[R] Lack of independence in anova()

[Spencer Graves]

Hi, Duncan & Göran:

	  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 Göran Broström 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).
>>>
>>>Göran Broström
>>
>>
>>But true if both X  and Y have positive probability of being
>>non-zero, n'est-pas?
>>
>>Tut, tut, Göran!
> 
> 
> 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 at 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

spencer.graves at pdf.com
www.pdf.com <http://www.pdf.com>
Tel:  408-938-4420
Fax: 408-280-7915