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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.

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By linearity of expectation, the expected number of returns is the sum of the probabilities of return. Thus we can prove the claim by induction.

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

\begin{align} E[N_{2(n+1)}]&=E[N_{2n}]+P_{00}^{(2(n+1))}\\ &=(2n+1) \dbinom{2n}{n} \left(\frac12\right)^{2n}-1+\binom{2(n+1)}{n+1}\left(\frac12\right)^{2(n+1)}\\ &=(2n+1) \dbinom{2n}{n} \left(\frac12\right)^{2n}-1+\binom{2(n+1)}{n+1}\left(\frac12\right)^{2(n+1)}\\ &=\left(\frac{(n+1)^2}{2(n+1)}\cdot2^2+1\right)\binom{2(n+1)}{n+1}\left(\frac12\right)^{2(n+1)}-1\\ &=(2(n+1)+1)\binom{2(n+1)}{n+1}\left(\frac12\right)^{2(n+1)}-1\;. \end{align}

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  • $\begingroup$ Thank you so much. Can you explain the properties you used to get the second to last line? $\endgroup$
    – kdrozd
    Mar 5 '16 at 22:45
  • $\begingroup$ @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 $$ \left(\frac12\right)^{2n}=2^2\left(\frac12\right)^{2(n+1)} \;. $$ $\endgroup$
    – joriki
    Mar 6 '16 at 7:41

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