Define sufficient statistic, and state the factorisation criterion for determining whether a statistic is sufficient. Show that a Bayesian posterior distribution depends on the data only through the value of a sufficient statistic.

Given the value $\mu$ of an unknown parameter $M$, observables $X_{1}, \ldots, X_{n}$ are independent and identically distributed with distribution $\mathcal{N}(\mu, 1)$. Show that the statistic $\bar{X}:=n^{-1} \sum_{i=1}^{n} X_{i}$ is sufficient for $\mathrm{M}$.

If the prior distribution is $\mathrm{M} \sim \mathcal{N}\left(0, \tau^{2}\right)$, determine the posterior distribution of $\mathrm{M}$ and the predictive distribution of $\bar{X}$.

In fact, there are two hypotheses as to the value of M. Under hypothesis $H_{0}$, $\mathrm{M}$ takes the known value 0 ; under $H_{1}, \mathrm{M}$ is unknown, with prior distribution $\mathcal{N}\left(0, \tau^{2}\right)$. Explain why the Bayes factor for choosing between $H_{0}$ and $H_{1}$ depends only on $\bar{X}$, and determine its value for data $X_{1}=x_{1}, \ldots, X_{n}=x_{n}$.

The frequentist $5 \%$-level test of $H_{0}$ against $H_{1}$ rejects $H_{0}$ when $|\bar{X}| \geqslant 1.96 / \sqrt{n}$. What is the Bayes factor for the critical case $|\bar{x}|=1.96 / \sqrt{n}$ ? How does this behave as $n \rightarrow \infty$ ? Comment on the similarities or differences in behaviour between the frequentist and Bayesian tests.