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Typical (or common) approaches to prove an estimator is consistent require finite mean and variance. The proofs usually follow from concentration bounds, e.g. Markov, Chebyshev, etc.

I'm wondering how one shows consistency of an estimator if the expectation is infinite?

Or alternately-and leading to a similar conclusion but perhaps more general-are there concentration of measure bounds when the mean is infinite? (please provide example, proof and/or links)

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    $\begingroup$ "the expectation is infinite" Whose expectation? the expectation of estimate? $\endgroup$ – user158565 Jul 4 at 3:06
  • $\begingroup$ @user158565 fair point. I suppose you could concoct examples w/an r.v. with infinite expectation but define a function/estimator of that random variable whose expectation is finite. I'd say both, i.e. infinite expectation of the random variable or infinite expectation of the estimator, although I'd think the latter case is less interesting b/c if the expectation of estimator is infinite, then so is the variance. $\endgroup$ – Lucas Roberts Jul 4 at 13:24
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    $\begingroup$ I think your question is meaningless. Suppose using $T(X)$ to estimate $\theta$. Then one of $E(T(X))$ and $\theta$ being infinite would result in the inconsistent of the estimate $T(X)$. $\endgroup$ – user158565 Jul 4 at 19:47
  • $\begingroup$ @user158565 that is why I said in the comment above, "although I'd think the latter case is less interesting b/c if the expectation of estimator is infinite, then so is the variance." Although I'm not sure how you conclude that $\theta$ is infinite, I did say that it was and you should assume that $\theta$ is finite. $\endgroup$ – Lucas Roberts Jul 5 at 0:09
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    $\begingroup$ Then under conditions that $\theta < \infty$ and $E(T(X)) = \infty$, then $T(x)$ cannot be the consistent estimator of $\theta$, $\endgroup$ – user158565 Jul 5 at 0:58

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