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?
Solved – Product of two independent random variables
distributionsmathematical-statisticsprobability
Related Solutions
The problem is ill-posed. There is no unique solution if you dont specify Z. Think about it. Pick any Y with any distribution you choose. X+Y will have some distribution that depends on what you chose for Y. So if you specify Z then Y is determined but two distinct choices for Y will give two different distributions for Z. There are infinitely many solutions to your problem.
Now given that of course your second problem is also ill-posed. Just knowing that X and Y a re independent tells you almost nothing about Z. Take X and Y normal and Z will be normal. If X and Y are identical distributed Cauchys Z will be Cauchy. Two independent chi squares lead to Z be chi-square. So there are three solutions to problem 2 and that is only scratching the surface
In problem 1 you need to know Z's distribution to determine Y's. In problem 2 you cant get Z without specifying the distributions of X and Y.
simplify the term in the integral to
$T=e^{-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)} y^{k/2-2} $
find the polynomial $p(y)$ such that
$[p(y)e^{-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)}]'=p'(y)e^{-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)} + p(y) [-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)]' e^{-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)} = T$
which reduces to finding $p(y)$ such that
$p'(y) + p(y) [-\frac{1}{2}((\frac{\frac{z}{y}-\mu_x}{\sigma_x} )^2 -y)]' = y^{k/2-2}$
or
$p'(y) -\frac{1}{2} p(y) (\frac{z \mu_x }{\sigma_x^2} y^{-2} \frac{z^2}{\sigma_x^2} y^{-3} -1)= y^{k/2-2}$
which can be done evaluating all powers of $y$ seperately
edit after comments
Above solution won't work as it diverges.
Yet, some others have worked on this type of product.
Using Fourrier transform:
Schoenecker, Steven, and Tod Luginbuhl. "Characteristic Functions of the Product of Two Gaussian Random Variables and the Product of a Gaussian and a Gamma Random Variable." IEEE Signal Processing Letters 23.5 (2016): 644-647. http://ieeexplore.ieee.org/document/7425177/#full-text-section
For the product $Z=XY$ with $X \sim \mathcal{N}(0,1)$ and $Y \sim \Gamma(\alpha,\beta)$ they obtained the characteristic function:
$\varphi_{Z} = \frac{1}{\beta^\alpha }\vert t \vert^{-\alpha} exp \left( \frac{1}{4\beta^2t^2} \right) D_{-\alpha} \left( \frac{1}{\beta \vert t \vert } \right)$
with $D_\alpha$ Whittaker's function ( http://people.math.sfu.ca/~cbm/aands/page_686.htm )
Using Mellin transform:
Springer and Thomson have described more generally the evaluation of products of beta, gamma and Gaussian distributed random variables.
Springer, M. D., and W. E. Thompson. "The distribution of products of beta, gamma and Gaussian random variables." SIAM Journal on Applied Mathematics 18.4 (1970): 721-737. http://epubs.siam.org/doi/10.1137/0118065
They use the Mellin integral transform. The Mellin transform of $Z$ is the product of the Mellin transforms of $X$ and $Y$ (see http://epubs.siam.org/doi/10.1137/0118065 or https://projecteuclid.org/euclid.aoms/1177730201). In the studied cases of products the reverse transform of this product can be expressed as a Meijer G-function for which they also provide and prove computational methods.
They did not analyze the product of a Gaussian and gamma distributed variable, although you might be able to use the same techniques. If I try to do this quickly then I believe it should be possible to obtain an H-function (https://en.wikipedia.org/wiki/Fox_H-function ) although I do not directly see the possibility to get a G-function or make other simplifications.
$M\lbrace f_Y(x) \vert s \rbrace = 2^{s-1} \Gamma(\tfrac{1}{2}k+s-1)/\Gamma(\tfrac{1}{2}k)$
and
$M\lbrace f_X(x) \vert s \rbrace = \frac{1}{\pi}2^{(s-1)/2} \sigma^{s-1} \Gamma(s/2) $
you get
$M\lbrace f_Z(x) \vert s \rbrace = \frac{1}{\pi}2^{\frac{3}{2}(s-1)} \sigma^{s-1} \Gamma(s/2) \Gamma(\tfrac{1}{2}k+s-1)/\Gamma(\tfrac{1}{2}k) $
and the distribution of $Z$ is:
$f_Z(y) = \frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} y^{-s} M\lbrace f_Z(x) \vert s \rbrace ds $
which looks to me (after a change of variables to eliminate the $2^{\frac{3}{2}(s-1)}$ term) as at least a H-function
what is still left is the puzzle to express this inverse Mellin transform as a G function. The occurrence of both $s$ and $s/2$ complicates this. In the separate case for a product of only Gaussian distributed variables the $s/2$ could be transformed into $s$ by substituting the variable $x=w^2$. But because of the terms of the chi-square distribution this does not work anymore. Maybe this is the reason why nobody has provided a solution for this case.
Best Answer
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 $\phi$