# Expected # of Returns in a Symmetric Simple Random Walk

The problem involves a 1-D symmetric simple random walk starting from the origin.

Let $N_{n}$ denote the the number of returns by time n. Show that:

$$E[N_{2n}]=(2n+1) \dbinom{2n}{n} (\frac{1}{2})^{2n}-1$$

I know that the Probability of being at zero after 2n steps is $P_{00}^{(2n)} = \dbinom{2n}{n} (\frac{1}{2})^{2n}$, but I'm not sure how to use this to solve for $E[N_{2n}]$. Any help would be greatly appreciated.

Base case: For $n=0$, $E[N_0]=0$, as required.
Induction step: Given $E[N_{2n}]=(2n+1) \dbinom{2n}{n} \left(\frac12\right)^{2n}-1$, we have
• @kdrozd: I used $$\binom{2n}{n} = \frac{(2n)!}{n!^2} = \frac{(n+1)^2}{(2n+1)(2n+2)}\frac{(2n+2)!}{(n+1)!} = \frac{(n+1)^2}{(2n+1)(2n+2)}\binom{2(n+1)}{n+1}$$ and $$\left(\frac12\right)^{2n}=2^2\left(\frac12\right)^{2(n+1)} \;.$$ Mar 6 '16 at 7:41