For a simple $n$-step random walk, how to show that the mean number of visits to the point $b$ is $\sum_{n=1}^\infty P(S_1 \cdots S_n \neq 0, S_n=b)$?

Question

I'm reading Grimmett and Stirzaker's Probability and Random Processes and stuck on one the claims.

In the scenario where $S_n$ is a simple random walk starting at $S_0 = 0$, the authors make the claim that, letting "$\mu_b$ be the mean number of visits of the walk to point $b$ before it returns to its starting point," $$ \mu_b = \sum_{n=1}^\infty P(S_1 \cdots S_n \neq 0, S_n=b). $$

This claim is on the bottom of page 79, which can be found here: https://books.google.com/books?id=G3ig-0M4wSIC&pg=PA79&lpg=PA79&dq=mean+number+of+visits+of+the+walk+to+the+point+b+grimmett+probability&source=bl&ots=BGliWRNRI2&sig=e6Mc2yLjAReDDMIeFLrJ12ivwkY&hl=en&sa=X&ved=0ahUKEwj159O_tM3SAhXE4iYKHemQCQEQ6AEIIjAB#v=onepage&q&f=false

I'd appreciate the help!

Answer 1

This is linearity of expectation and monotone convergence. \begin{align*} E(\text{# visits to $b$ before returning to origin}) \\ = E(\sum_{i=1}^n\{\text{visit at time $n$ to $b$ before returning to origin}\})\\ = \sum_{n=1}^\infty P(S_1 \cdots S_n \neq 0, S_n=b) \end{align*}

Answer 2

The number of visits $N: \Omega \rightarrow \mathbb{N}$ of the random walk to $B$ before returning to it's origin (taken to be $0$) is given by the random variable

$$ N = \sum_{k = 1}^{+\infty} \mathbb{I}\{S_{1}...S_{k-1} \neq 0\}.\mathbb{I}\{S_{k}=B\} = \lim_{n \rightarrow +\infty} \, \, \sum_{k = 1}^{n} \mathbb{I}\{S_{1}...S_{k-1} \neq 0\}.\mathbb{I}\{S_{k}=B\} $$

I use $\mathbb{I}$ as the indicator function here (can't seem to produce the correct symbol). Hence, by monotone convergence:

$$ E(N) = E(\lim_{n \rightarrow +\infty} \, \, \sum_{k = 2}^{n} \mathbb{I}\{S_{1}...S_{k-1} \neq 0\}.\mathbb{I}\{S_{k}=B\}) = \sum_{k = 1}^{+\infty}P(S_{1}...S_{k-1} \neq 0, S_{k} =B) $$