distribution function

Question

Suppose that $X$ and $Y$ are two random variables such that:

$$E \left(\frac{a}{a+X} \right)=E\left(\frac{a}{a+Y}\right)< \infty \qquad\forall a > \pi.$$ Can we conclude that $X$, $Y$ have the same distribution?

Answer 1

This is not a complete solution, but presents an idea that along with some more data could be made into a solution.

By dividing the top and bottom of the fraction by $a$ and letting $\epsilon = \frac{1}{a}$ we have: $$E\left(\frac{1}{1+\epsilon X}\right) = E\left(\frac{1}{1+\epsilon Y}\right) < \infty \hspace{20pt} \forall \epsilon < 1/\pi$$

By linearity of $E$ we have then: $$E\left(\frac{\frac{1}{1+\epsilon X} - 1}{\epsilon}\right) = E\left(\frac{\frac{1}{1+\epsilon Y} - 1}{\epsilon}\right) \hspace{20pt} \forall \epsilon < 1/\pi$$

Taking the limit $\epsilon \to 0$ now gives:

$$\lim_{\epsilon \to 0} E\left(\frac{\frac{1}{1+\epsilon X} - 1}{\epsilon}\right) = \lim_{\epsilon \to 0} E\left(\frac{\frac{1}{1+\epsilon Y} - 1}{\epsilon}\right) \hspace{20pt}$$

Now if we had some additional information and we could do two things:

1. Justify moving the limit $\epsilon \to 0$ from outside the $E$ to inside the $E$
2. Justify that the limit is finite

Then we would have $E(X) = E(Y)$. What we did above amounted to taking a derivative with respect to $\epsilon$ evaluated at $\epsilon = 0$. If we could justify the change of limit for TWO derivatives, we would have $E(X^2) = E(Y^2)$. If we could do this for any number of derivatives, we would have $E(X^n) = E(Y^n)$ for every $n$ and we would be in a good position to use the method of moments to conclude $X$ and $Y$ have the same distribution (again I think we would need a bit more information on $X$ and $Y$ to conclude this)

Answer 2

If $\mu$ is the distribution of $X$, then $\mathbb E\left(\frac{a}{a+X}\right)=S_\mu\left(-\frac1a\right)$, where $S_\mu$ is the Stieltjes transform of $\mu$. To get an idea of the reasons why supplementary conditions are needed for $S_\mu$ to determine $\mu$, see Carleman's condition and Krein's condition.