This question is related to my previous question which was partially answered my @MichaelHardy.

Let $X$ and $Y$ be two independent standard normal random variables. Now, suppose that $g:\mathbb{R}^2\mapsto \mathbb{R}$ is a symmetric smooth function such that $g(X,Y)$ is normally distributed. Intuitively, $g$ should be in the form $g(X,Y)=a(X+Y)+b$ for some $a,b\in \mathbb{R}$ since the other "algebraic operations" except sum will never give normal distribution again. However, how will we prove this conjecture precisely?

  • $\begingroup$ With symmetric you mean that $g(x,y)=g(y,x)$, right? $\endgroup$ – Bernhard Aug 3 '15 at 11:46
  • $\begingroup$ @Bernhard: Yes, exactly. $\endgroup$ – Jlamprong Aug 3 '15 at 11:47

I think your assertion is not correct. Take the following example:

The product $X\cdot Y$ has a product normal distribution with cumulative distribution function $F_{PN}$. Hence, as a consequence, $F_{PN}(X\cdot Y)$ has a uniform distribution by the Inverse Transform Sampling Method. Similarly, if $Z$ has normal distribution with cumulative distribution function $F_N$, then $F_N(Z)$ has a uniform distribution. Hence, $F_N^{-1}(F_{PN}(X\cdot Y))$ has normal distribution, too.


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