I try to understand why the risk of the James-Stein estimator is always greater than the risk of the so-called called oracle estimator.
The JS-estimator is defined as follows $$\hat{\vartheta}_{\text{JS}}(X)=\Big(1-\frac{p-2}{||X||^2}\Big)X=a_1X, $$ with $X\sim \mathcal{N}(\vartheta,\sigma^2I_p)$, $\vartheta\in\mathbb{R}^p$ and $I_p$ the p-dimensional identity matrix. Its risk function with respect to scaled squared error loss is given by $$ R(\hat{\vartheta}_{\text{JS}},\vartheta)=1-\frac{(p-2)^2}{p}\mathbb{E}_\vartheta\big[||X||^{-2}\big]<1$$.
Now im comparing this risk with the risk of the oracle estimator given by $$\hat\vartheta_{\text{OC}}(X)=\frac{||\vartheta||^2}{||\vartheta||^2 +p}X=a_2X. $$ It's risk is $$ R(\hat\vartheta_{\text{OC}},\vartheta)=\frac{||\vartheta||^2}{||\vartheta||^2+p}<1.$$ So I want to show, that $$R(\hat\vartheta_{\text{OC}},\vartheta)\leq R(\hat{\vartheta}_{\text{JS}},\vartheta).$$
It seems quite logical that the risk of the JS-estimator must be greater, as its pre-factor $a_1$ estimates the pre-factor of the oracle estimator $a_2$, but I don't know how to show it mathematically . My only suggest is, as $X\sim \mathcal{N}(\vartheta,\sigma^2I_p)$, $||X||^2$ is noncenteral $\chi^2$ distributed with $p$ degrees of freedom and non-centrality parameter $\lambda=||\vartheta||^2$.
Thanks for your help!