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For $X\sim N_p(\theta ,I)$ and $p\ge 4$, this estimator preserves the sample mean and applies James–Stein shrinkage in its orthogonal $(p-1)$-dimensional contrast subspace. Its risk under quadratic loss is $p-(p-3{)}^2{\mathbb{E}}_{\theta }\Vert X-\bar{X}{1}_p{\Vert }^{-2}$. The noncentrality is $\Vert \theta -\bar{\theta }{1}_p{\Vert }^2$, so the reciprocal moment of a noncentral chi-squared variable makes the risk minimal exactly when all true components coincide; the minimum is $3$. A negative realized shrinkage factor reverses contrast signs, and its magnitude can exceed one for sufficiently small contrast norm.

Put $P=I-p^{-1}{1}_p{1}_p^T$, $Z=PX=X-\bar{X}{1}_p$, $Q=\Vert Z{\Vert }^2$ and $a=p-3$. Componentwise,Thus the James–Stein shrinkage toward the sample mean leaves the average unchanged and multiplies every contrast about it by the same data-dependent factor $1-a/Q$. If $Q>a$ the contrasts shrink without changing sign; if $Q<a$ their signs reverse, and if $Q<a/2$ their magnitudes increase. The event $Q=0$ has probability zero, and the estimator can be assigned $\bar{X}{1}_p$ there.

The sample mean is a linear normal statistic with mean $\bar{\theta }$ and variance $1/p$, soThe error splits into orthogonal mean and contrast components:Since the second term is perpendicular to $1_p$, its cross product with the first is zero for every sample. Taking squared norms and expectations, $p\mathbb{E}(\bar{X}-\bar{\theta }{)}^2=1$, which yields the risk function

To compute the contrast risk, choose an orthonormal basis matrix $U$ for the range of $P$, with $U^{T}U=I_m$, $U{U}^T=P$ and $m=p-1\ge 3$. Then $W=U^{T}X\sim N_m(\eta ,I)$ with $\eta =U^T\theta$, $Q=\Vert W{\Vert }^2$, and the contrast norm is the norm of $(1-a/Q)W-\eta$. Expand it:For $g_j(w)=w_j/\Vert w{\Vert }^2$, the Gaussian Stein identity givesThis application at a singularity is valid: near zero the reciprocal squared radius is integrable in dimension $m>2$, since its radial integral is ${\int }_0^1{r}^{m-3}dr$. Apply Gaussian integration by parts outside a ball of radius $\epsilon$; the inner boundary contribution is bounded by a constant times ${\epsilon }^{m-2}$ and vanishes. The Gaussian tail eliminates the boundary at infinity. Therefore all required expectations are finite and the displayed identity holds.

Since $a=m-2=p-3$, the Stein risk identity now gives contrast risk $m-a^2\mathbb{E}(1/Q)$. Adding the unit mean risk proves

Finally $Q$ has the noncentral chi-squared distribution ${\chi }_m^2(\lambda )$ with $\lambda =\Vert P\theta {\Vert }^2$. To establish where its reciprocal expectation is largest, complete the square in the normal representation $Q=\Vert W{\Vert }^2$ to obtain its Laplace transform:Since $1/q={\int }_0^{\infty }{e}^{-tq}dt$, the Tonelli theorem gives the reciprocal moment of a noncentral chi-squared variableIts integrand is positive and strictly decreasing in $\lambda$ for every $t>0$. The expectation is therefore strictly largest at $\lambda =0$, whereBecause this expectation is subtracted in the risk, the minimizing set is exactly $P\theta =0$, that is, all coordinates of $\theta$ equal. On this line the risk is $p-(p-3)=3$: