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Could you provide an example of a random variable $X$ such that $\mathbb{E}(e^X)<\infty$ but $\text{Var}(e^X)=\infty$?

Related: "Random variable with finite logarithmic first moment, infinite logarithmic variance"
Motivated by "Is (covariance) stationarity preserved under log or exponential transformation?".

Richard Hardy
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  • Your question appears predicated on a mistake: $E[(e^X)^2] = E[e^{2X}],$ not $E[e^{X^2}].$ In light of that, please indicate which version of the question you intended to ask. Note that if it's the first interpretation, then for any positive variable $Y$ with finite mean but infinite variance (such as the absolute value of a Student $t_2$), $X=\log(Y)$ satisfies your requirements. – whuber Mar 04 '21 at 18:03
  • @whuber, thank you for spotting the mistake! I have edited the question, hopefully it can be reopened now. I was suspecting the answer to be something similar to what you say (I indicated this in a comment in the linked thread), but I would like to get it "officially", i..e posted as an answer, maybe with some proof or at least intuition added. – Richard Hardy Mar 04 '21 at 18:09
  • What else needs to be said? My comment merely restated your criterion in a simpler way and pointed out there are readily available examples, of which you can find many here on CV: https://stats.stackexchange.com/search?q=finite+mean+infinite+variance. – whuber Mar 04 '21 at 18:15
  • @whuber, I see you have marked it as a duplicate. I disagree that the linked thread answers the question. The essence of my question is showing how exponentiation affects moments, not showing that there exist variables with finite first moment and finite second moment. Constructing an example the way you did is nontrivial (at least for some people like me), so it would be instructive and helpful to discuss the case. Could you please reopen the question? – Richard Hardy Mar 04 '21 at 18:31
  • I do not concur, because the issue comes down to the one-to-one correspondence between positive random variables $Y$ and their logarithms $\log(Y).$ This is a matter of arithmetic--whether you call such a variable "$Y$" or "$e^X$", it's merely a way to rename a given variable--and so has nothing of value to illustrate concerning random variables. – whuber Mar 04 '21 at 18:33
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    @whuber, I continue do disagree as what is obvious for a true expert like you can be far from obvious for many other users. I do see value in an illustration and an explanation. But I understand I cannot do much about the closing decision here, so at least I can thank you for your clear and timely explanation. – Richard Hardy Mar 04 '21 at 18:41

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