jim clark wrote:
> Hi
>
> On Fri, 26 Mar 2004, David Jones wrote:
>
>> jim clark wrote:
>>> No, the two ways of computing r^2 should agree, no matter how
>>> many predictors you have.
>>
>> Unfortunately/importantly, this is not true. They only agree if the
>> predictor is of a specific form:
>> Y2 = a + b*Y3
>> where Y3 is another predictor, and a and b are effectively fitted
by
>> least squares (possibly at the same time Y3 is fitted). If the form
>> of the predictor Y2 is not such as to have the "free" parameters a
>> and b of this type, then Pearson's r^2 will give a different answer
>> because it essentially does allow for these extra free parameters
>> (because it is the correlation).
>>
>> If you do use Pearson's r^2 as a measure of "fit" you are
>> effectively allowing for further adjustment of your predictor,
beyond
>> what you originally fitted ... thus this r^2 (correlation) will be
at
>> least as large as R^2 (calculated from the errors).
>
> Below is what I wrote in its entirety. I would really like to
> see an example where the correlation between the original y and
> the predicted y does NOT equal the multiple R defined as
> sqrt(SSpred/SStotal).
>
> -----------------------------------------------------------
> No, the two ways of computing r^2 should agree, no matter how
> many predictors you have. Just to be clear,
>
> if y^ = b0 + b1*x1 + b2*x2 ....
>
> , where one or more xs could be polynomial predictors,
>
> then
>
> R^2 = SSy^ / SSy (just another variation of your formula)
>
> will equal r(yy^)^2
> -----------------------------------------------------------
Well, the original question mentioned explicitly the case of
non-linear least squares,
for example
y^= exp(a*X1)+ exp(b*X2)
... but the problem also arises in the case of a linear model fitted
by least squares, where the intercept term is omitted ("forcing line
through the origin").
David Jones
.
.
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