Lack of independence in anova()

Hi, G??ran:  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

G??ran Brostr??m 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).

G??ran Brostr??m

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Lack of independence in anova()

Thread (27 messages)