2
$\begingroup$

Are there examples of estimators that are minimax, but not efficient? Perhaps, to be more concrete, any of the following:

  1. (strong) Estimator sequence for each $n$ that is minimax but does not match the Cramer-Rao lower bound for any $n$
  2. (in-between) Estimator sequence for each $n$ that is minimax but does not match the Cramer-Rao lower bound asymptotically ($n\to\infty$)
  3. (weak) Estimator sequence that is asymptotically minimax but does not match the Cramer-Rao lower bound asymptotically ($n\to\infty$)
  4. Converses: Estimator sequence for each $n$ that matches the Cramer-Rao lower bound for each $n$, but is not minimax for any $n$ (or asymptotically)
$\endgroup$
3
  • 1
    $\begingroup$ What do you mean by "minimax optimal"? $\endgroup$
    – Xi'an
    Commented Feb 2, 2023 at 17:26
  • 1
    $\begingroup$ Is there more than one usage of this term? I mean: en.wikipedia.org/wiki/Minimax_estimator $\endgroup$
    – JohnA
    Commented Feb 2, 2023 at 17:45
  • $\begingroup$ Ah ha! I see what you mean. Fair enough, sorry for the confusion. You are right that "minimax estimator" is the correct formal terminology. $\endgroup$
    – JohnA
    Commented Feb 2, 2023 at 23:23

1 Answer 1

0
$\begingroup$

1 and 4. That Wikipedia link has the example $$\hat\delta^M= \frac{x+0.5\sqrt{n}}{n+\sqrt{n}}$$ for estimating $\theta$ in $Bin(n,\theta)$. It's not finite-sample efficient because that's one of the models where the MLE, $x/n$, is finite-sample efficient.

  1. The local asymptotic minimax theorem says that an asymptotically efficient estimator is locally asymptotically minimax (over neighbourhoods of size $n^{-1/2}$) for any bowl-shaped loss function, so if there's a counterexample here I think it would have to be a bit pathological.
$\endgroup$
1
  • 1
    $\begingroup$ I think you may have answered as I was editing my question: I believe your answer should now read "1 and 4" and 3 instead of 2. Can you please confirm/edit your answer before I accept. Good answer, btw, and my apologies for the confusion. $\endgroup$
    – JohnA
    Commented Feb 7, 2023 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.