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Convolution of two Gaussians

  • Thread starter nshiell
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  • #1
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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.
 

Answers and Replies

  • #2
Does anyone know a book or website or wants to type out the answer for me?
Nicholi.
Could you not write out the convolution explicitly (even for two standard gaussians) ?
 
  • #3
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int(exp(-x^2)*exp(-(r-x)^2))dx

there is the limits are +/- infinity
 
  • #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.
 
  • #5
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Thanks!
 
  • #6
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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.
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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