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It's always very surprising to learn that some of the entities one has been assiduously studying actually represent negligibly tiny minorities (e.g. continuous functions vis-à-vis all functions)...


Now, the Central Limit Theorem, for one, holds only for probability distributions with a finite variance.

How common are such distributions in the space of all probability distributions?


More formally, let $U$ be the set of all probability distributions on $\mathbb{R}$ (say), and $F \subset U$ be the subset of those probability distributions that have a finite variance. I expect that, of the cardinalities $|F|$ and $|U\setminus F|$, one will be a strictly larger infinity than the other, but I have no intuition as to which.


(I guess this is a "meta-measure theory" question.)

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    $\begingroup$ What's stopping you there -- why not asking about those with an expectation? $\endgroup$ – Clement C. Aug 22 '17 at 11:58
  • $\begingroup$ @ClementC: the motivation for my question was wanting to gauge the CLT's generality, but you bring up an excellent point. $\endgroup$ – kjo Aug 22 '17 at 12:01
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    $\begingroup$ A small note, but you'll want to consider probability distributions up to a.e. equality at most, because there are $\mathfrak c$-sized subsets of measure zero on which you can choose any function value you want, so that any probability distribution already has $\mathfrak{c}^\mathfrak{c}$-many probability distributions a.e. equal to it. $\endgroup$ – Mees de Vries Aug 22 '17 at 12:08
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    $\begingroup$ @kjo a different, but related point: $L_2$ is dense in $L_1$. $\endgroup$ – Clement C. Aug 22 '17 at 12:10
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I presume that you speak about distributions on the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$. In terms of cardinality, there are quite few of them, namely, $\mathfrak{c} = |\mathbb{R}|$. Indeed, each probability distribution $\mathcal B(\mathbb R)$ is uniquely determined by its values on the intervals $(-\infty,q)$ with $q\in\mathbb{Q}$. Therefore, the number of probability measures does not exceed the number of sequences of real numbers, which is $\mathfrak{c}$. Since, obviously, there are both $\mathfrak{c}$ distributions with finite and infinite variance, so your guess is wrong.

There are other ways to compare sets, e.g. in terms of the Baire categories. If we consider the weak convergence topology, then both finite variance and infinite variance sets are easily shown to be second category, so another draw here.

Overall, I see no reason to believe that there are fewer finite variance (or finite expectation or finite exponential moment) distributions than others.

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    $\begingroup$ Here is an unfinished thought: it is very easy to find an injection from finite-variance into infinite-variance distributions: just add any fixed infinite-variance one to each finite-variance one. Are there any easy injections the other way around? $\endgroup$ – Mees de Vries Aug 22 '17 at 12:20
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    $\begingroup$ @MeesdeVries, yes, there are plenty. Take $\arctan$ of any infinite variance variable to get a finite variance one. But this is a good point. $\endgroup$ – zhoraster Aug 22 '17 at 12:30
  • $\begingroup$ @zhoraster It's early here, so I may very well spout nonsense. But in your latest comment: if $X$ has an infinite variance due to non-integrability at $0$, wouldn't $\arctan X$ have exactly the same problem? $\endgroup$ – Clement C. Aug 22 '17 at 12:49
  • $\begingroup$ @ClementC, you are right, this is nonsense. $\endgroup$ – zhoraster Aug 22 '17 at 13:44
  • $\begingroup$ @zhoraster I now see why. But thanks for your... politeness, I guess. $\endgroup$ – Clement C. Aug 22 '17 at 13:48

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