I'm having problems understanding one inequality for one proof. The assumptions given are the following:
Let $S_t$ be the $(p-q)$-biased lazy random walk on $\mathbb{Z}$ and $\mu_t=\frac{t(p-q)}{2}$. Let $X_t=S_t \mod{n}$ with $X_0=x_0$ and let $\rho$ be the distance on the cycle. Set $A_t=\{k: \rho(k, \lfloor x_0+\mu_t \rfloor \mod{n}) \ge n/4\} $.
Why is it that:
$$P(X_t\in A_t)\le P(|S_t-\mu_t|\ge n/4)$$
If $X_t$ is in $A_t$, then, $S_n$ is not within $n/4$ of $\mu_t$ in mod n, but that is a stronger statement than saying that $S_n$ is not within $n/4$ of $\mu_t$. (I think it should say $n/4 + 1$ in the definition of $A_t$ which you can see by letting n = 4, and $x_0 + \mu_t = 0.5$, then, when $S_n = 1$, LHS includes it but RHS does not).