Re: The sum of square deviations: A THEOREM
- From: "Kevin E. Thorpe" <kevin.thorpe@xxxxxxxxxxx>
- Date: 11 Jul 2006 17:49:18 -0700
I would have replied sooner, but with being on vacation and
spending the afternoon on the golf course, I am not online
all the time.
"Luis A. Afonso" wrote:
Since, apparently, you are willing to have a calm and polite discussion, I follow-up.
__1__The best way is not to take into account what is said in this thread and only to focussing the basis: the Theorem I
I derived my myself which is a very well known since at lot of time ago.
Notation
__S = sum of squares deviations from the Population mean
__S0 = idem from the sample mean (obtained from the Population.
____I show that S>=S0. (go to my Jul 10, 2006 4:55 AM and check it.)
Do not forget that the theorem concerns sum of squares deviations.
The problem from my perspective is defining the sum
of squared deviations relative to the POPULATION mean for
a SAMPLE from the POPULATION. So, sum[(xi - mu)^2]/N
is not really an estimate of the population variance. If you
substitute the sample mean for mu, you get the MLE (at least
for normally distributed data).
So, in my opinion, the comparison you are trying to make
does not make sense.
__5__ In conclusion
We should not, in any instances (IMO), make confusion
between the Sample Variance and the UNBIASED
ESTIMATE of the Population variance - are rather different things.
I find this statement odd. The SAMPLE variance s^2 IS the
unbiased estimate of the POPULATION variance.
--
Kevin E. Thorpe
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
.
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