# Deconvolution of sum of two random variables

Let $Z = X + c \cdot Y$ where $X$ and $Y$ are independent random variables drawn form the same distribution given by the pdf $g()$ and $0 < c < 1$

I have observations of $Z_i$'s and thus can approximate the discrete pdf $f()$ which is the distribution of Z.

Thus:

$f(x) = (g \ast g^\prime)(x) = \sum_{d \in D} g(d) g((x-d)c)$

where $g^\prime(x) = g(xc)$

How to calculate $g()$ based on $f()$?

• Shouldn't it be $\sum_{d\in D} g(d)g((x-d)/c)$ ? – Thomas Rippl Feb 14 '14 at 13:02
• you have an identification issue since if e.g. X,Y normal mean 0 the result is normal, mean 0 variance $\sigma^2( 1 + c^2)$, so more than 1 $c, \sigma$ pair can correspond to the dist. of Z. – mike Feb 14 '14 at 13:13
• @mike: If $c$ is fixed this problem disappears. – Jochen Wengenroth Feb 14 '14 at 13:39
• At least in principle you can calculate all moments $E(X^n)$ by a recursion from the moments of $Z$. – Jochen Wengenroth Feb 14 '14 at 13:40
• Have you looked at the Wikipedia articles "deconvolution" and "blind deconvolution"? Sorry I can't post this as a comment either. [converted to comment -- mods] – user46979 Feb 14 '14 at 22:26

Looking at your equation $f(x)=(g \ast g')(x)$ relating the densities, wouldn't a Fourier transform do the job? Taking Fourier on both sides, we get $\cal{F}f (\omega) = \cal{F}g(\omega) \cal{F}g'(\omega)$ and ${\cal F}g'(\omega)=1/c{\cal F}g(\omega/c)$. Now I am unsure since I don't understand the problem. Is $c$ known to you? Also, I am unclear as to whether $g$ is a discrete or continuous density. If discrete, how to make sense of $g(xc)$ for any $0<c<1$? I would agree with Thomas that dividing by $c$ makes more sense here, provided that $D$ is defined appropriately (i.e. range of possible values of $X$) and given the model equation.

We can take logarithms on both sides to obtain, using your notation ($\hat{f}$ for Fourier of $f$), and ignoring still the fact that it should be divided by $c$, not multiplied): $$\log(c) +\log \hat{f}(x) = \log(\hat{g}(x))+\log(\hat{g}(x/c))$$ where $x \in D$ a discrete set. This is a linear system of equations that can be solved efficiently.

• Forier transformation is of course a good idea. However, how do you solve the equation $c\hat{f}(x) =\hat{g}(x)\hat{g} (x/c)$ in an efficient way to calulate $g$ by means of the Fourier inversion formula? – Jochen Wengenroth Feb 14 '14 at 19:31
• ${\cal F}g'(\omega)=1/c{\cal F}g(\omega/c)$ does only hold for linear functions, right? So this approach only works if g() is linear. – Philip Feb 18 '14 at 16:50
I found this paper Rates of convergence for constrained deconvolution problem which describes how to estimate the distribution of $Z$ from observations of a random process of the form $Z = \alpha X + \beta Y$.