James–Stein estimator
= James–Stein estimator
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For one observation $X\sim N_p(\theta,I_p)$ with $p\geq3$, the James–Stein estimator is
$$
\delta_{\rm JS}(X)=
\left(1-\frac{p-2}{\lVert X\rVert^2}\right)X.
$$
Under <quadratic loss>, Stein's risk identity gives
$$
R(\theta,\delta_{\rm JS})
=p-(p-2)^2\mathbb E_\theta\frac1{\lVert X\rVert^2}<p,
$$
so it dominates the usual estimator $X$.