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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.

The James–Stein estimator in the specified unit-variance model iswith an arbitrary value at $X=0$, a probability-zero event. The risk of $X$ is $p$. Gaussian Stein identity states $E_{\theta }[(X_i-{\theta }_i)g_i(X)]=E_{\theta }[{\partial }_i{g}_i(X)]$ for suitably integrable differentiable functions; it extends to the present singular function by truncation, since $p>2$ makes $E\Vert X{\Vert }^{-2}$ finite.

Put $c=p-2$ and $g(x)=-cx/\Vert x{\Vert }^2$. Then $\Vert g{\Vert }^2=c^2/\Vert x{\Vert }^2$ and $\mathrm{div}g=-c(p-2)/\Vert x{\Vert }^2$. Expanding the quadratic risk and applying Stein's lemma yieldsThus domination is strict for every finite $\theta$.

To identify the worst-case risk, let $r=\Vert \theta \Vert \to \infty$. On $\Vert X\Vert >r/2$, the reciprocal square is at most $4/r^2$. On the complementary ball, the Gaussian density is at most $(2\pi {)}^{-p/2}{e}^{-r^2/8}$, so its reciprocal-square integral is bounded by a constant times $e^{-r^2/8}{r}^{p-2}$, tending to zero. Therefore $E_{\theta }\Vert X{\Vert }^{-2}\to 0$, and $R(\theta ,\hat{\theta })\to p$ from below. HenceThe second supremum is approached at infinity and need not be attained at a finite parameter.

An admissible estimator is one for which there is no estimator $\delta$ satisfyingfor every $\theta$, with strict inequality for at least one $\theta$. For $p\ge 3$, the James–Stein estimatorhas riskIt strictly dominates $X$, and hence