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# Convolution of Two Gaussians is a Gaussian

Posted on 2005-05-17
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Hey, I'm looking for a proof of this

int G(y-a,sig1)G(y-b,sig2)dy = G(a-b,sig1+sig2)

where G(y-x,sig) is a gaussian with covariance matrix sig and vector y-x.

Or in words, the proof that the convolution of two gaussians centered at a and b is a gaussian centered on a-b with covariance equal to sig1 + sig2
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Question by:DanJW

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Accepted Solution

This is a long messy proof  and an amazing result, the proof requires a Fourier transdform and the convolution theorem.  This is the best I could find

http://www.phys.psu.edu/~lezon/gaussians.html
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LVL 3

Assisted Solution

Dan,

More succinctly: as in Gwyn's link, we will use the convolution theorem of the Fourier transform, which says

FT{f*g} = FT{f} FT{g}

where * denotes the convolution operation, and f and g are arbitrary functions.

First I'll show you how it goes in one dimension:

The Fourier transform of a Gaussian is also Gaussian (but with the inverse variance), i.e.

FT{exp(-x^2/(2 sigma^2))} = exp(-2 sigma^2 q^2)

where q is the conjugated variable with respect to x in Fourier space.

FT{G1 * G2} = FT{G1} FT{G2}
= exp(-2 sigma1^2 q^2) exp(-2 sigma2^2 q^2)
= exp(-2 (sigma1^2 + sigma^2) q^2)
= FT{ exp(-x^2/(2 (sigma1^2+sigma2^2)) }

Now we Fourier transform back on both sides and we are done.

For higher dimensions:  You can always get a diagonal covariance matrix by making a linear transformation of the coordinates.  Then you can factor into one-dimensional Gaussians and use the above approach.

mathbiol
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