# Expectation of stopping time for random walk with probability $\frac{1}{2}$.

Continuing this post,

We consider a simple symmetric random walk $(S_n)_{n \in \mathbb{N}}$ on $\mathbb{Z}$ which starts at 1 : $S_0 = 1$ and there exists an iid sequence $(X_n)_{n \geq 1}$ such that $\mathbb{P}(X_1 = -1) = \mathbb{P}(X_1 = 1) = 1/2$ and $\forall n \in \mathbb{N}$ $S_{n+1} = S_n + X_{n+1}$ And we look at the stopping time $T = inf \{ n \geq 0, S_n = 0 \}$

I am curious how one computes $E(T)$, or even show it is finite/infinite? (I seem to remember I have read that $E(T)=\infty$.)

My idea:

$E(\tau) = \int_{\tau=n} \tau(\omega) \, dP = \sum n P(\tau =n) =\sum P(\tau \ge n)$. By Borell Cantelli's, if the latter is infinite,then $P( \{\tau \ge n\} \, i.o. )=1$, contradicting $P(\tau < \infty)=1$. But this holds only if the events are independent, which are clearly not...

• In order to show that $\mathbb{E}(T)=\infty$ one can apply Walds identities. If the expectation was finite, then it would follow from Walds identities that $S_{\tau}=1$ which is clearly not true. – saz May 20 '18 at 16:18