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