Product of two independent random variables

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 $\phi$