10 ms·
(I cannot speak for the original article, I’ve not put the effort to fully understand it so I won’t categorically say it’s wrong but it didn’t seem right to me.
by kgwgk 2y ago
(I cannot speak for the original article, I’ve not put the effort to fully understand it so I won’t categorically say it’s wrong but it didn’t seem right to me.)
The “paradox” is that it can truly be arbitrary! Pick a random point. Shrink your least-squares estimator. You got yourself a “better” estimator - without having any additional information.
That’s why the “Inadmissibility of the Usual Estimator for the Mean of a Multivariate Normal Distribution” paper had the impact that it had.
- mitthrowaway2 2y agoThen you'll have to clarify what you mean by "random" when you say "pick a random point". Unless you mean that every point on a spherical surface centered on x would have a lower expected squared error than x itself?
- kgwgk 2y agoWe may be talking about different things. Let's say that you have a standard multivariate normal with unknown mean mu = [a, b, c]. The usual maximum-likelihood estimator of the unknown mean when you get an observation is to take the observed value as estimate. If you observe [x, y, z] the "naive" estimator gives you the estimate mû = [x, y, z]. For any arbitrary point [p, q, r] you can define another estimator. If you observe [x, y, z] this "shrinkage" estimator gives you an estimate which is no longer precisely at [x, y, z] but is displaced in the direction of [p, q, r]. For simplicity let's say the resulting estimate is mû' = [x', y', z']. Whatever the choice you make for [p, q, r] the "shrinkage" estimator has lower mean squared error than the "naive" estimator. The expected value of (x'-a)²+(y'-b)²+(z'-c)² is lower than the expected value of (x-a)²+(y-b)²+(z-c)².