I was reading a proof on the lower bound of the mixing time of the lazy $(p − q)$-biased walk, which remains in its current position with probability 1/2, moves clockwise with probability $p/2$, and moves counterclockwise with probability $q/2$. Here $p + q = 1$. The proof goes as follows
where $\pi$ is the stationary distribution. I don't understand two steps in the proof.
- why is $\pi(A_t) \geq \frac{1}{2}$
- Why do we have the inequality $P(X_t \in A_t) \leq P(|S_t-\mu_t| \geq n/4)$
For the first one my guess is that the stationary distribution might be the uniform distribution but in that case shouldn't this be a equality? And for the second one I am confused how we can get rid of the $mod \ n$?