# James-Stein phenomenon: What does it mean that a James-Stein estimator beats least squares estimator?

Background James-Stein estimator and Stein's phenomenon, as described in Wikipedia are rather counterintuitive and amazing.

It is claimed that if one wants to estimate the mean $\Theta$ of
Gaussian distributed vector $y$~$N(\Theta, \sigma^2 Id)$, then the naive estimation - (i.e. just take $y$ as an estimation) is not good for size of vector greater or equal 3.

"Not good" means that (quote Wikipedia) "James–Stein estimator always achieves lower (Mean squared error (MSE) than the least squares estimator".

Question please clarify the sentence above. I wonder the following - usually when we calculate MSE we need some distribution on the estimated parameter $\Theta$ and averaging in MSE is taken over this distribution also.

So what distribution is assumed ? Or may be for ANY distribution it holds true ?

PS

Stein's example is more general (quote Wikipedia):

Stein's example (or phenomenon or paradox), in decision theory and estimation theory, is the phenomenon that when three or more parameters are estimated simultaneously, there exist combined estimators more accurate on average (that is, having lower expected mean-squared error) than any method that handles the parameters separately. This is surprising since the parameters and the measurements might be totally unrelated.

Popular articles have appeared hailing the James-Stein estimator a paradox; one should use the price of tea in China to obtain a better estimate of the chance of rain in Melbourne!

(Quote from Deane Yang's suggested page http://jmanton.wordpress.com/2010/06/05/comments-on-james-stein-estimation-theory/ )

• This seemed helpful to me: jmanton.wordpress.com/2010/06/05/… – Deane Yang Apr 11 '12 at 9:30
• @AlexanderChervov I provide a more mathematical explanation in addition to your OP. Hope helps. – Henry.L Apr 11 '17 at 19:57

This has always bothered me. "One should use the price of tea in China to obtain a better estimate of the chance of rain in Melbourne" is not a good characterization at all. One should use the price of tea in China and the chance of rain in Melbourne to obtain a better estimate of the vector which includes both the average price of tea in China and the chance of rain in Melbourne. The Stein result only obtains if you care about a vector-valued parameter; that is the observations are assumed independent probabilistically but clearly interact with one another via the loss function being use.

The idea behind the quote is that you can hedge your bets on any given coordinate dimension by "shrinking" back towards the "global" mean (across all elements of the mean vector). But observe that the "shrinkage" need not in fact be towards the overall mean for the result to hold...you can shrink back towards any value at all and still get the result, which has to do with the definition of admissibility used. My favorite description of what is going on is here.

• In fact, I thought the link I provided made the same point. – Deane Yang Apr 11 '12 at 14:02
• @Deane Yang, I'm sure it did, but I took the opportunity to rant :) – R Hahn Apr 11 '12 at 14:09
• Yes, it's a good rant. I'm glad you made the same point explicitly. – Deane Yang Apr 11 '12 at 15:30

There is an excellent issue of Statistical Science that address the James-Stein phenomena from various aspects. https://www.jstor.org/stable/i23208816

Question What does it mean that a James-Stein estimator beats least squares estimator?

It means that the JS estimator has a smaller risk than LSE w.r.t. a prescribed risk function $R(\delta)=E_{\theta}L(\delta,\theta)$; which is equivalent to say that if we choose $L^2$ loss function, then when the dimension is higher than 3, the LSE for the mean is no longer admissible. The only admissible estimator should be JS estimator.

The original JS estimator assumes multi-dimensional Gaussian distribution. But later with consideration of admissibility, there are various cases where shirkage estimator(JS estimator) actually beats the LSE(also MLE) and other frequentist estimators. And "beating" also depends on the $L^2$ loss function you choose, and hence the risk function $R$ you choose for this decision problem. To illustrate this point, we consider Gaussian case with $L^2$ loss function in the explanation below.

There are two approaches that seem very intuitive to me.

(1)Linkage between JS estimator and the Diffusion process.

Brown [1] pointed out that under the framework of decision theory, it could be shown that the admissibility and the recurrence of Brownian motion is equivalent. Moreover, Brown shown in his major theorem that via a variational minimization problem, the admissibility of an estimator can be discussed using recurrence of corresponding Brownian motion. Such a variational approach can be extended to various other situations.

Theorem 5.1 in [1], modified. A necessary condition for $\delta$ to be admissible with given risk $R$ is that there exists a non-negative measure $F$ s.t. its density $f^{*}<\infty$ and $\delta(x)=\delta_{F}(x)$(generalized Bayes rule w.r.t. F) for almost all $x\in E^{m}$ w.r.t. Lebesgue measure. Furthermore

(A) If $\{Z_{t}\}$ is transient then $\delta$ is inadmissible.

(B) If $\{Z_{t}\}$ is recurrent and the risk set w.r.t. $R$ is bounded uniformly then $\delta$ is inadmissible.

where $\{Z_{t}\}$ is the diffusion process in $E^{m}$ along with its infinitesimal generator with local mean $\nabla(logf^{*})=\delta_{F}(x)-x$ and a local variance covariance matrix $2I$.

Therefore since 1d and 2d Brownian motions are recurrent, so is the mean estimator; but when it goes to 3d, Brownian motions are transient and hence mean estimators are no longer admissible and beaten by JS estimator.

This is probably one of the most celebrated results derived from Bayesian statistics and it integrated the stochastic process so seamlessly that I believe it even lead to later MCMC simulation developments.

(2)Geometric interpretation of JS estimator.(taught by Prof.M.P. :)

Zhao and Brown [3] pointed out that if we restricted ourselves to spherically symmetric estimators, then the naive geometric optimal estimator derived from simple Euclidean geometry $$\delta_{NGO}(Z)=\left[1-\frac{p-1}{\left\Vert Z\right\Vert ^{2}}\right]Z$$ is providing exact amount of shrinkage as JS estimator did. This approach is intuitive in sense that spherical symmetric estimators of the form $\delta(X)=\tau(\|X\|)X$ with a scalar function $\tau$, the key of using this class of estimators is that (i) By admissibility consideration, estimators that are not spherically symmetric are inadmissible (ii) By structure of the spherical estimator, it provides a natural geometry that is isometric to Euclidean geometry. By (i)(ii), the problem of choosing an admissible estimator becomes a "compass-and-ruler" problem of geometry. Amazing!

Reference

[1]Brown, Lawrence D. "Admissible estimators, recurrent diffusions, and insoluble boundary value problems." The Annals of Mathematical Statistics 42.3 (1971): 855-903. http://projecteuclid.org/euclid.aoms/1177693318

[3]Brown, Lawrence D., and Linda H. Zhao. "A Geometrical Explanation of Stein Shrinkage." Statistical Science (2012): 24-30.

To some of the specific points raised in this question, Probability puzzle blog is a write up that actually shows the James Stein estimator in action. Specifically, there is R code you can use to verify it yourself for various distributions (Normal and Poisson). A key point to note is that the estimator does return closer "true" variables than MLE. Hope that helps