From: Duncan Murdoch <murdoch_at_stats.uwo.ca>

Date: Thu 07 Jul 2005 - 11:10:29 EST

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.html Received on Thu Jul 07 11:14:17 2005

Date: Thu 07 Jul 2005 - 11:10:29 EST

Spencer Graves wrote:

> Hi, Göran: I'll bite:

*>
**> (a) I'd like to see your counterexample.
**>
**> (b) I'd like to know what is wrong with my the following, apparently
**> defective, proof that they can't be independent: First consider
**> indicator functions of independent events A, B, and C.
**>
**> P{(AC)&(BC)} = P{ABC} = PA*PB*PC.
**>
**> But P(AC)*P(BC) = PA*PB*(PC)^2. Thus, AC and BC can be independent
**> only if PC = 0 or 1, i.e., the indicator of C is constant almost surely.
**>
**> Is there a flaw in this?
*

I don't see one.

> If not, is there some reason this case

> cannot be extended the product of arbitrary random variables X, Y, and

*> W=1/Z?
*

Because you can't? The situations are different?

If C is a non-trivial event independent of A, then AC is strictly a subset of A. However, as the example I just posted shows (with constant 1), you can have a non-trivial random variable W where XW has exactly the same distribution as X.

Duncan Murdoch

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.html Received on Thu Jul 07 11:14:17 2005

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