I have very limited probability background, but I came across a problem in an engineering application:
Is there a formula that computes the average number of steps taken for a particle beginning at the origin to return to the origin if it has equal probabilities of moving left and right on a 1d chain, given that at some point n/2 to the right the particle gets reflected (P = 1 to left), and some point -n/2 to the left the particle also gets reflected (P = 1 to the right)?
In a Markov chain with limiting distribution $\vec{\pi}$, it takes $\frac1{\pi_i}$ expected time to return to state $i$ starting from that state. So it's enough to compute the limiting distribution of this Markov chain.
Rather than have reflective barriers at $-\frac n2$ and $\frac n2$, it is equivalent to make the Markov chain periodic modulo $n$, so that state $n$ is the same as state $0$. When you're at state $\frac n2$ in the periodic Markov chain, you can move to $\frac n2 -1$ or $\frac n2 + 1$, but either way you get $1$ step closer to $0$ - same as in the reflective Markov chain.
In the periodic Markov chain, it is easy to see that $\pi_0 = \frac1n$ by symmetry: there are $n$ indistinguishable states. Therefore the average return time is $\frac1{\pi_0} = n$.