# Divergent mean first passage time

We are going to construct an example of a first passage time without a mean.

Consider a stochastic process $X_t$ on the real line defined by placing a point particle with unit mass at the origin and setting it in motion with constant velocity $V$. The velocity $V$ is drawn from the Maxwell-Boltzmann distribution corresponding to a fixed temperature $\beta^{-1} > 0$.

More explicitly, a trajectory of this process is characterized by \begin{equation*} X_t = t V, \quad \text{where} \quad V \sim N(0, \beta^{-1}) \end{equation*}

Consider the interval $\Omega = (-1, 1) \subset \mathbb{R}$. The first passage time (FPT or exit time) from $\Omega$ is the non-negative random variable defined by $$T = \inf \{ t \ge 0 \mid X_t \not\in \Omega \}$$ or, in other words, the first instant at which the process crosses the boundary of $\Omega$.

We see that $T = 1 / |V|$ and, consequently, its probability density function is $$f(t) = \sqrt{\frac{2 \beta}{\pi}} t^{-2} \e^{-\frac{\beta}{2} t^{-2}}.$$

The figure above could be interpreted as being the result of repeating the experiment many times with many initial velocities all sampled from the Maxwell-Boltzmann distribution. Early on, it is impossible for the atom to move from its initial position at the origin to the boundary of $\Omega$, so the probability of exiting the interval is almost zero for a little while. Then, the probability grows as particles arrive at the boundary and it finally tapers off as the slowest atoms exit the interval.

Now, we can see that the distribution of FPTs has a heavy tail. How heavy? So much so that the mean goes to $+\infty$! Recall that the initial velocities were normally distributed with mean zero. So there are many arbitrarily slow trajectories that contribute to the distribution of the FPT.

Indeed, the mean FPT is $$\mathbb{E}[T] = \int_0^\infty t \, f(t) \, \mathrm{d} t = \int_0^\infty \sqrt{\frac{2 \beta}{\pi}} t^{-1} \e^{-\frac{\beta}{2} t^{-2}} \, \mathrm{d} t.$$ From the fact that the integrand is non-negative and the inequality $\e^x \ge 1 + x$, valid for all $x \in \R$, we see that \begin{align*} \int_0^\infty t^{-1} \e^{-\frac{\beta}{2} t^{-2}} \, \mathrm{d} t &= \int_0^\epsilon t^{-1} \e^{-\frac{\beta}{2} t^{-2}} \, \mathrm{d} t + \int_\epsilon^\infty t^{-1} \e^{-\frac{\beta}{2} t^{-2}} \, \mathrm{d} t \\ &\ge \int_\epsilon^\infty \left( \frac{1}{t} - \frac{\beta}{2 t^3} \right) \, \mathrm{d} t, \end{align*} for any $\epsilon > 0$, and we conclude that $\E[T] = +\infty$.

Published: November 23, 2022.