0
$\begingroup$

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

$$Q=\frac{X+Y}{|X-Y|}?$$

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:

$$R=\frac{(X+Y)^2}{(X-Y)^2}?$$

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 $R$?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$X+Y$ 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.

$\endgroup$
  • $\begingroup$ 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

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.