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Let $\bar{\pi }$ be the normalized target and ${\bar{\pi }}_{-i}$ its marginal distribution. A coordinate-$i$ kernel $P_i$ leaves $x_{-i}=x_{-i}^{\prime }$ and has the joint old-new measureThis is symmetric in the old and new state: the common remaining coordinates are fixed and the two conditional factors exchange places. Thus each coordinate kernel satisfies detailed balance. Averaging them with the fixed weights $1/p$ provesIntegrating out the old state proves invariance of $\bar{\pi }$. The common normalization cancels, so detailed balance can also be written with the unnormalized target. This establishes detailed balance of a random-scan Gibbs sampler and a reversible Markov chain. Fresh independent random choices make the update a Markov chain, as explained above; invariance is not an assertion that an arbitrarily initialized chain starts in stationarity.

For $k\in {\mathcal{N}}_i$, put $a_k^{(i)}={\sum }_{j\in {\mathcal{N}}_k\setminus \{i\}}{s}_j$ and $H_i={\sum }_{j\in {\mathcal{N}}_i}{s}_j$. Removing all factors independent of $s_i$ from the posterior distribution gives, for $u\in \{-1,1\}$,A stable implementation uses the difference of the log weights. The conditional log odds areso ${\pi }_i(+1\mid s_{-i},x)=1/(1+e^{-L_i})$, a logistic function. For very large $|L_i|$, evaluate this logistic expression using the sign of $L_i$ to avoid exponential overflow.

For Gibbs sampling for a finite hidden spin field, initialize any spin configuration. At every step choose $i\in D$ uniformly, draw a fresh uniform variable, set spin $i$ to $+1$ with the probability above and to $-1$ otherwise, and leave all remaining spins fixed. This Random-scan Gibbs sampler has the stated posterior distribution invariant by detailed balance of a random-scan Gibbs sampler. All conditional probabilities are strictly between zero and one. For the one-site case, the conditional probabilities are both $1/2$.