 Solution

Assume the known variances are positive and write $\bar{x}=n^{-1}{\sum }_i{x}_i$. The likelihood times the normal prior distribution is proportional toCollecting the quadratic and linear terms in $\theta$ completes the square with precision $v^{-1}=n/{\sigma }^2+1/{\tau }^2$ and mean $m=v(n\bar{x}/{\sigma }^2+\mu /{\tau }^2)$. Thus normal-normal conjugacy givesFor quadratic loss, the posterior risk of reporting $a$ is $\mathbb{E}[(a-\theta {)}^2\mid x]=v+(a-m{)}^2$, uniquely minimized at $a=m$. For absolute-error loss, differentiating the posterior risk gives $2F_{\theta \mid x}(a)-1$, so its minimizer is a posterior median. The normal posterior is symmetric about $m$ and has a strictly increasing distribution function, hence that median is uniquely $m$. Therefore both optimal point estimates equal $m$, the posterior mean.