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

From: Spencer Graves <spencer.graves_at_pdf.com>
Date: Thu 07 Jul 2005 - 12:04:26 EST

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:

```>>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
>>>
>>>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
```--
Spencer Graves, PhD
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