Prove the expected number of visits to $k$ before the first return to 0 is 1 Let $(S_n)_{n\geq0}$ be a simple symmetric random walk with $S_0 =0$ and $k\in \mathbb{Z}$ \ $ \{0\}$ and $N_k$ be the number of visits to $k$ before the first return to $0$. Then $E[N_k] =1$
I need to prove this. 
So far:
For $k>0$, I can write $N_k$ as a sum of indicator functions of the events $A_{k,n} = \{S_1>0,S_2>0,...,S_{n-1}>0, S_n=k\}$
 A: Let us suppose the random variable $N_{i,k}$ denotes the number of returns to $k$ before hitting $0$, starting at (and not including landing at) $i$. It should be obvious that if $i$ and $k$ have opposite signs, then $N_{i,k}$ is the zero random variable. But by the same reasoning, if $i$ is on the opposite side of $k$ from $0$, then $E[N_{i,k}]$ is independent of $i$. (Essentially, if you’re on the bad side of $k$, you have to run into $k$ eventually to get to $0$, and the system is memoryless so right when you hit $k$, nothing else matters.)
So then we set up equations: if $a_i:=a_{i,k}=E[N_{i,k}]$ (and WLOG $k>0$) then we get the system
$\begin{align}a_{-1}=a_{-2}=\ldots&=0\\
a_0&=\frac{a_1}{2}\\
a_1&=\frac{a_2}{2}\\
a_2&=\frac{a_3+a_1}{2}\\
&\ldots\\
a_{k-1}&=\frac{a_{k-2}+(a_k+1)}{2}\\
a_k&=\frac{a_{k-1}+a_{k+1}}{2}\\
a_{k+1}&=\frac{(a_k+1)+a_{k+1}}{2}\end{align}$
This is tedious to solve, but the solution ends up being
$a_0=1, a_1=2, a_2=4, a_3=6, \ldots a_{k-1}=2k-2, a_k=2k-1, a_{k+1}=2k.$
So the expected number of times you hit $k$ will always be $1$ starting at $0$.
A: Consider the random process $S_{\tau(n)}$ where $0=\tau(0)<\tau(1)<\tau(2)<\dots$ are the times $t$ such that $S_t\in\{0,k\}.$ Since $\mathbb Z$ is recurrent, this is well defined except on a set of measure zero.
$S_{\tau(n)}$ is a Markov chain with two states, $0$ and $k.$ It's ergodic, and by symmetry  the stationary distribution is the uniform distribution. A fundamental result of discrete Markov chains is that the mean recurrence time for a state is the reciprocal of the stationary probability, so it's $2.$ This means the expected time spent in state $k$ between visits to $0$ is $2-1.$
