Product of two independent random variables I have a sample of about 1000 values​​. These data are obtained from the product of two independent random variables $\xi \ast \psi $. The first random variable has a uniform distribution $\xi \sim U(0,1)$. The distribution of the second random variable is not known. How can I estimate the distribution of the second ($ \psi $) random variable?
 A: We have, Assuming $\psi$ has support on the positive real line,
$$\xi \,\psi = X$$ Where $X \sim F_n$ and $F_n$ is the empirical distribution of the data.
Taking the log of this equation we get,
$$ Log(\xi) + Log(\psi) = Log(X) $$
Thus by Levy's continuity theorem, and independance of $\xi$ and$\psi$
taking the charactersitic functions:  
$$ \Psi_{Log(\xi)}(t)\Psi_{Log(\psi)}(t) = \Psi_{Log(X)}$$
Now, $ \xi\sim Unif[0,1]$$, therefore $$-Log(\xi) \sim Exp(1) $
Thus,
$$\Psi_{Log(\xi)}(-t)= \left(1 + it\right)^{-1}\,$$
Given that $\Psi_{ln(X)} =\frac{1}{n}\sum_{k=1}^{1000}\exp(itX_k) ,$
With $ X_1 ... X_{1000}$ The random sample of $\ln(X)$.
We can now specify completly the distribution of $Log(\psi)$ through its characteristic function:
$$ \left(1 + it\right)^{-1}\,\Psi_{Log(\psi)}(t) = \frac{1}{n}\sum_{k=1}^{1000}\exp(itX_k)$$
If we assume  that the moment generating functions of $\ln(\psi)$ exist and that $t<1$ we can write the above equation in term of moment generating functions:
$$ M_{Log(\psi)}(t) = \frac{1}{n}\sum_{k=1}^{1000}\exp(-t\,X_k)\,\left(1 - t\right)\,$$
It is enough then to invert the Moment generating function to get the distribution of $ln(\phi)$ and thus that of $\phi$
