For $f,g\in L^2(\pi)$, <detailed balance> says that the measure $\pi(dx)K(x,dy)$ is invariant under exchanging $x$ and $y$. Therefore
$$
\begin{aligned}
\langle f,Kg\rangle_\pi
&=\int f(x)g(y)\,\pi(dx)K(x,dy)\\
&=\int f(y)g(x)\,\pi(dx)K(x,dy)
=\langle Kf,g\rangle_\pi.
\end{aligned}
$$
This is precisely the defining identity for a <self-adjoint operator>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 216 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-216/3/b/solution