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I'm trying to prove that MLE from the proof of one theorem in Lehmann's "Theory of point estimation" (the theorem is below) is a measurable function. I know that under some regularity conditions (e.g. stats.stackexchange.com/questions/430954/example-of-a-non-measurable-maximum-likelihood-estimator
or https://math.stackexchange.com/questions/1251393/when-is-the-maximum-likelihood-estimator-measurable ) MLE is measurable, but it didn't help me to prove that the closest root from the proof is a measurable function.

The problem is as follows. By definition, an estimate is a measurable function. Even the existence of a measurable version of MLE is not so obvious, but here an extremum arises over the set of MLE. It is unlikely that a theorem from a classical book can be wrong, but how to prove it?

The theorem is here:

enter image description here

I think that in $(3.14)$ the set $|\hat{\theta}_n - \theta_0|$ may be not measurable (but I don't know counterexample).

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  • $\begingroup$ I wonder whether there is any difference between saying "The true parameter value is an interior point of the sample space" and saying "The parameter space contains an open set of which the true parameter value is an interior point." $\endgroup$
    – Michael Hardy
    Commented Oct 23, 2021 at 21:12
  • $\begingroup$ @MichaelHardy If there's a typo in the first phrase and we compare "The true parameter value is an interior point of the parameter (not sample) space" and "The parameter space contains an open set of which the true parameter value is an interior point" then there's no difference, but how could it help? $\endgroup$
    – Botnakov N.
    Commented Oct 23, 2021 at 21:21
  • $\begingroup$ Maybe it doesn't help with your question, but the author's phrasing makes the matter superficially appear more complicated than it is. $\endgroup$
    – Michael Hardy
    Commented Oct 23, 2021 at 21:29
  • $\begingroup$ @MichaelHardy, I agree. $\endgroup$
    – Botnakov N.
    Commented Oct 23, 2021 at 21:32

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In general, a maximum likelihood estimator (MLE) does not have to be measurable. For instance, suppose that $f(x|\theta)=g(x-\theta)$, where $g(x)=1(0<x<1)$. Then, for any $(x_1,\dots,x_n)\in\mathbb R^n$, any number $\hat\theta(x_1,\dots,x_n)\in(\max_i x_i-1,\min_i x_i)$ is a maximizer of the likelihood $f(x_1|\theta)\cdots f(x_n|\theta)$ (in real $\theta$), and it is easy to make the resulting function $\hat\theta$ non-measurable. One can also use a mollifier to make $g$ and hence the likelihood however smooth, but with the non-measurability of an MLE preserved.

However, there is no reason to worry about this. As noted on page 1163 of this paper, in such situations one can simply use "the corresponding outer and inner measures".

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  • $\begingroup$ Thank you! So if $MLE$ is non-measurable then we may define consistensy of non-measurable estitamor as convergence of inner $P_{\theta}$-measure of $\{ |\hat{\theta}_n - \theta| < \delta \}$ to $0$, right? Sounds interesting! I have not seen this approach. $\endgroup$
    – Botnakov N.
    Commented Oct 24, 2021 at 11:41
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    $\begingroup$ @BotnakovN. : The consistency can be generally defined as the convergence of $(P_\theta)^∗(|\hat\theta_n−\theta|>\delta)$ to $0$ for each real $\delta>0$. I think this will hold in all existing theorems about MLE consistency, and this will be at least formally stronger than the statement that $P_\theta(|\hat\theta_n−\theta|>\delta)\to0$ for all measurable versions of $\hat\theta_n$ and for each real $\delta>0$. $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2021 at 13:17
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    $\begingroup$ @BotnakovN. : The condition that we should take the root closest to $\theta_0$ makes no sense statistically, because $\theta_0$ is unknown, and then $\hat\theta_n$ will not be an estimator in any sense, be it measurable or not -- because an estimator cannot depend on an unknown parameter. However, indeed it appears harder to get a counterexample with this additional "closest" condition. But, given what has been said here, do we really need such a counterexample? Anyhow, generally, I do not think Lehmann's book is quite rigorous. It is a statistical book, not really mathematical. $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2021 at 13:22
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    $\begingroup$ @BotnakovN. : Also, in Lehmann's proof, it is said "Let $\theta_n^*$ be the root [of (3.12)? -- I.P.] closest to $\theta_0$. [This exists because the limit of a sequence of roots is again a root by the continuity of of $l(\theta)$.]" Here one can say the following: (i) for the closest root to $\theta_0$ of $l'(\theta)=0$ to exist, one needs -- not the continuity of $l(\theta)$ -- but continuity of $l'(\theta)$; (ii) it is unclear why such a closest root must be a maximizer of $l(\theta)$; (iii) it is unclear why such a closest root is unique. $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2021 at 14:26
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    $\begingroup$ Previous comment continued: So, there are a number of problems with Lehmann's proof -- which I think are worse than the possible non-measurability. $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2021 at 14:27

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