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For a probability density $\pi$ and threshold $c>0$, this region includes precisely the points at or above the threshold. It is a Bayes act for loss $c|S|-1_S(\theta )$, since expected loss is ${\int }_S(c-\pi (\theta ))d\theta$. The minimal value is $-\int (\pi -c{)}_{+}d\theta$; adding or removing threshold-equality points does not change it.

Write $R(\theta ,\delta )={\mathbb{E}}_{\theta }L(\theta ,\delta (X))$ and $r(\pi ,\delta )=\int R(\theta ,\delta )d\pi (\theta )$. An equaliser rule has constant risk $R(\theta ,\delta )=c$. An Extended Bayes rule has, for every $\epsilon >0$, a proper prior $\pi$ such that $r(\pi ,\delta )\le r^{\ast }(\pi )+\epsilon$, where $r^{\ast }(\pi )={\inf }_{d}r(\pi ,d)$. If a rule has both properties, every competitor $d$ satisfiesLetting $\epsilon \to 0$ proves that its maximum risk $c$ is the minimax risk.

For the given normal distribution model, completing the square in prior times likelihood gives the posterior $N(m,H^{-1})$, with $H=h_0+nh$ and $m=(h_0{m}_0+h{\sum }_i{x}_i)/H$. Another completion of the square givesThis is maximized uniquely at $a=m$. Consequently the Bayes act and posterior posterior expected loss areThe posterior loss is independent of $x$, so averaging over the predictive distribution gives the same value for the Bayes risk of this Bayes decision rule.

The sample mean obeys $\bar{X}-\theta \sim N(0,(nh{)}^{-1})$, so its risk is the constant $c=1-\sqrt{nh/(nh+k)}$. Take any fixed $m_0$ and let the proper prior precisions $h_0\downarrow 0$. The optimal Bayes risks just calculated tend to $c$, while the integrated risk of $\bar{X}$ is exactly $c$ for every prior. Thus its excess Bayes risk tends to zero: it is extended Bayes as well as equaliser. The preceding argument proves

In the normal form of a Bayesian decision problem, choose a measurable decision rule $\delta$ before seeing the data to minimize the prior Bayes risk $\mathbb{E}L(\Theta ,\delta (X))$. In the extensive form, after observing $X=x$, choose an action minimizing posterior expected loss. Conditional expectation makes these equivalent, subject to the usual measurable selection conditions. If ${\Pi }_x$ is the posterior distribution and $R_0(\Pi )={\inf }_a\int L(\theta ,a)\Pi (d\theta )$ is the best no-data loss, thenThe inequality follows by ignoring the data.

For a measurable region $S$, posterior or prior expected loss is ${\int }_S(c-\pi (\theta ))d\theta$. It is minimized by including exactly the locations where the integrand is negative; including equality points is harmless. Thus the highest density region $S^{\ast }=\{\pi \ge c\}$ is a Bayes act andFor $N(\mu ,{\sigma }^2)$, if $c\ge 1/(\sigma \sqrt{2\pi })$ the answer is zero. Otherwise put $z_c=\sqrt{2\log [1/(c\sigma \sqrt{2\pi })]}$. The optimal interval is $[\mu -\sigma z_c,\mu +\sigma z_c]$, andFor the specified prior and $c=1/2$, the prior density is everywhere below $c$, so $R_0=0$. Completing the square gives $\Theta \mid X=x\sim N(9x/10,1/10)$. Its peak is $\sqrt{10/(2\pi )}>1/2$, so a positive-measure region has posterior density strictly above $c$ and its optimal expected loss is strictly negative. That loss is independent of its mean, hence independent of $x$. Therefore $R_1<0=R_0$. Explicitly, with $z_c=\sqrt{2\log (2\sqrt{10/(2\pi )})}$, $R_1=z_c/\sqrt{10}-2\Phi (z_c)+1$.