Distribution of X+YX−Y\frac{X+Y}{X-Y}

It is given that X and Y are independent Gaussian random variables with mean 0 and variance μ\mu.

The distribution of X+YX−Y\frac{X+Y}{X-Y} is asked.

Z = X+Y
W = X-Y

Then Z and W are also independent ( because cov(Z,W) = 0 ).
How to proceed further?



2 Answers


The distribution of X+YX+Y is N(0,2μ)\mathcal N(0,2\mu) and the distribution of X−YX-Y is the same N(0,2μ)\mathcal N(0,2\mu). As you pointed out, X+YX+Y and X−YX-Y are independent, so we have a ratio of two independent normal random variables.

When UU and VV are two independent normally distributed random variables with expected value 00 and variance 11, then the ratio U/VU/V has the standard Cauchy distribution.

We have that

Hence, the distribution of the ratio is the standard Cauchy distribution.



Isn’t there a √2\sqrt2 missing ?
– Yves Daoust
2 days ago



@YvesDaoust It is indeed. Thank you for pointing out my mistake!
– Cm7F7Bb
2 days ago



For other problems of this type, where one wonders sometimes how and why Cauchy distribution is present, have a look at the interesting paper (raco.cat/index.php/Questiio/article/viewFile/27030/200222):
– JeanMarie
2 days ago