Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
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
3  
Surely there exists a title which is more descriptive? –  Mariano Suárez-Alvarez Mar 16 '10 at 15:23
1  
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
2  
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

3 Answers 3

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?

share|improve this answer
    
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?

share|improve this answer
    
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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.