$X_{1}$ and $X_{2}$ are independent, chi-square distributed random variables. I've managed to show that $Z=X_{1}+X_{2}$ is independent of $\frac{X_{1}}{Z}$. How do I conclude then that $Z=X_{1}+X_{2}$ is independent of $(\frac{X_{1}}{Z},\frac{X_{2}}{Z})$?
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$\begingroup$ Are you sure these assertions are true? $\endgroup$– Michael R. ChernickOct 1, 2012 at 1:19
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$\begingroup$ i doubt it. Can you show me why? $\endgroup$– Michael R. ChernickOct 1, 2012 at 1:25
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$\begingroup$ Is it due to the fact that $X_{2} = Z - X_{1}$? $\endgroup$– Ken DunnOct 1, 2012 at 1:27
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$\begingroup$ I don't see why Z would be independent of X1/Z since Z and X1 are clearly dependent and Z is negatively correlated to 1/Z. Before you try to show that Z is independent of X2/Z how did you manage to show Z and X1/Z are independent? $\endgroup$– Michael R. ChernickOct 1, 2012 at 1:43
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1$\begingroup$ Okay so then if I assume you that you worked that out for the chi square distributions then as you pointed out since X2=Z-X1 X2/Z=1-X1/Z which is just a constant minus a random variable that is independent of Z and hence is independent of Z. $\endgroup$– Michael R. ChernickOct 1, 2012 at 1:57
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1 Answer
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We have $$\left(\frac{X_1}Z,\frac{X_2}Z\right)=\left(\frac{X_1}{X_1+X_2},\frac{X_2}{X_1+X_2}\right)=(0,1)+\left(\frac{X_1}{X_1+X_2},-\frac{X_1}{X_1+X_2}\right),$$ hence you just have to show that $Z$ is independent of $\frac{X_1}Z(1,-1)$, which is the case by what you showed.
answered Sep 30, 2012 at 19:37