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.


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

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

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

  • $\begingroup$ Shouldn't it be $\sum_{d\in D} g(d)g((x-d)/c)$ ? $\endgroup$ – Thomas Rippl Feb 14 '14 at 13:02
  • $\begingroup$ 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. $\endgroup$ – mike Feb 14 '14 at 13:13
  • $\begingroup$ @mike: If $c$ is fixed this problem disappears. $\endgroup$ – Jochen Wengenroth Feb 14 '14 at 13:39
  • $\begingroup$ At least in principle you can calculate all moments $E(X^n)$ by a recursion from the moments of $Z$. $\endgroup$ – Jochen Wengenroth Feb 14 '14 at 13:40
  • $\begingroup$ 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] $\endgroup$ – 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.

Edited to answer Jochen comment:

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.

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  • $\begingroup$ 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? $\endgroup$ – Jochen Wengenroth Feb 14 '14 at 19:31
  • $\begingroup$ I see I can add comments here -- I edited the above already in regards to your comment. $\endgroup$ – Robert Feb 15 '14 at 0:36
  • $\begingroup$ c is known to me. g() and f() are continuous in principle. However, since i only have observations of Zi's i can only calculate a discrete pdf. $\endgroup$ – Philip Feb 16 '14 at 20:24
  • $\begingroup$ ${\cal F}g'(\omega)=1/c{\cal F}g(\omega/c)$ does only hold for linear functions, right? So this approach only works if g() is linear. $\endgroup$ – Philip Feb 18 '14 at 16:50
  • $\begingroup$ No -- in fact, if g() is linear, it will not define a density at all. The above holds for a larger class of functions. You can easily reserve it (just write down the definition of Fourier). It holds for Lebesgue integrable functions. $\endgroup$ – Robert Feb 18 '14 at 16:55

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 $Z = \alpha X + \beta Y$.

The trick is to use the characteristic function of Z and it really works out quite nicely.

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You don't really have the distribution of Z but rather a sampling from it and perhaps some noise on top of that. There is no closed form solution to this kind of problem as you can trade off between smoothness and goodness of fit.

In practice, I'd avoid Fourier methods as they amplify the noise. Use a parametric method. I'd suggest modeling the distribution by a sum of Gaussians for which scaling and convolution keep things in the family of Gaussians. A single Gaussian is trivial. Two or more Gaussians requires you to use the principle of maximum likelihood which turns this into an optimization problem.

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