= Product stationarity under single-coordinate Markov updates
{title2=$\pi^{\otimes m}$}
Let $K$ be a <Markov kernel> with <Stationary distribution> $\pi$ on one coordinate. On $m$ coordinates, update coordinate $j$ with <Markov kernel>[kernel] $K$ while leaving the others unchanged. The <product measure>[product distribution] $\pi^{\otimes m}$ is invariant: its updated coordinate still has law $\pi$, and <independence>[independent transition randomness] leaves it <independent> of the untouched coordinates. Choosing the updated coordinate independently of the current configuration, with any fixed selection <probabilities>, forms a mixture of these invariant <Markov kernel>[kernels] and preserves the same product law.
For a two-state <Markov kernel>[kernel] with $K(1,1)=a$ and $K(0,1)=b$, the stationary <Probability> of state one solves $p=pa+(1-p)b$. If $1-a+b>0$, then
$$
p=\frac{b}{1-a+b}.
$$
Under the <product measure>[product stationary distribution], the occupied-coordinate count has <binomial distribution> with parameters $m,p$, <mean> $mp$ and <variance> $mp(1-p)$. In the exceptional case $a=1,b=0$, the <Markov kernel>[kernel] is the identity and every $p\in[0,1]$ is stationary; the displayed fraction is undefined and must not be used.
State-dependent coordinate selection can invalidate the conclusion. On two bits, let each single-coordinate <Markov kernel>[kernel] refresh to a fair bit, but select coordinate one whenever its current value is zero and coordinate two otherwise. Starting with <independent random variables>[independent fair bits], the <Probability> that the first bit equals one after a step is $1/2+(1/2)(1/2)=3/4$, so the uniform product law is not invariant. Invariance of each coordinate-update <Markov kernel>[kernel] alone is insufficient when the mixture weights depend on the state.
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