Independence of Beta ratios of Gamma variates

If $$X= x_1/(x_1+x_2+x_3)$$ and $$Y= x_2/(x_1+x_2+x_3)$$ where $$x_1, x_2, x_3$$ are independent $$\chi^2$$-distributed random variables with d.f. $$-n_1,n_2, n_3$$ respectively. Are $$X$$ and $$Y$$ independent?

I am familiar with the usual joint distribution method to check for independence but I am looking for a shorter approach.

• @Jarle Tufto They both seem different to me. But I may be wrong. Would you mind having a go at it anyway? – Jor_El Jan 14 '19 at 10:48
• This is a duplicate of your question at stats.stackexchange.com/questions/386961/…. – whuber Jan 14 '19 at 15:46
• @whuber how will i establish independence in the above case when both X and Y have three variables in the denominator? – Jor_El Jan 14 '19 at 16:02
• @whuber Is it correct that it is a duplicate? – wolfies Jan 14 '19 at 16:31
• @Wolfies Good point -- I had read the first denominator as including only $x_1+x_2$ (as in the predecessor question), because otherwise the issue becomes trivial. – whuber Jan 14 '19 at 17:45

The question does not really have anything to do with the Chisquared distribution: it is about the relationship between the new ratios $$X$$ and $$Y$$ and whether that structure creates dependency. One can always easily check this oneself by generating some random values for each of the $$X_i$$, then generate $$X$$ and $$Y$$, and then plot $$X$$ vs $$Y$$.

Or better if you wish to think about it ...

With 2 variables:

If $$X = \frac{X_1}{X_1+X_2}$$ and $$Y = \frac{X_2}{X_1+X_2}$$, then $$X$$ and $$Y$$ are two parts of the same ratio that must sum to 1 (i.e. $$X+Y=1$$), and the relationship between $$X$$ and $$Y$$ could be plotted as:

With 3 variables:

With the 3rd variable $$X_3$$ added, we have $$X = \frac{X_1}{X_1+X_2 + X_3}$$ and $$Y = \frac{X_2}{X_1+X_2+X_3}$$. If we define a third ratio $$Z = \frac{X_3}{X_1+X_2 + X_3}$$, then $$X + Y + Z = 1$$ (i.e. that $$X + Y <1$$) and thus that the domain of support for $$X$$ and $$Y$$ could be plotted as the shaded space below the line:

If you were to generate pseudorandom data for $$X$$ and $$Y$$ and plot it, it would look something like this:

• Thanks for replying. But I was looking for a rigorous solution. – Jor_El Jan 14 '19 at 7:38
• What part of the above is not rigorous? – wolfies Jan 14 '19 at 18:28
• I meant a mathematically rigorous solution like we do in a typical mathematical statistics problem. – Jor_El Jan 14 '19 at 18:55
• "Like we do in a typical mathematical statistics problem" is not a useful description. Any proof using the standards of, say, Principia Mathematica would be millions of lines long! In my experience, most such problems--like most mathematics problems generally--are not solved in any fully rigorous manner. Indeed "rigor" is a matter of determining what elements of a mathematical explanation you expect your audience to understand (or supply themselves) and what elements you need to be explicit about. Sometimes, just a picture or diagram can be more "rigorous" than any amount of "proving." – whuber Jan 15 '19 at 13:22

If$$X= x_1/(x_1+x_2+x_3)\quad\text{and}\quad Y= x_2/(x_1+x_2+x_3)\qquad x_i\stackrel{\text{ind}}{\sim}\chi^2_{n_i}\quad i=1,2,3$$then, as shown by whuber $$(X,Y,1-X-Y)\sim\text{Dir}_3\left(n_1,n_2,n_3\right)$$a Dirichlet distribution with density $$\frac{\Gamma(n_1+n_2+n_3)}{\Gamma(n_1)\Gamma(n_2)\Gamma(n_3)}x^{n_1-1}y^{n_2-1}(1-x-y)^{n_3-1}$$ Therefore the density of $$(X,Y)$$ does not separate into a function of $$x$$ and a function of $$y$$, ergo, they are dependent. Of course, a simpler and correct argument is that, since $$0\le X+Y\le 1$$ the supports of $$X$$ given $$Y=y$$ is $$(0,1-y)$$ hence depends on the realised value of $$Y$$.