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?
>
As far as I can see, this is correct.
> If not, is there some reason this case
>cannot be extended the product of arbitrary random variables X, Y, and
>W=1/Z?
>
>
Yes. Random variables are independent if all events which can be defined
in terms of them are independent.
If Z is non-constant, it must be some event defined by Z with
probability strictly beteween 0 and 1
and the above argument cannot be used.
Kjetil
> Thanks,
> spencer graves
>
>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
>>
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>>
>
>
>
--
Kjetil Halvorsen.
Peace is the most effective weapon of mass construction.
-- Mahdi Elmandjra
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