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

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.