Gaussians are making Gaussians

Let f and g be Gaussian distributions.

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Go ahead, add them. You don’t get another Gaussian distribution.

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Well, of course not. They don’t have the same mean. So set the means equal.

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That’s no better. The sum of f and g is still very much not Gaussian.

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So, that’s no good. But of course it failed — just look at those visuals!

What about multiplication? Here’s what the product of two Gaussian distributions with equal means looks like.

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That looks much better!

In fact this is true: the product of two Gaussians distributions remains a Gaussian function. The only proofs I know of dive into some algebra — I like this one — but the core idea is that multiplying exponents is additive. That’s what keeps it all in the Gaussian family.

So consider two Gaussian functions, one with a mean \mu and the other with a mean at 0 (for a touch of simplicity):

f(x) = \frac{1}{\sqrt{2\pi}\sigma_f} e^{\frac{x^2}{2\sigma^2_f}}

g(x) = \frac{1}{\sqrt{2\pi}\sigma_g}  e^{\frac{(x -\mu)^2}{2\sigma^2_g}}

Their product will look like this:

f(x)g(x) = \frac{1}{2\pi\sigma_f\sigma_g} e^{\frac{x^2}{2\sigma^2_f}+\frac{(x-\mu)^2}{2\sigma^2_g}}

Making common denominators and adding through:

f(x)g(x) = \frac{1}{2\pi\sigma_f\sigma_g} e^{\frac{\sigma^2_g x^2 +\sigma^2_f(x -\mu)^2}{2\sigma^2_f\sigma^2_g}}

Might as well expand that exponent a bit and summarize:

f(x)g(x) = \frac{1}{2\pi\sigma_f\sigma_g} e^{\frac{(\sigma^2_f +\sigma^2_g) x^2 -2\sigma^2_f x \mu + \sigma^2_f \mu^2}{2\sigma^2_f\sigma^2_g}}

And then you can divide the numerator and denominator by (\sigma^2_f +\sigma^2_g) and you’ll end up with a quadratic trinomial in x. You can always express that quadratic trinomial as (x - M)^2 somehow or another.

(Brief but important nit-picky note: that would make the product of two Gaussian function, but the scale factor on the left side of the expression is off, so it’s not a Gaussian distribution. You’d have to scale the product of two Gaussian distribution in order to get another Gaussian distribution.)

Is this useful? Is this significant, in some way? I don’t know. Apparently it’s useful in applying Bayes’ Theorem, but I know nothing about that.

One thing I do know is that it makes for some fun visuals.

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One thought on “Gaussians are making Gaussians”

  1. This made me wonder if the set of Gaussian distributions under multiplication form a group. Is there an identity Gaussian e that doesn’t change any other distribution when multiplied by it? For any Gaussian distribution f, is there another Gaussian distribution g such that f × g = e? It looks like multiplication of Gaussian distributions is commutative and associative.

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