Use the single-edge <heat-bath Markov chain>: choose an edge uniformly and resample it from its conditional random-cluster law. If its endpoints are already connected without that edge, its conditional open probability is $p$; otherwise opening it merges two components and the probability is
$$
\frac p{p+2(1-p)}=\frac p{2-p}.
$$
Detailed balance makes $\phi_{p,2}$ stationary. Driving this chain and Bernoulli heat-bath chains by the same update edges and uniforms gives the stochastic domination
$$
\operatorname{Ber}\left(\frac p{2-p}\right)^{\otimes E}
\preceq\phi_{p,2}\preceq
\operatorname{Ber}(p)^{\otimes E}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 209 / 3 / b / Solution

A heat-bath Markov chain repeatedly chooses one coordinate and redraws it from its conditional distribution given all other coordinates. The target distribution is reversible and stationary for these updates.


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