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A decision rule is admissible if no other rule has risk function no greater at every parameter and strictly smaller at one parameter. Parameter estimation is one decision problem; its admissible rules are admissible estimators.

A prior on a parameter, a conditional observation distribution, an action space and a loss specify a Bayesian decision problem. The normal form minimizes prior expected loss over decision rules; the extensive form minimizes posterior expected loss after each observation. Conditional expectation gives the same Bayes risk for these formulations, provided measurable optimal or approximate choices can be made.

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.

A loss function $L(\theta ,a)$ quantifies the cost of taking action $a$ when the parameter is $\theta$. A decision rule $d$ is a measurable map from the observation space to the action space (or, for randomized rules, a conditional action distribution). Its risk function is $R(\theta ,d)={\mathbb{E}}_{\theta }L(\theta ,d(X))$. For a prior probability distribution $\pi$, its Bayes risk isA minimax rule minimizes ${\sup }_{\theta }R(\theta ,d)$.

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

For a decision rule $\delta$ and loss $L$, its risk function is $R(\theta ,\delta )={\mathbb{E}}_{\theta }L(\theta ,\delta (X))$. It is an admissible estimator if there is no rule with risk everywhere no larger and somewhere strictly smaller. A Bayes decision rule minimizes $\int R(\theta ,\delta )\pi (d\theta )$ for a specified prior $\pi$. A minimax decision rule minimizes ${\sup }_{\theta }R(\theta ,\delta )$ over all rules.

Here the maximum-likelihood estimator is ${\delta }_0(X)=X$, and its squared-error risk isWrite the proposed estimator as $\delta (X)=X+g(X)$ with $g(x)=-x/\Vert x{\Vert }^2$ for $x\ne 0$, and assign any value at zero, an event of probability zero. Expanding the squared loss givesStein's lemma for a unit-covariance normal vector says $\mathbb{E}[(X_i-{\theta }_i)g_i(X)]=\mathbb{E}[{\partial }_i{g}_i(X)]$ when the relevant derivatives and expectations are integrable. With $r=\Vert x\Vert$,The singularity causes no difficulty in three dimensions: $r^{-2}$ is locally integrable against the volume element $r^{2}drd\Omega$, and the normal tails control infinity. More formally apply Stein's lemma first to $g_{ϵ}(x)=-x/(r^2+ϵ)$, then let $ϵ\downarrow 0$; its squared norm and derivative magnitudes are bounded by constants times $r^{-2}$, so dominated convergence justifies the limit. ConsequentlyThe expectation is finite and strictly positive. Thus this James–Stein estimator strictly dominates the maximum likelihood estimator, which is inadmissible under the specified squared-error loss.

Suppose one observation comes from either $P$ or $Q$, each with prior probability $1/2$. A decision rule that chooses $P$ on an event $E$ and $Q$ otherwise has success probabilityOn a finite set, choosing $E=\{x:P(x)>Q(x)\}$ therefore gives $\frac{1}{2}(1+d_{\mathrm{TV}}(P,Q))$. This is the equal-prior version of minimum sum of errors in a simple hypothesis test; allowing a statistical test to use an auxiliary random variable cannot improve it, because for each outcome the better of the two decisions already maximizes its contribution.

For the four-outcome example, choose $P$ after $A$ and $Q$ after $B$ or $C$, with either decision allowed after $D$. Choosing $Q$ after $D$ yields $\frac{1}{2}(0.4+0.9)=0.65$. The best success probability is 65%, with a 35% error probability. A smaller total variation distance means the two sources are harder to distinguish from one observation. Unequal prior probabilities, unequal costs, or multiple observations change this particular decision problem; the $65\%$ figure is not a universal bound for every experiment.