# Gaussians are making Gaussians

Let $f$ and $g$ be Gaussian distributions. Go ahead, add them. You don’t get another Gaussian distribution. Well, of course not. They don’t have the same mean. So set the means equal. That’s no better. The sum of $f$ and $g$ is still very much not Gaussian. 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. 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. ## 1 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.