# Convolution of two Gaussians

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.

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) ?

int(exp(-x^2)*exp(-(r-x)^2))dx

there is the limits are +/- infinity

CompuChip
Homework Helper
You can use another method for this:

$$\exp\left[ -x^2 \right] \exp\left[ -(r-x)^2 \right] = \exp\left[ -2 x^2 + 2 r x - r^2 \right] = \exp\left[ -2 (x + r/2)^2 \right] \exp\left[ - r^2 / 2 \right]$$
where the last term does not depend on x and the first one is just another Gaussian, but centered around a different point.

Thanks!

You can use another method for this:

$$\exp\left[ -x^2 \right] \exp\left[ -(r-x)^2 \right] = \exp\left[ -2 x^2 + 2 r x - r^2 \right] = \exp\left[ -2 (x + r/2)^2 \right] \exp\left[ - r^2 / 2 \right]$$
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.