= Reversible density-ratio propagation
{title2=$d(K^t\nu)/d\pi=K^t(d\nu/d\pi)$}
For a <reversible Markov chain> with <stationary distribution> $\pi$ and $\nu\ll\pi$, <detailed balance> gives
$$
\frac{d(K^t\nu)}{d\pi}(x)=\int K^t(x,dy)\frac{d\nu}{d\pi}(y).
$$
To prove this, integrate either side against a bounded test <function> and exchange the two endpoints using <detailed balance>. If $\pi=h/Z$, the output <probability density function> is $q_t(x)=h(x)\mathbb E_{B\sim K^t(x,\cdot)}[\nu(B)/h(B)]$. A reverse simulation can therefore estimate this <density ratio> without evaluating a transition <probability density function>.
Back to article page
