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If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where $x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$ respectively, are $X$ & $Y$ independent?

I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?

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It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing \begin{align*} x_1&=\{\varrho \sin(\theta)\}^2\\ x_1&=\{\varrho \cos(\theta)\}^2\\ \end{align*} we get that $$X=\frac{\{\varrho \sin(\theta)\}^2}{\{\varrho \sin(\theta)\}^2+\{\varrho \cos(\theta)\}^2}=\sin(\theta)^2$$ and $$Y=\frac{\varrho^2}{\varrho^2+x_3}$$ are indeed functions of different variates (although the independence between $\varrho$ and $\theta$ has to be established).

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A geometrical interpretation/intuition

You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points

In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.

The distribution of $\frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $\frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".


You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $\sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $\sqrt{x_1+x_2}$. The result should only depend on the fraction $X=\frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.

See here a similar (but much simpler) calculation.

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