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Let 0 < p < 1 and let $S_n$ be the simple random walk with step probabilities p, 1 − p.

In other words $S_n = X_1 + · · · + X_n$ and the {$X_i$} are i.i.d. random variables with distribution $$P(X_i =1)=p \qquad and \qquad P(X_i =−1)=1−p$$ Fix a positive integer b > 0, and let T be the first hitting time of the point b: T(w) = inf{n ≥ 1 : $S_n(w)$ = b}. If the walk never hits b, in other words the set {n : $S_n$(w) = b} is empty, then T(w) = $\inf$.

How could I find ET?

Matt Smith
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  • Think averages. On average the walk moves $2p-1$ up. If $2p-1\le 0$, $E(T)=\infty$. Otherwise, $(2p-1)E(T)=b$. – A.S. Dec 09 '15 at 01:35
  • Can you elaborate a little. – Matt Smith Dec 09 '15 at 01:36
  • $M_n=S_n-n(2p-1)$ is a martingale. If $ET_b<\infty$ then $E(M_{T_b})=0$. You are left to note that $S_{T_b}=b>0$. – A.S. Dec 09 '15 at 01:41
  • @A.S. Cmiiw, but to say in the first place that $(2p-1)E[T] = b$ assumes that $E[T] < \infty$ ? – BCLC Dec 12 '15 at 20:53
  • Matt Smith, I'm trying to answer that problem too. I think we should use this proposition: https://books.google.com.hk/books?id=e9saZ0YSi-AC&pg=PA233&lpg=PA233&dq=%22what%20always%20stands%22%20%22probability%22&source=bl&ots=AWEio5s6Xq&sig=6kAyfQZorHRSFhYEVjQ83V7XGU4&hl=en&sa=X&ved=0ahUKEwje6t-I0MTJAhUleaYKHUa7AfIQ6AEIGjAA#v=onepage&q=%22what%20always%20stands%22%20%22probability%22&f=false – BCLC Dec 12 '15 at 21:00

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