# Why is the James-Stein estimator called a “shrinkage” estimator?

I have been reading about the James-Stein estimator. It is defined, in this notes, as

$$\hat{\theta}=\left(1 - \frac{p-2}{\|X\|^2}\right)X$$

I have read the proof but I don't understand the following statement:

Geometrically, the James–Stein estimator shrinks each component of $X$ towards the origin...

What does "shrinks each component of $X$ towards the origin" mean exactly? I was thinking of something like $$\|\hat{\theta} - 0\|^2 < \|X - 0\|^2,$$ which is true in this case as long as $(p+2) < \|X\|^2$, since $$\|\hat{\theta}\| = \frac{\|X\|^2 - (p+2)}{\|X\|^2} \|X\|.$$

Is this what people mean when they say "shrink towards zero" because in the $L^2$ norm sense, the JS estimator is closer to zero than $X$?

Update as of 22/09/2017: Today I realized that perhaps I am over-complicating things. It seems like people really mean that once you multiply $X$ by something that is smaller than $1$, namely, the term $\frac{\|X\|^2 - (p + 2)}{\|X\|^2}$, each component of $X$ will be smaller than it used to be.

$$E_\mu(\| \boldsymbol{\hat\mu}^{JS} - \boldsymbol{\mu} \|^2) < E_\mu(\| \boldsymbol{\hat\mu}^{MLE} - \boldsymbol{\mu} \|^2)$$
where $\boldsymbol{\mu} = (\mu_1,\mu_2,\dots,\mu_p)'$, $\hat\mu^{JS}_i$ is the Stein's estimator and $\hat\mu^{MLE}_i = x_i$, where both estimators are estimated on the $x_1,x_2,\dots,x_p$ sample. The proofs are given in the original papers and the appendix of the paper you refer to. In plain English, what they have shown is that if you simultaneously make $p > 2$ guesses, then in terms of total squared error, you'd do better by shrinking them, as compared to sticking to your initial guesses.
• @Tim I think Misakov argument is legitimate in that the James-Stein estimator brings the estimator of $\theta$ closer to zero than the MLE. Zero plays a central and centric role in this estimator and James-Stein estimators can be constructed that shrink towards other centres or even subspaces (as in George, 1986). For instance, Efron and Morris (1973) shrink towards the common mean, which amounts to the diagonal subspace. – Xi'an Sep 22 '17 at 11:13