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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 $z_c=\sqrt{2\log (2\sqrt{10/(2\pi )})}$, $R_1=z_c/\sqrt{10}-2\Phi (z_c)+1$.