4
$\begingroup$

Basic example: $X$ has a $p$-variate iid standard Normal distribution; the sample mean is not admissible if $p>2$ and is dominated by the Stein shrinkage estimator.

However, the Stein shrinkage estimator is also not admissible, and is dominated by the positive-part Stein shrinkage estimator. Which is also not admissible.

As far as I can tell, there is no known admissible estimator that dominates the sample mean. Is there a theoretical guarantee that there must be one, or could there just be an infinite sequence of inadmissible estimators, each slightly better than the last?

$\endgroup$
  • $\begingroup$ Wouldn't the limit of such a sequence (where each term dominates the one before it) be an admissible estimator by construction? $\endgroup$ – whuber Aug 22 '18 at 20:53
  • $\begingroup$ Replying to the comment by @whuber since I can't do comments: if the limit existed I think it would be an admissible estimator -- though I don't know if something that isn't finitely computable counts as an estimator. $\endgroup$ – Thomas Lumley Aug 22 '18 at 21:48
  • $\begingroup$ Please merge your accounts so that you can comment in your own question. $\endgroup$ – Glen_b Aug 23 '18 at 2:03
  • $\begingroup$ "Isn't finitely computable" is not a statistical concept and plays no role in standard statistical theories. If that's a concern to you, then please edit your question to explain why. $\endgroup$ – whuber Aug 23 '18 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.