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So, having discovered distribution convolution, which is a method for deriving the density of a sum of individual probability distribution densities,

$$S = X_{first\_distribution} + Y_{second\_distribution} + \dots,$$

my immediate question now is, is it statistically possible to derive the initial marginal distributions?

For example, if the new distribution, $S$, was specified definitely as the sum of 2 others, whose distribution forms we happened to know, $X_{normal} \ \text{and} \ Y_{gamma}$, is there a statistical method to recover the parameters of $X$ and $Y$ ($\mu, \sigma, k, \text{and} \ \theta$)?

A similar question was asked here, but it looks like the answers are specific to the simple situation of Gaussians being added to each other, whereas I am wondering about a more general principle.

Any advice on this question here would be greatly appreciated by me.

Coolio2654
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    It depends. *E.g.*, because the sum of two independent Normal variables with means $\mu_i$ and variances $\sigma_i^2$ is Normal with mean $\mu_1+\mu_2$ and variance $\sigma_1^2+\sigma_2^2,$ with either parameter you need to find two numbers that sum to a given one--and obviously there are infinitely many possibilities. In other cases there is a unique solution. (*E.g.,*, from the sum of a Uniform$[-1/4,1/4]$ variable and a Poisson($\lambda$) variable it is easy to recover the summands.) It would help, then, to focus your post on the particular problem that might have inspired this question. – whuber Jan 26 '20 at 20:21
  • That is a very useful thing to point out, whuber, that many convolutions are a form of addition, and thus trying to find a unique solution - what I was trying to do - is quite silly. My reason for posting this question is for understanding the general principle behind *de*-convolution - since almost nothing seems to be written about it - or to understand why there does not exist a general method for doing something that seems to fundamental. – Coolio2654 Jan 26 '20 at 23:34
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    It might pay to closely examine the assumptions made for deconvolution – Glen_b Jan 27 '20 at 00:43
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    There's nothing mysterious about deconvolution: it's carried out exactly as convolution is. The root of the problem is that given the (univariate) distributions of a random variable $X$ and a random variable $Z=X+Y,$ you cannot use the standard formula for deconvolution to recover $Y$ because $X$ and $Z$ (obviously) are not independent. Any formula would be tantamount to knowing the copula between $X$ and $Z,$ which is a very strong assumption. – whuber Jan 27 '20 at 14:48
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    If you fit the data with the convolution of a normal distribution and a gamma distribution as per answer [here](https://stats.stackexchange.com/a/330859/99274), you will indeed recover the parameters you desire. The problem with deconvolution is that it is very noisy. There are ways to diminish the noise, and the simplest is to fit the convolution itself, if that is known, and in this case, it is known. – Carl Jul 03 '20 at 08:48

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