Convolution of Two Gaussians is a Gaussian

Posted on 2005-05-17
Last Modified: 2012-08-13
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
Question by:DanJW
    LVL 31

    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
    LVL 3

    Assisted Solution


    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.


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