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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.

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  • $\begingroup$ Hint: what's the Fourier transform of the convolution of two functions? $\endgroup$
    – Zarrax
    Jan 23, 2011 at 20:33
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    $\begingroup$ The first assertion is not true. $\endgroup$
    – Shai Covo
    Jan 23, 2011 at 20:49
  • $\begingroup$ @Shai: Yes it is $\endgroup$
    – Amit
    Jan 23, 2011 at 21:28
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    $\begingroup$ As @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. $\endgroup$
    – Amit
    Jan 23, 2011 at 22:04
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    $\begingroup$ @Shai Covo: But "Gaussian" by itself just means the function, not random variables distributed according to the function. $\endgroup$
    – wnoise
    Apr 29, 2011 at 20:15

5 Answers 5

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  1. the Fourier transform (FT) of a Gaussian is also a Gaussian
  2. The convolution in frequency domain (FT domain) transforms into a simple product
  3. 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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Fourier Transform will help you out to conclude that the convolution is also a gaussian.

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    $\begingroup$ I 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? $\endgroup$
    – Amit
    Jan 23, 2011 at 20:55
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    $\begingroup$ yes :) [ some extra characters to reach the minimum ] $\endgroup$
    – Zarrax
    Jan 23, 2011 at 21:01
  • $\begingroup$ @Mait: Yes that is the best way out. $\endgroup$
    – user17762
    Jan 23, 2011 at 21:23
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See this for two common alternatives.

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  • $\begingroup$ That link is great. Thank you. $\endgroup$
    – Amit
    Jan 23, 2011 at 21:08
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I think this pdf file can help you.

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  • $\begingroup$ The link seems to be down. $\endgroup$
    – Ramanujan
    Jun 7, 2021 at 17:34
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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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  • $\begingroup$ It would be good of you give a synopsis of the proof here and then provide a link. $\endgroup$
    – Shailesh
    Oct 9, 2015 at 8:59

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