# Brownian Motion hitting time is finite yet has infinite expectation?

I've read that a hitting time of a Brownian motion (defined as $T_a = \inf\{t\ge0:W_t=a\}$ where $W_t$ is a standard Brownian Motion, i.e. a Wiener process), has the following two properties, which I am trying to understand:

$$\mathbb{P}(T_a \lt \infty ) = 1$$

$$\mathbb{E}[T_a]=\infty$$

These are from pg 6 of http://www.pstat.ucsb.edu/faculty/ludkovski/bmNotes.pdf, for example.

My question is what these mean and how they are not in conflict. I interpret the first one to say that with probability one the hitting time for any $a$ is finite. I interpret the second one to say that the expectation of the stopping time for any $a$ is infinite. This seems in conflict to me (i.e. saying that something is finite with probability one yet has infinite expectation?). Please help me gain some intuition here, and correct me if I am misunderstanding one or both of these properties.

• Similar situation, maybe simpler: assume $N$ is discrete, with positive integer values and such that $$P(N=n)=\frac1{n(n+1)}$$ for every $n\geqslant1$, then $N$ is almost surely finite because $$\sum_{n=1}^{\infty}P(N=n)=1$$ and its expectation is infinite because $$\sum_{n=1}^{\infty}n\,P(N=n)=+\infty.$$ – Did Apr 19 '16 at 7:00
• The St. Petersburg lottery is another example of something which is almost surely finite but has infinite expectation. Another (more closely related to Brownian motion) is the expected first return time of a simple $1$-D random walk – Henry Apr 19 '16 at 7:09
• But how would you prove this result? I tried using some stopping results and the thing I ended up with is that either $E[T_a]=\infty$ or $E[T_a] \in [0,1]$ – asdf Apr 22 '18 at 10:32