Independence of ratios of independent variates 
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?
 A: 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).
A: 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.
