Let be $x_{1},x_{2},x_{3}$ i.i.d. random variables following a normal distribution with $\mu=0$ and $\sigma=1$. I'm intrigued by the following random variable, which is a ratio of a cubic form and a quadratic form:
$$\frac{1}{\sqrt 2}\frac{x_1 x_2^2 + x_1 x_3^2 + x_2 x_1^2 +x_2 x_3^2 + x_3 x_1^2 + x_3 x_2^2 - 6x_1 x_2 x_3}{x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_1 x_3 - x_2 x_3 }$$
When sampling the variable, it looks very close to normal (1m samples, 100 bins):
I'm trying to understand how it can be so close, but the distribution of cubic forms in random variables isn't well documented or developed, and exact calculations for the ratio of forms seem to be very complicated, even in the case of two quadratic forms.
Is there any simple approach to see why the variable distribution might be close to a normal?
Cross with stats.SE
If we can show that the characteristic function of the distribution for $$ Q=\frac{1}{\sqrt 2}\frac{x_1 x_2^2 + x_1 x_3^2 + x_2 x_1^2 +x_2 x_3^2 + x_3 x_1^2 + x_3 x_2^2 - 6x_1 x_2 x_3}{x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_1 x_3 - x_2 x_3 } $$ is $\langle e^{it Q}\rangle= e^{-t^2/2}$, then $Q$ is normally distributed.
To do this, we come up with the change of variables (based on the eigenvectors of the quadratic form in the denominator) $$ \begin{align*} x_1&=\frac{1}{\sqrt{3}}z-\frac{1}{\sqrt 2} r\cos\theta -\frac{1}{\sqrt 6}r\sin\theta,\\ x_2&=\frac{1}{\sqrt{3}}z+\sqrt{\frac{2}{3}}r\sin\theta,\\ x_3&=\frac{1}{\sqrt{3}}z+\frac{1}{\sqrt 2} r\cos\theta -\frac{1}{\sqrt 6}r\sin\theta. \end{align*} $$ We then write up the integral for the characteristic function $$ \begin{align*} \langle e^{it Q}\rangle&= \int dx_1\,dx_2\,dx_3\frac{1}{(2\pi)^{3/2}}\exp\left[-\frac12(x_1^2+x_2^2+x_3^2)+itQ\right]\\ &=\int dz\, dr\, d\theta\, r \frac{1}{(2\pi)^{3/2}}\exp\left[-\frac12 (r^2+z^2)+it\left(\sqrt{\frac{2}{3}}z+\frac{1}{\sqrt{3}}r\sin3\theta\right)\right]. \end{align*} $$ In the $\theta$-integral, we are integrating over a whole period and so we can change $\sin3\theta\rightarrow\sin \theta$ for free. Then by changing the polar coordinates back to Cartesian as $$ \begin{align*} x&=r\sin\theta,\\ y&=r\cos\theta, \end{align*} $$ the integral becomes $$ \langle e^{it Q}\rangle= \int dx\,dy\,dz\frac{1}{(2\pi)^{3/2}}\exp\left[-\frac12(x^2+y^2+z^2)+it\left(\frac{1}{\sqrt{3}}y+\sqrt{\frac{2}{3}}z\right)\right]=e^{-t^2/2}. $$