My first reaction to Duncan's example was "Touché -- with apologies to Göran for suspecting on over-trivial example"! I had not thought long enough about possible cases. Duncan is right; and maybe it is the same example as Göran was thinking of.
Regarding Spencer's argument below, in Duncan's statement he says "Z is supported on +/- A" (i.e. Z = A or Z = -A), so P(|Z| < 1) = 0 and so Spencer's 1-2z = 0 and z=1/2 (but Spencer stipulates that Z is symmetric). In general, suppose P(Z = A) = p > 0 and P(Z = -A) = q = 1-p. Since X and Y are symmetric, X/A has the same distribution as X/(-A) and similarly for Y; hence for any v and w, P(X/Z <= v | X = z) is independent of z = +/- A, therefore = P(X/Z <= v); and similarly for Y. Also X/A, Y/A are independent, and so are X/(-A) and Y/(-A). Hence P(X/Z <= v and Y/Z <= w) = p*P(X/Z <= v | Z = A)*P(Y/Z <= w | Z = A) + q*P(X/Z <= v | Z = -A)*P(Y/Z <= w | Z = -A) = (p + q)*P(X/Z <= v)*P(Y/Z <= w) = P(X/Z <= v)*P(Y/Z <= w) so X/Z and Y/Z are independent. 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). The original issue also was that, in R, there might be a bug in anova(). However, one can, in R and independently of the behaviour of anova(), demonstrate this positive correlation: C<-numeric(10000); for(i in (1:10000)){ X<-rchisq(1000,5)/5 Y<-rchisq(1000,5)/5 Z<-rchisq(1000,20)/20 C[i]<-cor(X/Z,Y/Z) } hist(C) which shows that all 10000 correlations are positive. Best wishes to all, Ted. On 07-Jul-05 Spencer Graves wrote: > Hi, Duncan & Göran: > > Consider the following: X, Y, Z have symmetric distributions with > the following restrictions: > > P(X=1)=P(X=-1)=x with P(|X|<1)=0 so P(|X|>1)=1-2x. > P(Y=1)=P(Y=-1)=y with P(|Y|<1)=0 so P(|Y|>1)=1-2y. > P(Z=1)=P(Z=-1)=z with P(|Z|>1)=0 so P(|Z|<1)=1-2z. > > Then > > P(X/Z=1)=2xz, P(Y/Z=1)=2yz, and > P{(X/Z=1)&(Y/Z)=1}=2xyz. > > Independence requires that this last probability is 4xyz^2. This is > true only if z=0.5. If z<0.5, then X/Z and Y/Z are clearly dependent. > > How's this? > spencer graves > > Duncan Murdoch wrote: > >> (Ted Harding) wrote: >> >>>On 06-Jul-05 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 >>> >>> >>>But true if both X and Y have positive probability of being >>>non-zero, n'est-pas? >>> >>>Tut, tut, Göran! >> >> >> If X and Y are independent with symmetric distributions about zero, >> and >> Z is is supported on +/- A for some non-zero constant A, then X/Z and >> Y/Z are still independent. There are probably other special cases >> too. >> >> Duncan Murdoch >> >> ______________________________________________ >> 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 > > -- > Spencer Graves, PhD > Senior Development Engineer > PDF Solutions, Inc. > 333 West San Carlos Street Suite 700 > San Jose, CA 95110, USA > > [EMAIL PROTECTED] > www.pdf.com <http://www.pdf.com> > Tel: 408-938-4420 > Fax: 408-280-7915 > > ______________________________________________ > 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 -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 07-Jul-05 Time: 11:18:04 ------------------------------ XFMail ------------------------------ ______________________________________________ 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