Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 6H / Solution

The likelihood contributes precision $n$ and precision-weighted mean $n\overline{X}$, while the prior contributes precision ${\tau }^2$ and precision-weighted mean ${\tau }^2\theta$. By normal-normal conjugacy with known observation variance, the posterior isThus the posterior mean isWhen $\theta =0$, its expectation under true parameter $\mu$ is $n\mu /(n+{\tau }^2)$ and its variance is $n/(n+{\tau }^2{)}^2$. Hence its frequentist mean squared error is