= Solution
For <probability distributions> on a finite <set>, the <total variation distance> is
$$
\|\mu-\nu\|_{\mathrm{TV}}=\max_{A\subseteq S}|\mu(A)-\nu(A)|=\frac12\sum_{z\in S}|\mu(z)-\nu(z)|.
$$
The equality follows by taking $A=\{z:\mu(z)\geq\nu(z)\}$: the positive and negative parts of $\mu-\nu$ have the same mass, since its total mass is zero.
Use the <stationary distribution> identity $\pi P^t=\pi$. For each initial state $x$,
$$
P^t(x,\cdot)-\pi=\sum_y\pi(y)\bigl(P^t(x,\cdot)-P^t(y,\cdot)\bigr).
$$
The <triangle inequality> for <total variation distance>, followed by the definition of the <pairwise mixing diameter>, gives
$$
\|P^t(x,\cdot)-\pi\|_{\mathrm{TV}}\leq\sum_y\pi(y)\bar d(t)=\bar d(t).
$$
Taking the maximum over $x$ proves $\boxed{d(t)\leq\bar d(t)}$. This particular inequality only needs a <stationary distribution>; it does not require an <irreducible Markov chain> or an <aperiodic Markov chain>.
Back to article page
