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Let ${\textstyle \{X_{1},\ldots ,X_{n}}\}$ be a sequence of independent and identical random variables. The distribution of $X_n$ is unknown.

Assuming that we know the distribution of the following summation: $${S}\equiv \sum_{n=1}^{\infty}\frac{X_n}{n^2}$$

Would it be possible to find the distribution of $X_n$ from $S$ ?

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  • $\begingroup$ In case $X$ and $Y$ are independent with distribution densities $f_X$ and $f_Y$, their has a density distribution of $f_X* f_Y$. So in your case you'll have a lot of convolutions, and the question is whether it's possible from the result of convolution to deduce its source. Sounds like a tough procedure. $\endgroup$
    – SBF
    Jan 3 at 20:02
  • $\begingroup$ In the original problem, all $X_n$ are independent and identical. Will this make the problem easier ? $\endgroup$
    – david
    Jan 3 at 22:47

1 Answer 1

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Partial answer

The characteristic function of $X$ satisfies a functional equation, namely $$\prod_{n=1}^{+\infty} \phi_X(t/n^2) = \phi_S(t).$$ Solving this kind of equation is not simple.

If the distribution of $X$ is completely determined by its moments, or equivalently by its cumulants, then it is determined by the cumulants of $S$ since for every integer $d \ge 1$, $$\kappa_d(S) = \sum_{n=1}^{+\infty}\kappa_d(X/n^2) = \zeta(2d)\kappa_d(X).$$

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  • $\begingroup$ +1 for using cumulants! that indeed may provide a good answer to the OP $\endgroup$
    – SBF
    Jan 6 at 13:58
  • $\begingroup$ @Christophe Leuridan: Thank you. Just want to confirm, in your answer, $\zeta(2d)$ is Riemann Zeta function at $2d$, am I right ? $\endgroup$
    – david
    Jan 6 at 18:33
  • $\begingroup$ @david Yes, since $\kappa_d(\lambda X) = \lambda^d \kappa_d(X)$. $\endgroup$
    – Christophe Leuridan
    Jan 6 at 18:41

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