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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 $p-(p-3)=3$:

The same affine transformation of the normal distribution givesIts variance stays one while its mean moves positively. This is the exact alternative distribution used to calculate statistical power. The square, if used instead, has a noncentral chi-squared distribution with one degree of freedom and noncentrality $n({\delta }^{\ast }{)}^2/{\sigma }^2$.

A dimension-two Bessel process starting at $r>0$ has the displayed marginal law, where $G_1,G_2$ are independent standard normal random variables. Take the norm of a planar Brownian motion starting at $(r,0)$. Planar Brownian motion avoids a fixed point, so the norm stays positive; Itô formula gives drift $1/(2R)$ and a martingale term of bracket $t$. The Lévy characterization of Brownian motion identifies that term as its driving Brownian motion. The squared radius divided by $t$ has a noncentral chi-squared distribution with two degrees of freedom and noncentrality $r^2/t$.