# Does infinite expectation imply regular variation?

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?

• 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. Oct 28, 2020 at 3:56
• Right, for positive random variables, that was unnecessary; edited. Oct 28, 2020 at 9:56
• 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 $1-F(x)\sim x^{-\alpha}l(x)$. Aug 18, 2023 at 7:43