Usbrth

Let $p geq 3$ and $X sim N(theta, I_p)$. I want to show that

$$mathbbE_thetabigg(frac1bigg) rightarrow 0 textrm as ||theta|| rightarrow infty.$$

I have the following proof, but it is not very straightforward, so I am wondering if there is a simple argument.

1. First note that $mathbbE_thetabig(frac1big)$ depends only on $||theta||$ by spherical symmetry. Let $g: [0, infty) rightarrow [0, infty]$ be such that $g(||theta||) = mathbbE_thetabig(frac1big).$ Note also that it is clear that $g$ is non-increasing.




2. Define the James-Stein estimator $delta^JS(X) = big(1 - fracp-2big)X$. It can be shown that

$$mathbbE_thetaBig([delta^JS(X) - theta]^2Big) = p - (p-2)^2mathbbE_thetabigg(frac1bigg).$$ Since this is $geq 0$, we must have that $mathbbE_thetabig(frac1big)$ is finite for all $theta in mathbbR^p$, so $g$ is finite.

3. Let $S$ be the unit sphere in $mathbbR^p$, and $E = mathbbR^p setminus S$. Then $g(r) = g_S(r) + g_E(r)$ where $g_S(r) = mathbbE_thetabig(frac1 1_Xin Sbig)$ and $g_E(r) = mathbbE_thetabig(frac1 1_Xin Ebig)$ where $||theta|| = r$.




4. It is not hard to show that $g_E(r) rightarrow 0$ as $r rightarrow 0$. Now, given arbitrary $epsilon > 0$, we can pick $r geq 1$ large enough that $f(x | rmathbfe_1) leq epsilon f(x | mathbfe_1)$ for all $xin S$, where $f(x | theta)$ is the p.d.f. of the $N(theta, I_p)$ distribution and $mathbfe_1 = (1, 0, ..., 0)$. Then $g_S(r) leq epsilon g_S(1)$. Since $g_S(1)$ is finite, we have that $inf_r geq 0 g_S(r) = 0$.

Thus, $inf_rgeq 0 g(r) = 0$. Since $g$ is non-increasing, its limit as $r to infty$ must be 0.

Note: this result shows in particular that, although the James-Stein estimator dominates the estimator $X$ under quadratic loss, they have the same maximal risk.

I don't know if this is fundamentally simpler, but it looks superficially different.

The quantity $R=|X|^2$ has a non-central chi squared distribution with $p$ degrees of freedom and non-centrality parameter $lambda=|theta|^2$. We can represent $R=S+T$ where $Sge0$ and $Tge0$ are independent, $S$ has an ordinary chi-squared distribution on $p$ degrees of freedom, and $T$ is a mixture of ordinary chi-square rvs with $2k$ degrees of freedom, where $k$ is Poisson distributed with parameter $lambda/2$. ($T$ is sometimes called a non-central chi-squared on zero degrees of freedom; see the wikipedia page cited above for details.)

We will use the dominated convergence theorem, with dominating integrand $1/S$, for which $1/Rle 1/S$. Note that $E(1/S)<infty$ because of the shape of the density function of $S$: you are integrating $int_0^infty s^p/2-1 exp(-s/2) (1/s),ds$, which is finite if $pge3$.

Let $lambda_ntoinfty$ and let $T_n$ be a random variable with non-centralality parameter $lambda_n$ and zero degrees of freedom. We want to see if $E(1/(S+T_n))to0$. Note that $1/T_n$ converges in distribution to $0$ as $ntoinfty$. By the Skorokhod theorem, we can pretend $1/T_n$ converges to $0$ with probability $1$. So now we use the DCT. We have $1/(S+T_n)le 1/S$ with probability $1$, we have $lim_ntoinfty1/(S+T_n)=0$ with probability $1$, and we have $E(1/S)ltinfty$. So the DCT tells us $E(1/(S+T_n))to0$ as desired.

Another way to prove the result is to notice that this chi-square with zero degrees of freedom stuff implies the family of distributions of $1/|X|^2$ is uniformly integrable. Then passage to the limit under the integral sign is justified.