 Solution

Put $a=p(x)q(x,y)$ and $b=p(y)q(y,x)$. For $a+b>0$, the off-diagonal probability flux of the proposed rule isIf $a=b=0$, define both acceptance probabilities to be zero; both fluxes are zero. The diagonal detailed balance equations hold automatically, including the probability of rejected proposals. Summing the detailed balance equations over $x$ gives stationarity:The alternative rule therefore preserves $p$. The unknown normalizing constant cancels after replacing $p$ by $r$. This Barker acceptance probability is no larger than the Metropolis–Hastings acceptance probability for the same proposed move: if $t=b/a>0$, then $t/(1+t)\le \min (1,t)$. Preservation of $p$ still requires separate Markov chain irreducibility and convergence checks before it guarantees useful sampling.