Proposition 7.8 -- the distinguishing statistic bound

Let $\mu$ and $\nu$ be two probability distributions on a finite set $V$, and let $f:V\to\mathbb R$ be any real-valued statistic. Write $E_\mu(f)=\sum_x f(x)\mu(x)$ and $\operatorname{Var}_\mu(f)=E_\mu(f^2)-E_\mu(f)^2$ for the mean and variance of $f$ under $\mu$, and likewise for $\nu$, and set

the average of the two variances. Assume the statistic actually tells the two distributions apart in the mean: $E_\mu(f)\ne E_\nu(f)$. The total variation distance is $\|\mu-\nu\|_{TV}=\max_{A\subseteq V}|\mu(A)-\nu(A)|$, the largest discrepancy the two distributions assign to any event.

The theorem (Levin–Peres–Wilmer, Proposition 7.8) asserts: if the statistic separates the two means by at least $r\ge0$ standard deviations,

then the distributions themselves are far apart:

This is the standard route to a mixing-time lower bound. To show a chain is far from stationarity at time $t$, one need not analyze the whole distribution: it is enough to find a single statistic whose mean shifts by many standard deviations between $P^t(x,\cdot)$ and $\pi$. The bound is scale-free — only the ratio "mean gap over standard deviation" matters — and it improves with $r$, approaching $1$ as $r\to\infty$; a separation of $r$ standard deviations already forces the two distributions to disagree on some event with probability at least $1-4/(4+r^2)$.

A note on the hypothesis $E_\mu(f)\ne E_\nu(f)$. It does not appear in the proposition as printed, but it is exactly what the proof assumes: the argument opens by assuming the two means are distinct and concludes through $\|\mu-\nu\|_{TV}\ge1-\sigma^2/(\sigma^2+M^2)$, where $M=|E_\mu(f)-E_\nu(f)|/2$. Without a non-degeneracy assumption the statement is false as written: if $f$ has zero variance under both distributions and the two means agree, then the hypothesis reads $r\cdot0\le0$ and holds for every $r$, while the conclusion demands a positive lower bound on a total variation distance that may be $0$ — take $\mu=\nu$ the point mass on a one-point space, $f=0$ and $r=5$. Distinct means also keep $\sigma^2+M^2$ strictly positive, which is what the final step of the proof divides by. Nothing is lost in applications: a statistic with equal means separates nothing, and the bound it would give at the only admissible $r$ is the vacuous $0\le\|\mu-\nu\|_{TV}$.