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Let $X_i$ denote $\chi_{r_i}^2$ i.i.d random variables, where $r_i$ is a (possibly distinct) positive integer for each $i$. I want to verify that $Y_1 = \frac{X_1}{X_2}$ and $Y_2 = X_1 + X_2$ are independent.

I know that there are several ways of showing this - one way would be to try to compute the joint pdf of $Y_1$ and $Y_2$ and marginalize one out and show that it is the product of both of them - but I'm having trouble writing the joint pdf. Another option would be to show that the conditional probabilities are equal to the individual probabilities. But this too requires the pdf and while I know that $Y_2$ is a chi-squared random variable, it seems like it would be messy to do it this way too.

Is there an more elegant way to show this?

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  • $\begingroup$ $r_i$ could be different for each i so $X_1$ and $X_2$ could have different parameters. I will update the question to clarify this. Sorry for the confusion! $\endgroup$ – user2668905 Sep 20 '17 at 4:27
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Let $X_1 = \xi_1^2 + \dots + \xi_n^2$, $X_2 = \xi_{n+1}^2 + \dots + \xi_{n+m}^2$, where $\xi_i$, $i=1,\dots,n+m$, are independent standard Gaussian variables. Since the distribution of $(\xi_1,\dots,\xi_{n+m})$ is radially symmetric, then, given $X_1 + X_2 = \xi_1^2 + \dots + \xi_{n+m}^2 = R$, the ratio $$\frac{X_1}{X_2} = \frac{X_1/R}{X_2/R}$$ has the same distribution as $$\frac{S_1^2 +\dots + S_n^2}{S_{n+1}^2 +\dots + S_{n+m}^2}, $$ where the vector $(S_1,\dots,S_{n+m})$ is uniformly distributed on a unit sphere in $\mathbb{R}^{n+m}$. So this distribution is independent of $R$, qed.

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  • $\begingroup$ Thanks. How can I see that $X_1 / R$ has the same distribution a $S_1 ^2 + ... + S_n ^2$? $\endgroup$ – user2668905 Sep 20 '17 at 5:33
  • $\begingroup$ @user2668905, the radial symmetry means that the vector $(\xi_1,\dots,\xi_{n+m})/\sqrt{R}$ is uniformly distributed on the unit sphere. $\endgroup$ – zhoraster Sep 20 '17 at 5:47
  • $\begingroup$ Thanks for the help! $\endgroup$ – user2668905 Sep 20 '17 at 5:49

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