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.

If

Z = X+Y

W = X-Y

Then Z and W are also independent ( because cov(Z,W) = 0 ).

How to proceed further?

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2 Answers

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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

X+YX−Y=(2μ)−1/2(X+Y)(2μ)−1/2(X−Y).

\frac{X+Y}{X-Y}=\frac{(2\mu)^{-1/2}(X+Y)}{(2\mu)^{-1/2}(X-Y)}.

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

P(X+YX−Y