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Suppose $X, Y$ are two positive random variables such that $\mathbb{E}[X^\alpha] = \mathbb{E}[Y^\alpha]$ for all $\alpha \in (0, 1/2)$. It is also known that the first moment exists for each of them, but a priori one does not know if the first moments are equal.

(It is also known that all negative moments exist, but if possible this assumption should be avoided.)

Is it possible to show that the two random variables have equal distribution?

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2 Answers 2

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The argument from the paper Carlo linked to (which, by the way, is essentially a classical result of Cramer's) can be adapted to your situation.

Consider the moment generating function $f(z) = Ee^{z\ln X}$ of the random variable $\ln X$. By your assumption, this is well defined for $0\le \textrm{Re}\, z<1/2$. Moreover, $f$ is continuous on this strip and holomorphic on the open strip. Therefore, if two such random variables $\ln X$, $\ln Y$ satisfy $f_X(z)=f_Y(z)$ for real $0\le z<1/2$, then the $f$'s also agree on $z=ix$, $x\in\mathbb R$. Since the Fourier transform determines the distribution, this gives the claim.

Of course, this argument works for any $c>0$ instead of $1/2$, so it establishes a slightly stronger statement.

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  • $\begingroup$ In fact before posting the question I was thinking about the same thing. The problem (which might be stupid, since I am not very strong in complex analysis) I had in mind was: how do I know $f$ is holomorphic on $\{z \in \mathbb{C}: 0 \le Re ~z < 1/2\}$? $\endgroup$ Apr 19, 2015 at 2:14
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    $\begingroup$ @random_person: I think the quickest way to see this is to use Fubini-Tonelli to confirm that for any closed curve $C$ inside the open strip, we have that $\int_C dz \int d\mu(t) e^{zt}=0$ (use that $\textrm{Re}\,z\le c<1/2$ for all $z\in C$). By Morera's theorem, $f$ is holomorphic. $\endgroup$ Apr 19, 2015 at 2:23
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The answer to your question is "Yes", as was shown in When Do the Moments Uniquely Identify a Distribution, by Carlos A. Coelho, Rui P. Alberto and Luis M. Grilo:

In this brief note we show that whenever $X$ is a positive random variable, if $E(X^h)$ is defined for $h$ in some neighborhood of zero then the moments $E(X^h)$ uniquely identify the distribution of $X$.

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    $\begingroup$ Unless I have misread it while skimming, this note doesn't quit settle the (main) question of the OP, since it requires moments for at least some $-1 < \alpha < 0$ to also exist. Correct? $\endgroup$
    – cardinal
    Apr 18, 2015 at 22:30

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