Let $P_\theta$ denote the distribution of the random variable $X$. The distribution depends on the parameter $\theta$ that lies in some parameter space $\Theta$. Consider a function $f(\theta)$ of $\theta$.

We say that $f(\theta)$ is identifiable if $f(\theta_1)\neq f(\theta_2)$ implies $P_{\theta_1}\neq P_{\theta_2}$ for every $\theta_1,\theta_2$ in $\Theta$. We say that $f(\theta)$ is estimable if $E_\theta g(X)=f(\theta)$ for all $\theta$ in $\Theta$ for some function $g$, where $E_\theta$ denotes the expected value computed from $P_\theta$.

It is easy to show that if $f(\theta)$ is estimable then it is identifiable. One can also find counterexamples where the converse is false. For example, suppose $X$ follows a Bernoulli distribution $P_\theta$ with parameter $\theta=P(X=1)\in(0,1)$, then nonlinear functions of $\theta$ are not estimable.

Can someone provide a simple example where a parameter is identifiable but has no estimator that is consistent for it? In the above example $\sqrt{\theta}$ has a consistent estimator, namely $\sqrt{n^{-1}\sum_{i=1}^nx_i}$ for a random sample $\{x_i\}_{i=1}^n$ from $P_\theta$.

  • $\begingroup$ You have given an inverse definition of identifiability and it corresponds to a tautology, since your description is equivalent to part of the definition of parameter: namely, a parameter has to be a definite property of the distribution. The remarks about estimability appear to be irrelevant to the question, making them somewhat puzzling. Please check over your post to verify that it correctly reflects what you want to ask. $\endgroup$
    – whuber
    Mar 25 '21 at 19:30
  • $\begingroup$ @whuber Thank you for your comment I made a few changes to my original post. What do you think now? The definitions are from "Statistical Models: Theory and Practice (Revised Edition)" by David A. Freedman from 2009. I do not see the tautology and in any case this is the question I want to ask. If someone believes the question itself does not make sense, then please do explain. $\endgroup$
    – Elias
    Mar 26 '21 at 10:44
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    $\begingroup$ @whuber Do you want to argue that I misunderstand what is written in the above quote? $\endgroup$
    – Elias
    Mar 26 '21 at 21:08
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    $\begingroup$ Also, I do not see the "error". The definition seems equivalent to the one on Wikipedia in the "Identifiability" article. en.wikipedia.org/wiki/Identifiability $\endgroup$
    – Elias
    Mar 26 '21 at 21:19
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    $\begingroup$ @kjetilbhalvorsen Done! (Needs to be accepted though, I think.) $\endgroup$
    – Elias
    Mar 31 '21 at 11:58

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