Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 3 / 2

The noise has the product Laplace distribution densitywhich is normalized because ${\int }_{\mathbb{R}}{e}^{-2|t|}dt=1$. By translation invariance of Lebesgue measure, the conditional law of $A(u)+N$ has densityHence an associated likelihood function isThe map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is continuous. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.

We may take the everywhere-defined representativeIt agrees with the likelihood from part a, hence certainly agrees ${\mu }_0\otimes {\lambda }_k$-almost everywhere. For every $u$, it is a strictly positive density in $y$ and is continuous in $y$. Moreoverso the constant function $h(u)=1$ is an integrable dominator for every probability measure ${\mu }_0$. All four sufficient assumptions from part 1d therefore hold, and the Bayesian inverse problem is well posed in total variation distance.