On 6/29/07, John Randall <[EMAIL PROTECTED]> wrote:
Raul Miller wrote:
>> > I can determine \sum(X_i-\bar X)^2 for each of those
>> > potential sample cases (0, 0.5, 0.5, 0) and then
>> > average them. For the fair coin, this average is
>> > 0.25. For that unfair coin, I get 0.1875 for this
>> > average.
>>
>> This precisely illustrates the point. The population variance is 3%8,
>> not 3%16. You can see in this case that using denominator n=2
>> gives the wrong answer, using denominator n-1 gives the correct
>> answer.
>
> I am not sure this is a valid point.
Why? This shows that the method you are using does not produce an
unbiased estimator of variance, but mine does.
This terminology "biased" vs. "unbiased" does not appear to be relevant
when talking about population variance.
> See
>
http://en.wikipedia.org/wiki/Standard_deviation#Estimating_population_standard_deviation_from_sample_standard_deviation
> as an example treatment which seems to conflict with the assertions
> you have advanced in this last paragraph.
>
This gives a biased estimator.
I should instead have referred you to
http://en.wikipedia.org/wiki/Standard_deviation#Standard_deviation_of_a_random_variable
> As I understand the wiki write up, even if you consider denominator
> n=1 as valid for samples which do not represent the population,
> you must still use denominator n=2 when you are dealing with
> the population distribution.
What is a sample which does not represent the population?
The issue seems to be completeness.
The "unbiased" estimator traditionally gets used when dealing with
variance for a sample which is not identical to the population. In
other words, if there's any possibility that the distribution of the
sample is different from the distribution of the population, it seems
to be traditional to use the "unbiased estimator" (n/n-1 when determining
RMS deviation rather than the "biased estimator" of 1 when determining
RMS deviation).
--
Raul
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