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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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Hint: what's the Fourier transform of the convolution of two functions? –  Zarrax Jan 23 '11 at 20:33
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The first assertion is not true. –  Shai Covo Jan 23 '11 at 20:49
    
@Shai: Yes it is –  Amit Jan 23 '11 at 21:28
    
@Amit: No (at least, for independent random variables). Why do you think it is? –  Shai Covo Jan 23 '11 at 21:32
    
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. –  Amit Jan 23 '11 at 22:04

4 Answers 4

up vote 2 down vote accepted

Fourier Transform will help you out to conclude that the convolution is also a gaussian.

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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? –  Amit Jan 23 '11 at 20:55
    
yes :) [ some extra characters to reach the minimum ] –  Zarrax Jan 23 '11 at 21:01
    
@Mait: Yes that is the best way out. –  user17762 Jan 23 '11 at 21:23
  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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I think this pdf file can help you.

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See this for two common alternatives.

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That link is great. Thank you. –  Amit Jan 23 '11 at 21:08

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