Understanding Hershel-Maxwell derivation of the normal distribution

Question

property 1: the probability density depends only on the distance from the origin

property 2: the $x$ and $y$ coordinates are independent from each other

let:

$$f(x,y) = f(r) = f(\sqrt{x^2+y^2}) = f(x)f(y)$$ let $h(x)$ be a helper function such that: $$h(x) = f(\sqrt{x}),$$ $$ h(x^2) = f(x)$$ then: $$h(x^2+y^2) = h(x^2)h(y^2)$$ $$h(x_1+...+x_n) = h(x_1)...h(x_n)$$ $$h(n) = b^n=e^{cx}$$ then: $$f(x) = e^{cx^2}$$ my questions are first why: $f(\sqrt{x^2+y^2}) = f(x)f(y)$ that propriety is true and

why the helper function also works with $n$ numbers? $h(x_1+...+x_n) = h(x_1)...h(x_n)$ since we are dealing with the equation of a circle?

Answer 1

You got confused because there is abuse of notation going on and even the first equality has a few steps missing. Here is a more rigorous presentation:

We seek a two-variable density function $f : \mathbb R^2 \to \mathbb R$ satisfying properties 1 and 2.

• property 1 rigorous version: "the probability density depends only on the distance from the origin" translates mathematically to: there is $g : \mathbb R \to \mathbb R$ such that $$ f(x, y) = g(\sqrt{x^2 + y^2}) $$ This is why $f(x, y) = f(\sqrt{x^2 + y^2})$ is abuse of notation; $f(\sqrt{x^2 + y^2})$ doesn't make sense because $f : \mathbb R^2 \to \mathbb R$ is a two-variable function whereas $f(\sqrt{x^2 + y^2})$ suggests it is a one-variable function.
• property 2 rigorous version: "the $x$ and $y$ coordinates are independent from each other" translates mathematically to: there are two $u : \mathbb R \to \mathbb R$, $v : \mathbb R \to \mathbb R$ such that $$ f(x, y) = u(x) v(y) $$ This is why $f(\sqrt{x^2 + y^2}) = f(x)f(y)$ is not only abuse of notation, it has several steps missing. Firstly, it should technically be $g(\sqrt{x^2 + y^2}) = cg(x)g(y)$ for some constant $c$. And secondly, you don't know a priori that $u(x)v(y) = cg(x)g(y)$. That is something you have to show and that is the correct version of your first question: why is $u(x)v(y) = cg(x)g(y)$?

Collecting all of these, your first equality should look like $$ f(x, y) = g(r) = g(\sqrt{x^2 + y^2}) = u(x)v(y) $$ and only now you can rigorously think about the problem.