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
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