Convolution of two Gaussians is a Gaussian

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.

• Hint: what's the Fourier transform of the convolution of two functions? – Zarrax Jan 23 '11 at 20:33
• 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
• 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
• @Shai Covo: But "Gaussian" by itself just means the function, not random variables distributed according to the function. – wnoise Apr 29 '11 at 20:15

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

• 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

See this for two common alternatives.

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