Past exam of the statistics course of the University of Oxford / 2015 / 3 / b

Take $\sigma ,\tau >0$ to be the positive standard deviations. Multiply the normal likelihood function by the normal prior density. Terms independent of $\theta$ can be absorbed into the normalizing constant:The identity ${\sum }_i(x_i-\theta {)}^2={\sum }_i(x_i-\overline{x}{)}^2+n(\theta -\overline{x}{)}^2$ makes the quadratic term explicit. Completing the square giveswhereThus $\theta \mid x\sim N({\mu }_n,v_n)$. This is normal-normal conjugacy: prior and data precisions add, and the posterior mean shrinks the sample mean towards the prior mean zero.

The symmetric continuous prior distribution gives $\mathrm{Pr}(\theta <0)=\mathrm{Pr}(\theta \ge 0)=1/2$, so the prior odds are one. Therefore the Bayes factor equals the posterior odds. Standardizing the normal posterior distribution, putThen $\mathrm{Pr}(H_0\mid x)=\Phi (-z)$ and $\mathrm{Pr}(H_1\mid x)=\Phi (z)$. The endpoint $\theta =0$ has zero posterior probability, so the strict/non-strict split causes no additional term. HenceIn particular, positive values of the sample mean make this Bayes factor smaller, favouring the positive-mean hypothesis.

The denominator $\Phi (z)$ is positive, soStrict increase of the standard normal distribution function turns this into $z>{\Phi }^{-1}(10/11)$. Since the coefficient multiplying $\overline{x}$ in $z$ is positive, the rejection rule is precisely $\overline{x}>c$, withThe quantile is approximately $1.33518$. Equivalently, the rule requires posterior probability of $H_0$ below $1/11$, because its prior odds are one. The calculation is an instance of the Bayes factor for the sign of a normal mean.