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Convolution of two Gaussians

  1. May 13, 2008 #1
    I've read on a bunch of websites that the convolution of two gaussians produces another gaussian however I have not seen this integration worked out. I am working on an integral which has a similar form as this convolution so it would be a great help too see. Does anyone know a book or website or wants to type out the answer for me? Thanks a lot.

    Nicholi.
     
  2. jcsd
  3. May 13, 2008 #2
    Could you not write out the convolution explicitly (even for two standard gaussians) ?
     
  4. May 13, 2008 #3
    int(exp(-x^2)*exp(-(r-x)^2))dx

    there is the limits are +/- infinity
     
  5. May 13, 2008 #4

    CompuChip

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    You can use another method for this:

    [tex]\exp\left[ -x^2 \right] \exp\left[ -(r-x)^2 \right] =
    \exp\left[ -2 x^2 + 2 r x - r^2 \right] =
    \exp\left[ -2 (x + r/2)^2 \right] \exp\left[ - r^2 / 2 \right]
    [/tex]
    where the last term does not depend on x and the first one is just another Gaussian, but centered around a different point.
     
  6. May 14, 2008 #5
    Thanks!
     
  7. Nov 22, 2010 #6
    This only shows c) here below.

    Another argument, besides direct computation, is by standard properties of Fourier transform:
    a) fourier transform of a gaussian is a gaussian
    b) fourier transform of convolution product = pointwise product of fourier transforms (up to choice of normalisation)
    c) pointwise product of gaussians is gaussian.
    d) inverse fourier transform of a gaussian is gaussian.
     
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