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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$\begingroup$ I presume that you meant independent gaussian RVs? $\endgroup$– Yemon ChoiCommented Mar 16, 2010 at 9:59
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$\begingroup$ Yes, the random variables are independent. $\endgroup$– user4606Commented Mar 16, 2010 at 14:57
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3$\begingroup$ Surely there exists a title which is more descriptive? $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 16, 2010 at 15:23
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1$\begingroup$ 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? $\endgroup$– Yemon ChoiCommented Mar 17, 2010 at 23:22
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2$\begingroup$ 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. $\endgroup$– fedjaCommented Apr 16, 2010 at 14:06
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3 Answers
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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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$\begingroup$ 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? $\endgroup$– user4606Commented Mar 21, 2010 at 5:47
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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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$\begingroup$ 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? $\endgroup$– user4606Commented Mar 21, 2010 at 5:53
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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.)