Show that the first passage time of Brownian motion has infinite expectation.

Question

Given $(W_{t})$ a standard Wiener process, define $T_{b}:=\inf\{t\geq 0:W_{t}=b\}$ to be its first passage time to a positive real number $b>0$. I'd like to show that $\mathbb{E}T_{b}=\infty$. There has been several posts about this:

Brownian Motion hitting time is finite yet has infinite expectation?

Paradox Brownian Motion : P(first passage time < infinite) = 1 yet E(first passage time) = infinite?

However, none of them solved this problem.

I have computed, using the reflection principle, that $T_{b}$ has density on $[0,\infty)$ that $$f_{T_{b}}(t)=\dfrac{b}{\sqrt{2\pi t^{3}}}e^{-\frac{b^{2}}{2t}}.$$ Then I tried to compute the expectation directly using this density, but it is clear that $$\int_{0}^{\infty}f_{T_{b}}(t)dt=1$$ for any $b>0$, so it is not $\infty$.

Which part of this argument is wrong? I should integral with respect to $db$? I am confused.

Thank you!

Answer 1

Note that for large $t$ we have $e^{\frac{-b^2}{2t}}$ approaches $e^{-0}=1$, so in particular there is a $T>0$ such that $e^{\frac{-b^2}{2t}}>1/2$ for all $t> T$. Thus

$$\int_0^\infty \frac{b}{\sqrt{2\pi t}}e^{\frac{-b^2}{2t}} dt\geq \int_T^\infty \frac{b}{\sqrt{2\pi t}}\frac{1}{2}dt=\frac{b}{\sqrt{2\pi }}\frac{1}{2} 2t^{1/2}|_T^{\infty}=\infty. $$

Answer 2

Assuming for simplicity that $b=1$, Wolfram Alpha shows that $$ \int_0^\infty \frac{e^{-1/(2t)}}{\sqrt{t^3}}\ dt = \sqrt{2\pi} \iff \int_0^\infty f(t) = 1. $$

Multiplying the integrand by $t$ should not cause issues around $t=0$ but may cause problems as $t \to \infty$. Not surprisingly, Wolfram Alpha also reports that $$ \int_0^\infty \frac{e^{-1/(2t)}}{\sqrt{t}}\ dt = \infty \iff \int_0^\infty tf(t) = \infty $$

In general, one expects $\int_1^\infty f(t) dt < \int_1^\infty tf(t) dt$.

Answer 3

Here's a "soft" approach. Let $T$ be the hitting time of $1$. (Assuming the Brownian motion starts at $0$.) Let $T_n=\min(T,n)$. Then $W_t^2-t$ is a martingale and $T_n$ is a bounded stopping time, so $\Bbb E[W_{T_n}^2]=\Bbb E[T_n]$. Moreover, $\Bbb E[T_n]$ increases to $\Bbb E[T]$. If the latter were finite then the sequence of random variables $\{W_{T_n}\}$ would be bounded in $L^2$, hence uniformly integrable. As the pointwise limit of $W_{T_n}$ is $W_T=1$ (a.s., because $T$ is finite a.s.), this uniform integrability would imply that $1=\Bbb E[W_T]=\lim_n\Bbb E[W_{T_n}]$. But $\Bbb E[W_{T_n}]=0$ for each $n$ because $W_t$ is a martingale with initial value $0$ and $T_n$ is a bounded stopping time.