have:

\begin{align} \mathbb{P}(\tau_0\ge t) &= \mathbb{P}(\min_{0\le s\le t}B_s\ge 0)\\ &= \mathbb{P}(\min_{0\le s\le t}(1-W_s)\ge 0)\\ &=\mathbb{P}(\max_{0\le s\le t}W_s\le 1)\\ \end{align}

The law of running maximum of standard BM is known as $$\max_{0\le s\le t}W_s\overset{d}{=}|W_t|$$ Then $$\mathbb{P}(\tau_0\ge t) =\mathbb{P}(|W_t|\le 1) = 2\cdot\Phi\left(\frac{1}{\sqrt t} \right)$$ where $$\Phi(\cdot)$$ is the CDF of the standard normal distribution.

Now, we calculate the law of $$\tau_0\wedge\tau_0'$$, given the fact that the two BM $$B_t$$ and $$B'_t$$ are independent: $$\mathbb{P}(\tau_0\wedge\tau_0\ge t) = \mathbb{P}(\tau_0\ge t)\cdot \mathbb{P}(\tau_0'\ge t) = 4\cdot\Phi^2\left(\frac{1}{\sqrt t} \right) \tag{1}$$

From $$(1)$$, it's easy to obtain the closed-form expression of the expectation of $$\tau_0\wedge\tau_0'$$: \begin{align}\mathbb{E}(\tau_0\wedge\tau_0\ge t) &= \int_0^{+\infty}t\cdot \frac{d}{dt}\left( 1- \mathbb{P}(\tau_0\wedge\tau_0\ge t)\right) dt \\ &= \color{red}{ 4\int_0^{+\infty} \frac{1}{\sqrt{t}}\cdot \Phi\left(\frac{1}{\sqrt t} \right)\cdot \varphi\left(\frac{1}{\sqrt t} \right) dt} \tag{2}\end{align} where $$\varphi(\cdot)$$ is the density function of the standard normal distribution.

Using Mathematica, I obtained the closed form expression of $$(2)$$: $$\color{red}{4\int_0^{+\infty} \frac{1}{\sqrt{t}}\cdot \Phi\left(\frac{1}{\sqrt t} \right)\cdot \varphi\left(\frac{1}{\sqrt t} \right) dt = \frac{2^{3/4} \sqrt \pi \Gamma\left(\frac{1}{4}\right) + \Gamma\left(-\frac{1}{4}\right) \cdot 2F1\left(\frac{1}{2},1,\frac{5}{4},-1\right)}{π}\approx 2.27}$$ where $$2F1(\cdot )$$ is the hypergeometric function and the numerical result can be found here.

As a consequence, the expectation of the minimum of two hitting times is finite.