For observed $v\in\mathbb R^N$, the Gaussian likelihood is
$$
\boxed{
\pi(v\mid u)
=(2\pi\sigma^2)^{-N/2}
\exp\left[-\frac1{2\sigma^2}\|v-G(u)\|_{\mathbb R^N}^2\right]}.
$$
Thus the <Bayesian inverse problem> is to determine the posterior distribution of $U$ given $V=v$. With
$$
\Phi(u;v)=\frac1{2\sigma^2}\|v-G(u)\|^2,
$$
Bayes' formula gives
$$
\boxed{
\frac{d\mu^v}{d\mu_0}(u)
=\frac1{Z(v)}e^{-\Phi(u;v)},
\qquad
Z(v)=\int_Xe^{-\Phi(u;v)}\,d\mu_0(u)}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-326/2/e/solution