Re: [R] Lack of independence in anova()

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

      + q*P(X/Z <= v | Z = -A)*P(Y/Z <= w | Z = -A)

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
>> 
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>
> --
> Spencer Graves, PhD
> Senior Development Engineer
> PDF Solutions, Inc.
> 333 West San Carlos Street Suite 700
> San Jose, CA 95110, USA
>
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>
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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
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