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For a posterior distribution with finite first moment, the posterior risk derivative at continuity points is $(\gamma +\delta )F(a)-\gamma$. Convexity therefore makes a posterior $\gamma /(\gamma +\delta )$-quantile a Bayes estimator. With atoms, the condition is $F(a^{-})\le \gamma /(\gamma +\delta )\le F(a)$. Increasing the cost of underestimation moves the optimal estimate upward.

A gamma prior with shape $\alpha$ and rate $\beta$ combined with $n$ independent observations from an exponential distribution of rate $\theta$ gives the posteriorUnder squared-error loss, the Bayes estimator is its posterior mean $(\alpha +n)/(\beta +{\sum }_i{X}_i)$.

A prior distribution assigns uncertainty to a parameter before observing data. Multiplying its density by the likelihood and normalizing gives the posterior distribution. A Bayes estimator minimizes the posterior expected loss function for each observed sample, and therefore minimizes integrated Bayes risk when the expectations exist.

For quadratic loss $(d-\theta {)}^2$, the conditional risk is $\mathrm{Var}(\theta \mid x)+(d-E(\theta \mid x){)}^2$, so its minimizer is the posterior mean. For absolute loss $|d-\theta |$, the one-sided derivatives or subgradients show that a minimizer is any posterior median, characterized by $P(\theta <d\mid x)\le 1/2$ and $P(\theta >d\mid x)\le 1/2$.

For the stated uniform distribution sampling model, let $x_{\min }={\min }_i{x}_i$, $x_{\max }={\max }_i{x}_i$. The likelihood is $2^{-n}$ on $x_{\max }-1<\theta <x_{\min }+1$ and zero elsewhere. Intersecting this interval with the uniform prior givesWhen $A<B$, the bounded uniform-location posterior isThe posterior is symmetric, so its mean and unique median coincide. If $A\ge B$, the data have zero marginal likelihood under the stipulated prior/sampling model and this posterior density cannot be normalized. An empty or singleton intersection is not an ordinary uniform posterior of positive width.

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.

A loss function $L(\theta ,a)$ measures the cost of taking action $a$ at parameter $\theta$. A decision rule $\delta$ selects an action from the data; its risk function is $R(\theta ,\delta )=E_{\theta }L(\theta ,\delta (X))$. A Bayes estimator minimizes the prior-averaged risk, or equivalently minimizes posterior expected loss at almost every observation. An admissible rule has no competing rule of no greater risk everywhere and strictly smaller risk somewhere. To construct a Bayes rule, form the posterior and minimize its expected loss over the allowed actions. Under squared error and finite posterior second moments the unique action is the posterior mean.

Put $S=|X{|}^2$. The given fourth moment and independence give $E_{v}S=nv$, $E_v{S}^2=n(n+2)v^2$. ThusEvery $\alpha \ne 1/(n+2)$ is strictly dominated at every $v>0$ by that coefficient. Hence it cannot be a finite-risk Bayes rule for any proper prior.

To make the second requested comparison, let $m_j=E_{\pi }{v}^j$ and assume finite quadratic Bayes risk. For the affine rules the integrated risk isWith $\alpha$ fixed, the minimum occurs at $\beta =(1-n\alpha )m_1$. Since $m_1>0$, this strictly improves the homogeneous rule unless $\alpha =1/n$, reducing risk by $(1-n\alpha {)}^2{m}_1^2$. Therefore a homogeneous rule with $\alpha \ne 1/n$ is not Bayes, as requested. Combining the two restrictions excludes every homogeneous coefficient.

For a proper prior without global finite moments, the same nonvacuous conclusion follows from the posterior-mean condition. LetFor $s>0$, $M_1^{\prime }(s)=-m(s)/2$. If the posterior mean were $\alpha s$, positivity requires $\alpha >0$ and $M_1=\alpha sm$. Differentiation forces $m(s)=C{s}^{-1-1/(2\alpha )}$. But the marginal density of $S$ would then be a constant times $s^{n/2-2-1/(2\alpha )}$, a pure power not integrable at both zero and infinity. This contradicts a proper prior's normalized predictive distribution. This is the proper-prior obstruction to homogeneous variance Bayes rules.

The finite-loss qualification matters: allowing all posterior losses to equal infinity makes every action a vacuous minimizer. For example, with $n=1$ and proper density $f_0(v)=(1+v{)}^{-2}$, the posterior second moment diverges at positive observations, so a literal infinite-loss definition would defeat the printed assertion. Ordinary meaningful Bayes rules exclude that convention.

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.

The posterior distribution has a long right tail. Its posterior mean is pulled upward by relatively rare large endpoints and is sensitive to the artificial upper cutoff; its untruncated posterior mean does not exist. The median is a central quantile that stays finite as the cutoff is removed and is much less sensitive to extreme draws. It is also the Bayes estimator under absolute-error loss, whereas a mean corresponds to squared-error loss when the relevant expectation exists.

The median better represents the typical posterior endpoint in this strongly right-skewed distribution. It need not be preferable for every decision problem, since the loss function matters. For comparison, exact summation of the truncated kernel at cutoff $5000$ gives median $195$ and mean about $397.67$; the displayed simulation summaries are Monte Carlo estimates, not exact posterior calculations.

Put $s=\sqrt{n}/2$. The posterior is $\mathrm{Beta}(X+s,n-X+s)$, andThe denominator is positive for $n\ge 1$. For $n>4$, both parameters exceed one and the mode simplifies to $(X+s-1)/(n+\sqrt{n}-2)$. For $n=4$ this is the uniform-prior mode, exactly $X/n$; for $n=1,2$ the endpoint or symmetric cases also make the mode rule coincide with $X/n$. Otherwise it is generally a different rule. The posterior mean agrees with $X/n$ only at $X=n/2$.

To find the minimax estimator under quadratic loss, consider the posterior mean ${\delta }_s=(X+s)/(n+2s)$. Its risk function isFor $s^2=n/4$, the numerator is identically $n/4$, soThis constant-risk binomial proportion estimator is the unique Bayes estimator for its proper beta prior under quadratic loss. A competing rule's maximum risk is at least its risk averaged over that prior, which is at least this Bayes risk. The constant-risk rule attains the bound, so the posterior mean in (iii) is the unique minimax rule. Consequently any coincident rule is minimax too: the mean in (i) is the same rule when $n=4$, and the mean in (ii) is the same rule when $n=1$. No posterior mode listed here equals that mean rule for every observation, since at $X=0$ its value is positive whereas their endpoint modes differ from it. This accounts for the small-$n$ coincidences as well as the general answer.

The specified gamma distribution uses shape $k$ and rate $\lambda$. Multiplying its density by the Poisson likelihood gives a posterior density proportional to ${\theta }^{k+S-1}{e}^{-(\lambda +n)\theta }$. Thus the conjugate prior calculation givesFor squared-error loss, the posterior expected loss isIt is minimized at the posterior mean, so the Bayes estimator isFor a fresh conditionally independent Poisson observation, the posterior predictive probability of zero is the posterior expectation of $e^{-\theta }$. Evaluating the gamma integral givesThe samples are independent conditional on $\theta$; integrating out the uncertain common parameter produces the predictive probability above.

For every estimator, its worst-case risk dominates its average risk under the uniform prior. Since the preceding Bayes estimator minimizes that average,The estimator $X/n$ has risk exactly $1/n$ throughout the interior, so it attains this lower bound. Under either endpoint convention discussed above its supremum over $[0,1]$ is also $1/n$. ThusThis constant-risk Bayes argument works also for $n=1$, when the only observed counts are endpoints and the posterior integrability argument still selects their corresponding decisions.

Condition on $X=k$. For $0<k<n$, the posterior expected loss is a strictly convex quadratic in the decision $d$. Its minimizer isThe beta density gives $\mathbb{E}[1/p\mid k]=(n+1)/k$ and $\mathbb{E}[1/(1-p)\mid k]=(n+1)/(n-k)$, so $d=k/n$. If $k=0$, every $d\ne 0$ has infinite posterior expected loss because of the nonintegrable $d^2/p$ term near zero, whereas $d=0$ has finite loss expectation. The analogous argument at $k=n$ forces $d=1$. Hence the Bayes estimator isIndeed $X/n$ is unbiased and has variance $p(1-p)/n$, which cancels the loss denominator. Its uniform-prior Bayes risk is $1/n$.

The printed loss is undefined at $p=0,1$, despite the stated closed parameter interval. A convention is necessary there. The natural extended loss assigns zero to the correct endpoint decision and infinity to an incorrect one; it gives endpoint risks zero for $X/n$. Alternatively a continuous extension of this estimator's risk gives endpoint values $1/n$. The interior result and the minimax value below are unchanged by either convention, and a uniform prior gives the endpoints measure zero. We do not identify an undefined $0/0$ with a value without specifying this choice.

With the uniform prior, the binomial likelihood gives the beta distribution posterior $\mathrm{Beta}(s+1,n-s+1)$. Under squared-error loss, the Bayes estimator is the posterior mean:

The production manager's prior gives posterior density proportional to $p^{s+1}(1-p{)}^{n-s}$. Minimizing $\mathbb{E}[(1-p)(\hat{p}-p{)}^2\mid X=s]$ gives the weighted posterior meanusing the supplied beta function. Cross multiplication shows ${\hat{p}}_P>{\hat{p}}_G$ exactly when $n>2s$, so it is greater unless $s\ge n/2$.

The probability of no recovery through month $M$ is the product of the conditional survival probabilities, soUnder quadratic loss, the Bayes estimator is the posterior mean. Posterior independence and the mean of a Beta distribution giveConsequently

The risk of a decision rule is its expected loss at a fixed parameter value:For a prior $\pi$, a Bayes estimator minimizes the integrated riskor equivalently minimizes posterior expected loss for almost every observation. A minimax decision rule minimizesover all decision rules.