I know that the product of two Gaussians is a Gaussian, and I know that the convolution of two Gaussians is also a Gaussian. I guess I was just wondering if there's a proof out there to show that the convolution of two Gaussians is a Gaussian.
Convolution of two Gaussians is a Gaussian
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$\begingroup$
probability
probability-distributions
normal-distribution
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0Hint: what's the Fourier transform of the convolution of two functions? – 2011-01-23
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2The first assertion is not true. – 2011-01-23
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0@Shai: Yes it is – 2011-01-23
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0@Amit: No (at least, for independent random variables). Why do you think it is? – 2011-01-23
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1As @sivaram suggested, taking the FT of both Gaussians, multiplying them, and IFTing the product yields the convolution of both Gaussians, which is a Gaussian in itself. That means that the FT of any Gaussian is a Gaussian (true), and that the product of both FTs (which are both Gaussians) is also a Gaussian, therefore, the product of any Gaussian is a Gaussian. – 2011-01-23
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0@Shai: And if you don't believe me, here's a link: https://ccrma.stanford.edu/~jos/sasp/Gaussians_Closed_Convolution.html – 2011-01-23
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0@Amit: I meant that it's not true for random variables. That is, if $X$ and $Y$ are independent normal rv's, then $XY$ is not a normal random variable. See, for example, http://mathworld.wolfram.com/NormalProductDistribution.html – 2011-01-23
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2@Shai Covo: But "Gaussian" by itself just means the function, not random variables distributed according to the function. – 2011-04-29
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0@wnoise: You are quite right. However, this is a source for confusion: http://newsgroups.derkeiler.com/Archive/Comp/comp.soft-sys.matlab/2009-03/msg04203.html – 2011-05-01
5 Answers
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Fourier Transform will help you out to conclude that the convolution is also a gaussian.
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1I know that the product of the two FT of Gaussians is also a gaussian, and that is also equivalent to the FT of the convolution of two gaussians. Do you think, to show that the convolution of two Gaussians is a gaussian, it would be easiest to take the FT of both, multiply, and take the IFT of the product? – 2011-01-23
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1yes :) [ some extra characters to reach the minimum ] – 2011-01-23
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0@Mait: Yes that is the best way out. – 2011-01-23
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- the Fourier transform (FT) of a Gaussian is also a Gaussian
- The convolution in frequency domain (FT domain) transforms into a simple product
- then taking the FT of 2 Gaussians individually, then making the product you get a (scaled) Gaussian and finally taking the inverse FT you get the Gaussian
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I think this pdf file can help you.
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There is a proof for product of multivariate Gaussian PDFs in here. Maybe this can help: http://www.tina-vision.net/docs/memos/2003-003.pdf
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0It would be good of you give a synopsis of the proof here and then provide a link. – 2015-10-09