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At crossover probability $p=1/2$, the two rows of the transition matrix of a binary symmetric channel coincide. Its output is then independent of its input and its capacity is zero.

Suppose an involution $R$ of a Markov chain satisfies $P(Rx,Ry)=P(x,y)$. Start one chain at $x$, set the other equal to its image under $R$ until the first hitting time of the fixed-point set of $R$, and subsequently use identical transitions. This is a coalescing coupling, because the symmetry preserves each marginal transition matrix and both chains coincide at that hitting time.

For a finite Markov chain with transition matrix $P$, define $\bar{d}(t)={\max }_{x,y}\Vert P^t(x,\cdot )-P^t(y,\cdot ){\Vert }_{\mathrm{TV}}$. If $\pi$ is a stationary distribution and $d(t)={\max }_x\Vert P^t(x,\cdot )-\pi {\Vert }_{\mathrm{TV}}$, then $d(t)\le \bar{d}(t)\le 2d(t)$. The first bound follows by writing $\pi ={\sum }_y\pi (y)P^t(y,\cdot )$ and using the triangle inequality; the second uses the triangle inequality through $\pi$.

The Markov property and the Chapman-Kolmogorov equation giveFor each intermediate state, one first uses the next-step transition probability and then the following one; the sampled history adds no information once $X_{2n}$ is known. Thus $W$ is a homogeneous Markov chain with transition matrix $P^2$ and initial law $\lambda$. In particular,If zero is not a state of $S$, this probability is zero.