Let both $X$ and $Y$ be independent and standard normal distributed random variables with mean $0$ and variance $1$. What is the distribution of:


I know that both $X+Y$ and $X-Y$ have the same distribution; they're both normal with mean $0$ and variance $2$. But I'm not sure how to describe the ratio of the two; furthermore, these normal random variables that we're taking the ratio of aren't independent either, since they're both composed of $X$ and $Y$. Finally, I'm not sure how the absolute value changes things (it makes the denominator a "folded normal" distribution, but I'm not sure how to work with that in this context).

Similarly, what is the distribution of:


I end up with the ratio of two separate, non-independent chi square random variables with 1 degree of freedom. How do I describe the distribution of that?

So what are the distributions of $Q$ and $R$?

  • $\begingroup$ These ratios are (or are constant multiples of) t (df=1) and F (df 1 and 1) distributions, respectively. If you have recently studied them, then check definitions of these two distribution families. Crucial point is numerator and denom. are indep. $\endgroup$ – BruceET Feb 20 '18 at 18:24
  • $\begingroup$ I understand that, the problem was that I didn't realize the numerator was independent of the denominator. $\endgroup$ – jippyjoe4 Feb 21 '18 at 1:35
  • $\begingroup$ Yes, $X + Y$ and $X - Y$ are uncorrelated, and because they are jointly normal, that makes them independent. // Also, for a random sample from a normal population, sample mean $\bar X$ and sample variance $S^2$ are independent. // Maybe look here. $\endgroup$ – BruceET Feb 21 '18 at 4:01

$X+Y$ and $X-Y$ are jointly normal and uncorrelated, therefore independent. So the distribution of $(X+Y)/|X-Y|$ is the same as the distribution of $X/|Y|$; by symmetry this is also the distribution of $X/Y$. This happens to be the standard Cauchy distribution.

  • $\begingroup$ I hadn't realized that. So then I assume that makes the second distribution a Fisher distribution, right? $\endgroup$ – jippyjoe4 Feb 20 '18 at 1:59

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