From: Göran Broström <gb_at_stat.umu.se>

Date: Thu 07 Jul 2005 - 23:56:04 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 Fri Jul 08 00:09:25 2005

Date: Thu 07 Jul 2005 - 23:56:04 EST

On Thu, Jul 07, 2005 at 11:18:09AM +0100, Ted Harding wrote:

> My first reaction to Duncan's example was "Touché -- with apologies

*> to Göran for suspecting on over-trivial example"!
*

No need to apologize; that was of course my first reaction to Thomas' statement.

> I had not thought

*> long enough about possible cases. Duncan is right; and maybe it is
**> the same example as Göran was thinking of.
*

On second thought it was not difficult to find: (X, Y) bivariate standard normal, P(Z = 1) = P(Z = -1) = 1/2.

[...]

> However, interesting though it maybe, this is a side-issue

*> to the original question concerning independence of the F-ratios
**> in an ANOVA. Here, numerators and denominator are all positive,
**> so examples like the above are not relevant.
**>
**> The original argument (that increasing Z diminishes both X/Z
**> and Y/Z simultaneously) applies; but it is also possible to
**> demonstrate analytically that P(X/Z <= v and Y/Z <= w) is
**> greater than P(X/Z <= v)*P(Y/Z <= w).
*

Maybe it is simplest to calculate Cov(X/Z, Y/Z), which turns out to be equal to E(X)E(Y)V(1/Z) (given total independence). So, a necessary condition for independence is that at least one of these three terms is zero. Which is impossible in the F-ratios case.

Göran

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 Fri Jul 08 00:09:25 2005

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