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Consider a positive random variable $X$ with $\text{E} X = \infty$. Is this sufficient to conclude $1-F_X(x) = \ell(x) x^{-\alpha}$, where $\alpha \in (0,1]$ and $\ell(x)$ is slowly varying at infinity (i.e. $\lim_{x\to\infty} \ell(x \lambda)/\ell(x) = 1$ for all $\lambda > 0$)?

If not, are there additional necessary and sufficient conditions on $X$ that make it so?

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    $\begingroup$ Your assumptions are stated rather strangely, as the "undefined" expectation you are referring to is when $\mathbb E[Y^+] = \mathbb E[Y^-] = \infty$, where $$ Y^+ := \max\{Y,0\},\quad Y^- = \max\{-Y,0\}. $$ The expectation is undefined because $\mathbb E[Y] := \mathbb E[Y^+] - \mathbb E[Y^-]$ is an indeterminate form of the type $\infty-\infty$. But if $\mathbb P(X\geqslant 0)=1$ then clearly $\mathbb E[X] = \mathbb E[X^+]$ which is a well-defined (extended) real number. I think the parenthesized part of the first sentence could be left out as it doesn't provide any additional information. $\endgroup$
    – Math1000
    Oct 28, 2020 at 3:56
  • $\begingroup$ Right, for positive random variables, that was unnecessary; edited. $\endgroup$
    – bodhi
    Oct 28, 2020 at 9:56
  • $\begingroup$ I am also very curious about this question. May I ask if you have solved this problem? Since I can not verify that for t-distribution with degree $\alpha\in(0,2]$, whether or not its tail probability $1-F(x)\sim x^{-\alpha}l(x)$. $\endgroup$
    – tt Chen
    Aug 18, 2023 at 7:43

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