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

The kernel $k(x,y)=e^{-|x-y|}$ is real and symmetric. For $f,g\in L^2[0,1]$, Fubini's theorem givesso $K$ is self-adjoint. It is linear, and because $k\in L^2([0,1{]}^2)$ it is a Hilbert-Schmidt operator, hence bounded, with

Suppose $Kf=\lambda f$. Differentiating the integral expression on the two sides of $y=x$ givesThereforeThe first boundary condition makesSubstitution into the second givesor

Extend $f$ by zero outside $[0,1]$. Since the Fourier transform of $e^{-|x|}$ is $2/(1+{\xi }^2)>0$,for $f\ne 0$. Thus every eigenvalue is positive. From the relation in part (b),The transcendental equation has its successive roots in intervals separated by the poles and zeros of $\tan \omega$, so ${\omega }_n$ grows linearly with $n$. Hence

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 seriesBecause $\mathbb{E}\Vert U-m{\Vert }^2={\sum }_n{\lambda }_n<\infty$, the series converges in $L^2(\Omega ;X)$ and almost surely. Its law is the Gaussian measure $N(m,K)$.

For observed $v\in {\mathbb{R}}^N$, the Gaussian likelihood isThus the Bayesian inverse problem is to determine the posterior distribution of $U$ given $V=v$. WithBayes' formula gives

The total variation distance isThe problem is a well-posed Bayesian inverse problem in total variation when every $v\in {\mathbb{R}}^N$ determines a unique posterior ${\mu }^v$ and

The heat solution operator at positive time is bounded from $L^2[0,1]$ to $C[0,1]$, so the finite sensor map $G:X\to {\mathbb{R}}^N$ is bounded and continuous. Therefore $\Phi (u;v)$ is jointly continuous and $0<e^{-\Phi }\le 1$. The normalizer satisfies $0<Z(v)\le 1$. If $v_n\to v$, the dominated convergence theorem gives bothand convergence in $L^1({\mu }_0)$ of the normalized posterior densities. Since total variation is one half of this $L^1$ distance for absolutely continuous measures, $d_{\mathrm{TV}}({\mu }^{v_n},{\mu }^v)\to 0$. Existence, uniqueness, and continuous dependence all follow.