Developing normal pdf from symmetry & independence

When I was in the 3rd grade of my middle school, I enjoyed my town bookstore as a standing library. There a series of six math-story books by Zhang Yuan-Nan impressed me a lot. I cited a case from one in my PPT when I taught the normal distribution -- the normal pdf can be derived from simple symmetry & independence conditions.

Today I can even google out an illegal pdf of its new edition to verify the case (2005, pp. 89). Actually I have bought the new edition series (now 3 books) and lent them to students. Those conditions are as instinctive as--

1. For white noise errors on 2-D, the independence means pdf at (x,y) is the product of 1-D pdf, that is, \phi\left(x\right)\phi\left(y\right) .

2. The symmetry means pdf at (x,y) is just a function of x^{2}+y^{2}, nothing to do with direction. That is, \phi\left(x\right)\phi\left(y\right)=f\left(x^{2}+y^{2}\right).

So, f\left(x^{2}\right)f\left(y^{2}\right)=f\left(x^{2}+0\right)f\left(0+y^{2}\right)=\phi^{2}\left(0\right)f\left(x^{2}+y^{2}\right).

For middle school students, the book stated a gap here to arrive at the final result f\left(x^{2}\right)=ke^{bx^{2}}, which is \phi\left(x\right)=\frac{1}{\phi\left(0\right)}f\left(x^{2}+0\right)=\frac{k}{\phi\left(0\right)}e^{bx^{2}}.

I think non-math graduate students with interests can close the gap by themselves with following small hints.

Denote \alpha=x^{2},\beta=y^{2}.
We have
\log f\left(\alpha\right)+\log f\left(\beta\right)=\log\phi^{2}\left(0\right)+\log f\left(\alpha+\beta\right),
\;\;\left[\log f\left(\alpha\right)-\log\phi^{2}\left(0\right)\right]+\left[\log f\left(\beta\right)-\log\phi^{2}\left(0\right)\right]   =\left(\log f\left(\alpha+\beta\right)-\log\phi^{2}\left(0\right)\right).
Denote g\left(\alpha\right)=\log f\left(\alpha\right)-\log\phi^{2}\left(0\right) .
That is, g\left(\alpha\right)+g\left(\beta\right)=g\left(\alpha+\beta\right).

Now to prove g\left(\frac{m}{n}\right)=\frac{m}{n}g\left(1\right),\forall m,n\in\mathbb{N}. With continuousness, it gets g\left(\alpha\right)=\alpha g\left(1\right),\forall\alpha\in\mathbb{R}.

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