How to get the distribution of a random variable from the distribution of their summation?

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 $$S$$ ?

• 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.
– SBF
Jan 3, 2023 at 20:02
• In the original problem, all $X_n$ are independent and identical. Will this make the problem easier ? Jan 3, 2023 at 22:47

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 $$d \ge 1$$, $$\kappa_d(S) = \sum_{n=1}^{+\infty}\kappa_d(X/n^2) = \zeta(2d)\kappa_d(X).$$
• @Christophe Leuridan: Thank you. Just want to confirm, in your answer, $\zeta(2d)$ is Riemann Zeta function at $2d$, am I right ? Jan 6, 2023 at 18:33
• @david Yes, since $\kappa_d(\lambda X) = \lambda^d \kappa_d(X)$. Jan 6, 2023 at 18:41