After integrating out the <multivariate normal distribution> $\beta$, the <marginal distribution> is
$$
Y\sim N\!\left(0,\sigma^2A\right),
\qquad A=XX^T+\alpha I_n.
$$
Up to terms independent of $\sigma^2$, the <log-likelihood> is
$$
\ell(\sigma^2)=-\frac n2\log\sigma^2
-\frac1{2\sigma^2}Y^TA^{-1}Y.
$$
Differentiating and setting the result to zero gives the <maximum marginal likelihood estimator>
$$
\widehat{\sigma}^2=\frac1nY^T(XX^T+\alpha I_n)^{-1}Y.
$$
This is an <Empirical Bayes method> because the estimated <hyperparameter> is then inserted into the prior and posterior distributions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 216 / 1 / b / Solution

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>.

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

{wiki}

A marginal distribution is the distribution of a selected coordinate or subvector of a <random vector>, obtained from the joint distribution by integrating or summing over the remaining coordinates.

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 Incoming links: Marginal distribution