Product stationarity under single-coordinate Markov updates (${\pi }^{\otimes m}$)

Let $K$ be a Markov kernel with Stationary distribution $\pi$ on one coordinate. On $m$ coordinates, update coordinate $j$ with kernel $K$ while leaving the others unchanged. The product distribution ${\pi }^{\otimes m}$ is invariant: its updated coordinate still has law $\pi$, and 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 kernels and preserves the same product law.

For a two-state kernel with $K(1,1)=a$ and $K(0,1)=b$, the stationary Probability of state one solves $p=pa+(1-p)b$. If $1-a+b>0$, thenUnder the 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 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 kernel refresh to a fair bit, but select coordinate one whenever its current value is zero and coordinate two otherwise. Starting with 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 kernel alone is insufficient when the mixture weights depend on the state.