Put
$$
S(x)=\sum_{i=1}^dp(x,i).
$$
The normalizing identity
$$
\int S(x)\nu(dx)
=\sum_i\int\nu(dx_{-i})\int\mu(dx_i\mid x_{-i})=d
$$
shows that
$$
\boxed{\pi(dx)=\frac{S(x)}d\nu(dx)}
$$
is a probability measure. If $x$ and $y$ differ only in coordinate $i$, the transition density from $x$ to $y$ is $p(x,i)\mu(y_i\mid x_{-i})/S(x)$. Therefore
$$
\begin{aligned}
\pi(x)K(x,y)
&=\frac1d\nu(x)p(x,i)\mu(y_i\mid x_{-i})\\
&=\frac1d\nu(x_{-i})
\mu(x_i\mid x_{-i})\mu(y_i\mid x_{-i}),
\end{aligned}
$$
which is symmetric in $x_i,y_i$. Thus detailed balance holds and the <Tempered Gibbs sampler> is $\pi$-reversible.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-216/3/a/solution