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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...

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    $\begingroup$ 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. $\endgroup$ – saz May 20 '18 at 16:18

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