Let $X_n$ be an aperiodic, discrete-time Markov chain so $S=\{1,...,N\}$ whose transition probability is symmetric. How can I show that for all $x \in S$ and all integers $n$, $P(S_{2n}=x|S_0=x) \ge \frac{1}{N}$. And would this hold in $2n$ is replaced with $2n+1$?
I'm really stuck and have no idea how to do this practice problem. Any help would be appreciated. Thanks
Let $p$ denote the transition matrix of the Markov chain, then $p$ is symmetric hence $p^n(x,y)=p^n(y,x)$ for every $x$ and $y$ in $S$ and, for every $x$ in $S$, $$ |S|\cdot p^{2n}(x,x)=|S|\cdot\sum_{y\in S}p^n(x,y)p^n(y,x)=\sum_{y\in S}1\cdot\sum_{y\in S}(p^n(x,y))^2. $$ By Cauchy-Schwarz inequality, the RHS is at least $$ \left(\sum_{y\in S}p(x,y)\right)^2=1.$$
No, as shown by chains of period $2$ in a neighborhood of radius $n+1$ around $x$.