From: Ted Harding <Ted.Harding_at_nessie.mcc.ac.uk>

Date: Thu 07 Jul 2005 - 20:18:09 EST

Z<-rchisq(1000,20)/20

C[i]<-cor(X/Z,Y/Z)

}

hist(C)

*>
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> Spencer Graves, PhD

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E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Thu Jul 07 20:42:49 2005

Date: Thu 07 Jul 2005 - 20:18:09 EST

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.

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

E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> 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 Received on Thu Jul 07 20:42:49 2005

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