= Solution
Let $K(t,\mathord\cdot)$ be the <Markov kernel> of the <Metropolis–Hastings algorithm>. Away from the possibility of remaining at $t$, its transition density is
$$
k(t,s)=q(s\mid t)\rho(t,s).
$$
Multiplication by the target density gives
$$
\begin{aligned}
\mu(t)k(t,s)
&=\mu(t)q(s\mid t)
\min\left\{\frac{\mu(s)q(t\mid s)}{\mu(t)q(s\mid t)},1\right\}\\
&=\min\{\mu(s)q(t\mid s),\mu(t)q(s\mid t)\}\\
&=\mu(s)k(s,t).
\end{aligned}
$$
This is <detailed balance> for every pair of distinct states.
Write
$$
r(t)=1-\int_\Theta k(t,s)\,ds
$$
for the rejection probability, so
$$
K(t,ds)=k(t,s)\,ds+r(t)\delta_t(ds).
$$
For every measurable $B\subseteq\Theta$, detailed balance and the <Tonelli theorem> give
$$
\begin{aligned}
\int_\Theta K(t,B)\mu(t)\,dt
&=\int_B\int_\Theta \mu(t)k(t,s)\,dt\,ds
+\int_B\mu(t)r(t)\,dt\\
&=\int_B\mu(s)\left[\int_\Theta k(s,t)\,dt+r(s)\right]ds\\
&=\int_B\mu(s)\,ds.
\end{aligned}
$$
\b[Hence $\mu$ is an <invariant distribution> of the chain.]
Back to article page
