Show two random variables have same distribution

Question

Let X, Y be two non-negative random variables satisfying the condition $\mathbb{E}[X^\alpha] = \mathbb{E}[Y^\alpha]$ for all $\alpha \in (0, 1/2)$.

How can one show that X and Y are equal in distribution?

Edit: (only if you find this helpful) $\mathbb{E}[X], \mathbb{E}[Y]$ also exist, but a priori one does not know whether they are equal or not.

If you believe that the claim is wrong, I would also be happy to see counterexamples, or at least some intuitive explanations.

Answer 1

It can be shown that nonnegative random variables $X$ and $Y$ have the same distribution so long as $\mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha]$ is finite for all $\alpha\in(a,b]$, any $0\le a <b$.

Setting $U=\log X$ and $V=\log Y$, define the functions $$ f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha V}], $$

These are defined for complex $\alpha$ with $0 < \Re[\alpha]< b$, as the terms inside the expectations are bounded by $\max(1,X^b)$ and $\max(1,Y^b)$ in absolute value. Furthermore, it can be seen that they are complex differentiable in this range. By assumption, they are equal for real $\alpha$ in $(a,b)$. Hence, by analytic continuation, they are equal on the domain $0 < \Re[\alpha] < b$.

Then, for any real $\omega$, dominated convergence gives, \begin{align} \mathbb{E}[1_{\{X > 0\}}e^{i\omega U}]&=\lim_{t\downarrow 0}f(t+i\omega)=\lim_{t\downarrow 0}g(t+i\omega)\\ &=\mathbb{E}[1_{\{Y > 0\}}e^{i\omega V}]. \end{align} Taking $\omega=0$ shows that $X$ and $Y$ have the same probability of being zero. Then, conditioning on $X$ and $Y$, respectively, being strictly positive we see that $U$ and $V$ have the same characteristic functions. Hence, they have the distribution and, therefore, so do $X$ and $Y$.