Gaussian Function Monoid

The Gaussian distribution is given by

$f(u,\sigma,x)=\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-u)^2}{2\sigma^2}}$

Where $u$ is the mean and $\sigma$ is the standard deviation. I can show numerically that

$\lim_{d\to 0}\frac{1}{d}\sum_{n=-\infty}^{\infty} f(u_1,\sigma_1,\frac{n}{d})\cdot f(u_2,\sigma_2,x-\frac{n}{d})=f(u_1 + u_2,\sqrt{\sigma_1^2 + \sigma_2^2},x)$

I'd love to see an analytic proof of this, but frankly I haven't been able to find a good way to approach it.

Using this definition we can show that the Gaussian distribution forms a monoid with the associative binary operation $\oplus$ defined as

$f(u_1,\sigma_1,x)\oplus f(u_2,\sigma_2,x) = f(u_1 + u_2,\sqrt{\sigma_1^2 + \sigma_2^2},x)$

With $u,\sigma, x\in\mathbb{R}$

And the identity is given by

$I=\lim_{\sigma\to 0}f(0,\sigma,x)=\delta(x)$

Where $\delta(x)$ is the Dirac delta function.

#Algebra

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Comments

The above sum can be fairly easily converted to an improper integral which shows that the sum is correct.

Alex Tremayne - 2 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$