if X and Y are Gauss distributed, what's the distribution of X^2-Y^2?

Question

X and Y are independent random variables with identical Gaussian distribution; for simplicity, the variance shall be 1. What's the distribution of Z=X^2+Y^2?

With a plus sign, it would be the chi-square distribution. The minus sign changes things completely, and makes the computation of the convolution cumbersome. Therefore I would appreciate a hint to relevant literature, or just the name of the joint distribution.

My own, machine-aided computation of the convolution integral leads to a lengthy expression that contains the modified Bessel function K_0. Does that ring something?

I assume X and Y are independent? – Nate Eldredge Dec 16 '13 at 16:48 — Nate Eldredge, Dec 16 '13 at 16:48
Yes, X and Y are independent. – Joachim W Dec 16 '13 at 16:56 — Joachim W, Dec 16 '13 at 16:56

André Nicolas · Answer 1 · 2013-12-16T17:24:37.963

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A Start: As an example, let $X$ and $Y$ be independent standard normal. Note that $X^2-Y^2=(X+Y)(X-Y)$ and $X+Y$ and $X-Y$ are independent normal. The distribution of the product is in the product normal family.

edited Dec 16 '13 at 17:24

answered Dec 16 '13 at 17:03

André Nicolas

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That's already enough to compute the variance, thanks! Beyond that, I do not yet see the advantage of going from a difference to a product to a sum of logarithms. – Joachim W Dec 16 '13 at 17:17
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No real advantage, one has to travel through the complex. I have deleted that part. My answer was meant to point to language under which your problem has been solved, there is a fair bit of literature on the product normal. – André Nicolas Dec 16 '13 at 17:26

if X and Y are Gauss distributed, what's the distribution of X^2-Y^2?

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