Probability that a symmetric random walk returns to $0$ exactly $k$ times in $2n$ steps

Question

I'm trying to find a formula to find the probability of exactly k returns in 2n steps of a symmetric random walk. More specifically, I am trying to show that the probability of 2 returns is exactly equal to the probability of return to 0 at 2n minus the probability of first return to 0 at 2n. I.e. probability of exactly two returns to 0 is $\ p _{0,0}^{2n}$ - $\ f _{2n}$ where $\ p _{0,0}^{2n}$ is the probability of return to 0 at 2n and $\ f _{2n}$ is the probability of first return to 0 at 2n

Answer 1

Stated slightly differently, the formula says that $$ P(R=2)=P(R\ge 2, X_{2n}=0), \quad\quad\quad\quad (1) $$ with $R$ denoting the number of returns and $X_k$ denoting the position of the random walk.

For any path with $R\ge 2$, focus on the portion after the second return to $0$. We again start this final portion of the path at $0$, and thus there are as many paths ending at $0$ as there are paths that avoid $0$ altogether. (This is a well know fact about random walks; for example, see Lemma 2.3 in my notes here.) If we count both types of paths in this way, we obtain (1).

More generally, the same argument shows that for any $k\ge 0$, we have $P(R=k)=P(R\ge k, X_{2n}=0)$.