The compact self-adjoint operator $K$ has an <orthonormal basis> of eigenvectors $e_n$. Part (c) gives $\lambda_n>0$ and $\sum_n\lambda_n<\infty$, so $K$ is positive and trace class, exactly the required condition for a <covariance operator of a Gaussian measure> on a Hilbert space.

For independent standard normal variables $\xi_n$, define the <Hilbert-space Gaussian series>
$$
\boxed{
U=m+\sum_{n=1}^\infty\sqrt{\lambda_n}\,\xi_ne_n}.
$$
Because $\mathbb E\|U-m\|^2=\sum_n\lambda_n<\infty$, the series converges in $L^2(\Omega;X)$ and almost surely. Its law is the <Gaussian measure> $N(m,K)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 / 2 / d / Solution

The covariance operator $K$ of a Gaussian measure on a real Hilbert space satisfies $\langle Kh,g\rangle=\operatorname{Cov}(\langle U,h\rangle,\langle U,g\rangle)$. A positive self-adjoint operator is the covariance of a Hilbert-space-valued Gaussian random variable exactly when it is trace class.


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 Incoming links: Covariance operator of a Gaussian measure