# Distribution under operations

Let $X$, $Y$, $Z$, and $W$ be i.i.d. copies of a standard gaussian variable, that is in distribution $\mathcal{N}\left(0,1\right)$, then what is the distribution of $\left|\frac{XY}{Z}-W\right|$?

Thanks!

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I presume that you meant independent gaussian RVs? – Yemon Choi Mar 16 '10 at 9:59
Yes, the random variables are independent. – user4606 Mar 16 '10 at 14:57
Surely there exists a title which is more descriptive? – Mariano Suárez-Alvarez Mar 16 '10 at 15:23
Is there any reason to expect there to be a nice answer? Also, why do you want to know the actual distribution, rather than (say) estimates on the moments or on the tails? – Yemon Choi Mar 17 '10 at 23:22
A side remark. You do not need to use your real name, of course, but I decided to ignore all the posts from uknown(yahoo)'s on the ground that the failure to set up some unique user name before asking a question shows a clear disrespect to the community. If you want to communicate to people, you should identify yourself in some way. – fedja Apr 16 '10 at 14:06

I don't think there's any reason for it to have a nice distribution. After fooling around with it a bit, the following seem to be true:

*It has infinite variance (Not too hard to show, X/Y has infinite variance, and the other random variables just increase it)

*Pretty sure it has an infinite mean (Empirically, the sample-mean seems to be stationary, and theoretically, there's a bounding argument with a Cauchy that I'm not positive of.

Any specific questions about the distribution?

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Asking distribution might be too much. Instead, let's just consider the tail estimate, as Yemon suggested. Can the probality of $\left|\frac{XY}{Z}-W\right|>t$ be bounded by 1/t up to a constant? – user4606 Mar 21 '10 at 5:47

Yes, the mean is infinite.

(If it was finite, since $W$ is integrable, $XY/Z$ would be integrable. Since $XY$ and $Z$ are independent and $XY$ is not identically $0$, $1/Z$ would be integrable. But now $Z$ is Gaussian, its density around $0$ is equivalent to a positive constant, hence $1/Z$ is not integrable.)

Why is this random variable, or its distribution, interesting?

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I was just curious. The tail estimate, as Yemon suggested, might be interesting. Can the probality of $\left|\frac{XY}{Z}-W\right|>t$ be bounded by 1/t up to a constant? – user4606 Mar 21 '10 at 5:53

Yes it can: the probability that $|(XY/Z)-W|\ge t$ is bounded by $c/t+o(1/t)$ with $c^2=2/\pi^3$, when $t\to\infty$. (By the way, Yemon Choi DID NOT suggest that the tail estimate might be interesting, he asked why you asked. And got no answer.)

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