 Solution

For the independent normal distribution observations, the log likelihood isIts derivative is ${\sum }_i(X_i-\theta )$ and its second derivative is $-n<0$, so the unique maximum-likelihood estimate isFor the prior $\theta \sim N(\mu ,{\tau }^{-1})$ with $\tau >0$, multiplying prior density by likelihood and completing the square givesThus normal-normal conjugacy gives the posterior distributionFor quadratic loss, the conditional expected loss of action $a$ is the posterior variance plus $(a-\mathbb{E}[\theta \mid X]{)}^2$. Its unique minimizer is the posterior mean, so the Bayes estimator isThis weights the sample and prior means by their respective precisions.