Let $S_n$ be a simple symmetric random walk. Show that $P(S_1S_2...S_{2n} \neq 0) = P(S_{2n} = 0)$

Solution 1:

Let $T_0=\min\{n\ge 1:S_n=0\}$. It can be shown that for $n\ge 1$, $$ \mathsf{P}(T_0=2n)=\frac{1}{2n-1}\mathsf{P}(S_{2n}=0)=\frac{1}{2n-1}\binom{2n}{n}2^{-2n}. $$ Thus, $$ \mathsf{P}(T_0>2n)=\sum_{j>n}\frac{1}{2j-1}\binom{2j}{j}2^{-2j}=\binom{2n}{n}2^{-2n}=\mathsf{P}(S_{2n}=0). $$

Alternatively, set $q_n:=\mathsf{P}(S_n=0)$ and $p_n:=\mathsf{P}(T_0>n)$, and for $j\le n$ let $$ A_j=\{S_{2j}=0,S_{2j+1}\ne 0,\ldots,S_{2n}\ne 0\}. $$ Then, $$ \mathsf{P}(A_j)=\mathsf{P}(S_{2j}=0)\times \mathsf{P}(T_0>2(n-j))=q_{2j}p_{2(n-j)}, $$ and since $A_0,\ldots,A_{2n}$ are disjoint, $$ \sum_{j=0}^n q_{2j}p_{2(n-j)}=\sum_{j=0}^n \mathsf{P}(A_j)=1, $$ which implies that $q_{2n}=p_{2n}$.