0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Jarle Tufto They both seem different to me. But I may be wrong. Would you mind having a go at it anyway? $\endgroup$ – Jor_El Jan 14 at 10:48
  • 1
    $\begingroup$ This is a duplicate of your question at stats.stackexchange.com/questions/386961/…. $\endgroup$ – whuber Jan 14 at 15:46
  • $\begingroup$ @whuber how will i establish independence in the above case when both X and Y have three variables in the denominator? $\endgroup$ – Jor_El Jan 14 at 16:02
  • $\begingroup$ @whuber Is it correct that it is a duplicate? $\endgroup$ – wolfies Jan 14 at 16:31
  • $\begingroup$ @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. $\endgroup$ – whuber Jan 14 at 17:45
2
$\begingroup$

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: enter image description here

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:

enter image description here

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

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for replying. But I was looking for a rigorous solution. $\endgroup$ – Jor_El Jan 14 at 7:38
  • 3
    $\begingroup$ What part of the above is not rigorous? $\endgroup$ – wolfies Jan 14 at 18:28
  • 1
    $\begingroup$ I meant a mathematically rigorous solution like we do in a typical mathematical statistics problem. $\endgroup$ – Jor_El Jan 14 at 18:55
  • 3
    $\begingroup$ "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." $\endgroup$ – whuber Jan 15 at 13:22
2
$\begingroup$

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 $Y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.