Given the pdf of Z = X+Y, what's the pdf of X if X&Y are IID

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In summary, the common problem in statistics is finding the pdf h(z) of Z = X + Y given the pdfs of X and Y. The solution is the convolution of the pdfs. The reverse problem, finding the pdf f(x) of X and Y given the known pdf h(z) of Z, can be solved using the Fourier transform or by calculating all the moments of the distribution. Additionally, there are ways to suppress high frequencies in the discretized version of the distribution. There is a theorem that describes the process of reconstructing a pdf from its moments, which involves approximating the pdf as a polynomial and using the first N moments to increase the accuracy of the reconstruction.
  • #1
ecneicScience
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A common problem in statistics is, "Given that X is distributed according to the pdf f(x) and Y is distributed according to the pdf g(y), what is the pdf h(z) of Z = X + Y?" The answer is the convolution of f(x) and g(y):

$$ h(z)=(f*g)(z)=\int_{-\infty}^\infty f(z-t)g(t) dt = \int_{-\infty}^\infty f(t)g(z-t) dt $$

In my research I have stumbled across the reverse problem given that X and Y are IID (identically and independently distributed): "Given that the random variable Z = X+Y has a known pdf h(z), what is the pdf f(x) if it is known that X and Y are IID?"

Does anyone have insight to how this problem could be solved? Is there such a thing as a deconvolution operator that could help me? Thanks in advance.
 
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  • #2
If you know Z with arbitrary precision, a Fourier transform should help.
Or calculate all the moments of the distribution.
 
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  • #3
mfb said:
If you know Z with arbitrary precision, a Fourier transform should help.
Or calculate all the moments of the distribution.

Exactly! I never actually thought I'd use those convolution theorems I learned in my FT class!

We know
$$ h = f*g $$
Taking the Fourier transform
$$ \mathcal{F}\{f*g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{g\} $$
We know f = g, so
$$ \mathcal{F}\{f*g\} = \mathcal{F}\{f\}^2 $$
Rearranging,
$$ f = \mathcal{F}^{-1} \bigg\{ \sqrt{ \mathcal{F} \{ f*g \} } \bigg\} $$
Putting in h,
$$ f = \mathcal{F}^{-1} \bigg\{ \sqrt{ \mathcal{F} \{ h \} } \bigg\} $$

Thank you for that insight. Hopefully my data is not too sparse so that I can perform a decent FFT.

Could you please elaborate how calculating moments could help?
 
  • #4
If you know all moments, you can reconstruct the function, and you can find relations between the moments of Z and X.

A discretized version can always have problems with high frequencies, but I guess there are ways to suppress them.
 
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  • #5
mfb said:
If you know all moments, you can reconstruct the function

That is not necessarily true. But I guess you can assume that the distribution of the OP is nice enough for it to be true.
 
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  • #6
For all functions where you have some chance to reconstruct it experimentally ;).
 
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mfb said:
If you know all moments, you can reconstruct the function, and you can find relations between the moments of Z and X.

A discretized version can always have problems with high frequencies, but I guess there are ways to suppress them.

Is there a theorem you could refer me to that describes this process of constructing a pdf from its moments?
 
  • #9
mfb said:

That article didn't seem that helpful or it went way over my head. What I had in mind when you said a pdf could be reconstructed from its moments, is that the pdf could be approximated as some polynomial, where each term is somehow a function of the first N moments, and that the reconstruction of became increasingly accurate with increasing N. I couldn't find anything that mentioned this in that article.
 
  • #10
The references given there might help.

Checking Wikipedia, the references given there, and google results with similar keywords is really not magic, and something that could be expected before asking questions.
 
  • #11
mfb said:
Checking Wikipedia, the references given there, and google results with similar keywords is really not magic, and something that could be expected before asking questions.

Thanks for the tip.
 
  • #12
Can you just look at this as computing the probability of X and Y, eg their union?
 

1. What is the meaning of IID in this context?

IID stands for "independent and identically distributed", which means that the two random variables X and Y are independent of each other and follow the same probability distribution.

2. What is the relationship between the pdf of Z and the pdf of X?

The pdf (probability density function) of Z is a convolution of the pdfs of X and Y. This means that the pdf of Z is derived from the joint distribution of X and Y, and can be expressed as the integral of the product of the pdfs of X and Y.

3. How can the pdf of X be calculated from the pdf of Z?

The pdf of X can be calculated by taking the partial derivative of the pdf of Z with respect to X. This will result in the marginal distribution of X, which is the pdf of X when Y is integrated out.

4. Can the pdf of X be different from the pdf of Y?

Yes, the pdf of X can be different from the pdf of Y, even if they are IID. This is because the pdf of Z is a convolution of the two pdfs, so the shape of the resulting pdf of X can be affected by the shape of the pdf of Y.

5. What assumptions are necessary for the pdf of X to be calculated in this scenario?

To calculate the pdf of X, we assume that X and Y are IID, meaning they are independent and follow the same probability distribution. We also assume that the pdf of Z exists, which means that X and Y are continuous random variables.

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