# Random walk mean number of visits to state before absorption

This is from Stirzaker's book Random Processes.

Suppose we have a simple random walk with probability going "up" p, "down" q. At time 0 it stats at 0, so

$$S_0 = 0$$

Now let $u_b$ be the mean number of visits to "state" b before returning to the origin. The following is given without proof:

$$u_b = \sum^n_{i=1}P(S_1S_2..S_n \neq 0, S_n = b)$$

where the the probability in brackets is the probability of random walk reaching b on the nth step AND not touching 0 during the trip. By the balot theorem, $P(S_1S_2..S_n \neq 0|S_n = b) = b/n$. But I still dont understand how the equality is found. Any help much appreciated.

• Which equation is the one you don't understand? What does $P(S_1S_2...S_n|S_n=b)$ mean? – pajonk Jun 20 '16 at 17:38
• The main equality ub = .... And thank you for pointing out the type, it should be S1S2..Sn not equal to 0 | Sn = b – Malin Jun 20 '16 at 17:57
• There must be a typo in your question. The sum is over an index $i$ that does not appear anywhere. – Michael Jun 20 '16 at 17:59
• Yes, sorry for all the typos, and thank you for answering the question! – Malin Jun 20 '16 at 18:09

## 1 Answer

There was likely a typo in your question, as the sum is over an index $i$ that does not appear anywhere. I think the true equation is intended to be: $$u_b = \sum_{i=1}^{\infty} P[S_1S_2\cdots S_i \neq 0, S_i=b]$$ You can prove that by defining $N_b$ as the random number of times we hit $b$ before returning to $0$ and then taking expectations of the identity: $$N_b = \sum_{i=1}^{\infty} 1\{S_i=b\}1\{S_1S_2\cdots S_i \neq 0\}$$ with $1\{A\}$ being an indicator function of the event $A$ (being 1 if $A$ occurs and $0$ else).