Substituting $\widetilde\theta=\tau^2y/(\sigma^2+\tau^2)$ and $\sigma_\theta^2=\sigma^2\tau^2/(\sigma^2+\tau^2)$ into parts c and d and simplifying gives
$$
\mathbb E_{\theta\mid y}\log L(\theta)
-D_{\mathrm{KL}}\{p(\theta\mid y)\Vert\pi\}
=-\frac12\log\{2\pi(\sigma^2+\tau^2)\}
-\frac{y^2}{2(\sigma^2+\tau^2)}
=\log Z.
$$
Thus the equality holds.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 4 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/4/e/solution