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.
