By the <stationary distribution> property, $X_1$ and $X_2$ have the same <marginal distribution> $\pi$. Expanding the square gives
$$
\begin{aligned}
\frac12\mathbb E[(f(X_2)-f(X_1))^2]
&=\frac12\left(2\langle f,f\rangle_\pi
-2\mathbb E[f(X_1)f(X_2)]\right)\\
&=\langle f,f\rangle_\pi-\langle f,Kf\rangle_\pi\\
&=\langle f,(I-K)f\rangle_\pi
=\mathcal E_K(f).
\end{aligned}
$$
This is the probabilistic representation of the <Dirichlet form of a Markov chain>.

Solution

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

 Solution

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