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Under loss $w(\theta )|\theta -a|$, assume $0<\mathbb{E}[w(\theta )\mid x]<\infty$. The posterior expected loss is a positive constant times ordinary absolute-error loss under the probability density proportional to $w(\theta )\pi (\theta \mid x)$. Its minimizers are the medians of that weighted posterior: slopes of the loss change sign where the weighted probability below $a$ crosses one half. In particular, relative absolute loss has $w(\theta )=1/\theta$ for positive parameters. For a Gamma distribution posterior of shape two and rate $b$, the weighted posterior is an exponential distribution of rate $b$, so the estimate is $(\log 2)/b$.

This relative absolute-error loss weights each parameter by $1/\theta$. Its posterior expected loss isThe effective density for the Bayes estimator under weighted absolute loss is proportional to $\pi (\theta \mid x)/\theta$, hence an exponential distribution of rate $b$. Its median solves $1-e^{-ba}=1/2$.

More explicitly, for $a\ge 0$, differentiating the split integrals at $\theta =a$ gives $R_x^{\prime }(a)=b(1-2e^{-ba})$ and $R_x^{\prime \prime }(a)=2b^2{e}^{-ba}>0$. For $a<0$, the loss has derivative $-b$, so its minimum cannot lie there. The unique global minimum is thereforeThis is the weighted posterior median; the ordinary median of the Gamma posterior would minimize unweighted absolute loss and would give a different answer.

Under squared-error loss, the posterior expected loss decomposes asIt is uniquely minimized by the posterior mean. Integrating the Gamma posterior gives $\mathbb{E}[\theta \mid x]=b^2{\int }_0^{\infty }{\theta }^2{e}^{-b\theta }d\theta =2/b$. Hence the Bayes estimator is

A Bayes estimator minimizes the integrated loss function under the prior and sampling distribution. For a nonnegative loss, Tonelli's theorem and the posterior distribution give the risk decompositionwhere $m$ is the marginal density. Thus choosing, for each observed $x$, a measurable minimizer of the posterior expected loss minimizes the integrated risk. Observations of marginal probability zero do not affect the rule.

Here the likelihood is $\theta e^{-\theta x}$ and the prior is $\mu e^{-\mu \theta }$ on positive parameters. Put $b=\mu +x$. Their product is $\mu \theta e^{-b\theta }$, and ${\int }_0^{\infty }\theta e^{-b\theta }d\theta =b^{-2}$. The marginal density and posterior are thereforeThis Gamma-exponential conjugacy calculation gives a Gamma distribution with shape $2$ and rate $b$. It is the posterior, rather than the prior alone, which determines both requested estimators.

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

Given $X=x$, the posterior expected loss from reporting $d$ isThis is a strictly convex quadratic in $d$ whenever the conditional inverse moment is finite and positive. Differentiating with respect to $d$ givesso the Bayes estimator under reciprocal weighted quadratic loss is