General description

Does an efficient estimator (which has sample variance equal to the Cramér–Rao bound) maximize the probability for being close to the true parameter $\theta$?

Say we compare the difference or absolute difference between the estimate and the true parameter $$\hat\Delta = \hat \theta - \theta$$

Is the distribution of $\hat\Delta$ for an efficient estimator stochastically dominant over the distribution of $\tilde\Delta$ for any other unbiased estimator?


I am thinking about this because of the question Estimator that is optimal under all sensible loss (evaluation) functions where we can say that the unbiased best estimator with respect to one convex loss function is also unbiased best estimator with respect to another loss function (From Iosif Pinelis, 2015, A characterization of best unbiased estimators. arXiv preprint arXiv:1508.07636). The stochastic dominance for being close to the true parameter seems to be the similar to me (it is a sufficient condition, and a stronger statement).

More precise expressions

The question statement above is broad, e.g. what type of unbiasedness is considered and do we have the same distance metric for negative and positive differences?

Let's consider the following two cases$^\dagger$ to make the question less broad:

Conjecture 1: If $\hat \theta$ is an efficient mean and median-unbiased estimator. Then for any mean and median-unbiased estimator $\tilde \theta$ $$\text{if $x>0$ then } P[\hat\Delta \leq x] \geq P[\tilde\Delta \leq x] \\ \text{if $x<0$ then } P[\hat\Delta \geq x] \geq P[\tilde\Delta \geq x]$$ where $\hat \Delta = \hat \theta - \theta$ and $\tilde \Delta = \tilde \theta - \theta$

Conjecture 2: If $\hat \theta$ is an efficient mean-unbiased estimator. Then for any mean-unbiased estimator $\tilde \theta$ and $x>0$ $$ P[\vert \hat\Delta \vert \geq x] \leq P[\vert \tilde\Delta \vert \geq x] $$

  • Are the above conjectures true?
  • If the propositions are too strong, can we adapt them to make it work?

$\dagger$The second is related to the first but drops the restriction for median-unbiasedness (and then we need to take both sides together or otherwise te proposition would be false for any estimator that has a different median than the efficient estimator).

Example, illustration:

Consider the estimation of the mean $\mu$ of the distribution of a population (that is assumed to be normal distributed) by (1) the sample median and (2) the sample mean.

In the case of a sample of size 5, and when the true distribution of the population is $N(0,1)$ this looks like

cdf example

In the image we see that the folded CDF of the sample mean (which is an efficient estimator for $\mu$) is below the folded CDF of the sample median. The question is whether the folded CDF of the sample mean is below the folded CDF of any other unbiased estimator as well.

Alternatively, using the CDF instead of folded CDFs we can ask the question whether the CDF of a the mean maximizes the distance from 0.5 at every point. We know that $$\forall \hat \theta : |F_{mean}(\hat \theta)-0.5| \geq |F_{median}(\hat \theta)-0.5| $$

do we also have this when we replace $F_{median}(\hat \theta)$ for the distribution of any other mean and median-unbiased estimator?

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    $\begingroup$ Check the Pitman nearness keyword, not that I find this criterion particularly sensible. $\endgroup$ – Xi'an Jan 9 at 15:10
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    $\begingroup$ From the conjecture, it would seem more reasonable to use median-unbiased estimators than mean-unbiased estimators. (Unbiased estimators exist in few settings and best unbiased in even fewer settings.) $\endgroup$ – Xi'an Jan 9 at 15:16
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    $\begingroup$ The 'Pitman closeness criterion' is indeed interesting. Based on the information on wikipedia I see it as "the probability for the absolute difference being closer". It is a bit different though. This Pitman closeness criterion might create interesting cases where some estimator has on average a smaller absolute difference but does not win according to this closeness criterion. $\endgroup$ – Martijn Weterings Jan 9 at 15:23
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    $\begingroup$ The criterion you propose is invariant by bijective monotone transforms, but mean-unbiasedness is not, while median-unbiasedness is. It is also incredibly strong in that the cdf of $\hat\theta$ has to be above the cdf of $\tilde\theta$ above $\theta$ and below the cdf of $\tilde\theta$ below $\theta$, for all values of the parameter $\theta$. $\endgroup$ – Xi'an Jan 9 at 15:32
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    $\begingroup$ @Xi'an I have added a visual example and now I get your comment about median-biasedness versus mean-biasedness. I have adjusted the question (although it is diverging from my original idea relating to the linked question which needs some more complex adjustments now). $\endgroup$ – Martijn Weterings Jan 9 at 17:15

Here is an experiment in a non-standard case, the location Cauchy problem, where non-standard means that there is no uniformly best unbiased estimator. Let us consider $(X_1,\ldots,X_N)$ a sample from a Cauchy $\mathcal{C}(\mu,1)$ distribution and the following four invariant estimators of $\mu$:

  1. $\hat{\mu}_1= \text{median}(X_1,\ldots,X_N)=X_{(N/2)}$
  2. $\hat{\mu}_2= \text{mean}(X_{(N/4)},\ldots,X_{(3N/4)})=\frac{2}{N}(X_{(N/4)}+\ldots+X_{(3N/4)})$
  3. $\hat{\mu}_3=\mu^\text{MLE}$ which is efficient
  4. $\hat{\mu}_4=\hat{\mu}_1+\frac{2}{N}\frac{\partial \ell}{\partial \mu}(\hat{\mu}_1)$

Then comparing the cdfs of the four estimators leads to this picture, where the cdfs of $\hat{\mu}_3$ (gold) and $\hat{\mu}_4$ (tomato) are comparable, and improving upon $\hat{\mu}_1$ (steelblue), itself improving upon $\hat{\mu}_2$ (sienna). enter image description here

A representation of the differences to the empirical cdf of the MLE makes it clearer:

enter image description here

Here is the corresponding R code:

  return(optimize(function(theta) -sum(dcauchy(x, 
    location=theta, log=TRUE)),  c(-100,100))$minimum)
for (t in 1:T){

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    $\begingroup$ Shouldn't the gold curve (the difference of the empirical MLE with itself) be zero in the plot of differences. $\endgroup$ – Martijn Weterings Jan 11 at 10:29
  • $\begingroup$ My bad, I changed the color codes: tomato is for the difference with the fourth, gold for the difference with Pitman, sienna for the difference with the trimmed mean, and blue for the difference with the median. $\endgroup$ – Xi'an Jan 11 at 10:41

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