From: Spencer Graves <spencer.graves_at_pdf.com>

Date: Thu 07 Jul 2005 - 12:04:26 EST

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> If X and Y are independent with symmetric distributions about zero, and

*> Z is is supported on +/- A for some non-zero constant A, then X/Z and
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*> Y/Z are still independent. There are probably other special cases too.
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*>
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*> Duncan Murdoch
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Date: Thu 07 Jul 2005 - 12:04:26 EST

Consider the following: X, Y, Z have symmetric distributions with the following restrictions:

P(X=1)=P(X=-1)=x with P(|X|<1)=0 so P(|X|>1)=1-2x. P(Y=1)=P(Y=-1)=y with P(|Y|<1)=0 so P(|Y|>1)=1-2y. P(Z=1)=P(Z=-1)=z with P(|Z|>1)=0 so P(|Z|<1)=1-2z. Then P(X/Z=1)=2xz, P(Y/Z=1)=2yz, and P{(X/Z=1)&(Y/Z)=1}=2xyz. Independence requires that this last probability is 4xyz^2. This istrue only if z=0.5. If z<0.5, then X/Z and Y/Z are clearly dependent.

How's this? spencer graves

>>On 06-Jul-05 Göran Broström wrote: >> >> >>>On Wed, Jul 06, 2005 at 10:06:45AM -0700, Thomas Lumley wrote: >>>(...) >>> >>> >>>>If X, Y, and Z are independent and Z takes on more than one >>>>value then X/Z and Y/Z can't be independent. >>> >>>Not really true. I can produce a counterexample on request >>>(admittedly quite trivial though). >>> >>>Göran Broström >> >> >>But true if both X and Y have positive probability of being >>non-zero, n'est-pas? >> >>Tut, tut, Göran!

> If X and Y are independent with symmetric distributions about zero, and

-- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA spencer.graves@pdf.com www.pdf.com <http://www.pdf.com> Tel: 408-938-4420 Fax: 408-280-7915 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.htmlReceived on Thu Jul 07 12:08:21 2005

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