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Topic: Expectation of the variance
Replies: 2   Last Post: Mar 23, 2011 6:28 PM

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bert

Posts: 280
Registered: 1/4/06
Re: Expectation of the variance
Posted: Mar 23, 2011 5:58 PM
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On Mar 23, 9:16 pm, Steven D'Aprano <steve
+comp.lang.pyt...@pearwood.info> wrote:
> I'm trying to demonstrate numerically (rather than algebraically) that
> the expectation of the sample variance is the population variance, but
> it's not working for me.
>
> Some quick(?) background... please correct me if I'm wrong about anything.
>
> The variance of a population is:
>
> ?^2 = 1/n * ?(x-?)^2 over all x in the population
>
> where ^2 means superscript 2 (i.e. squared). In case you can't read the
> symbols, here it is again in ASCII-only text:
>
> theta^2 = 1/n * SUM( (x-mu)^2 )
>
> If you don't have the entire population as your data, you can estimate
> the population variance by calculating a sample variance:
>
> s'^2 = 1/n * ?(x-?)^2 over all x in the sample
>
> where s' is being used instead of s subscript n.
>
> This is unbiased, provided you know the population mean mu ?. Normally
> you don't though, and you're reduced to estimating it from your sample:
>
> s'^2 = 1/n * ?(x-m)^2
>
> where m is being used as the symbol for sample mean x bar = ?x/n
>
> Unfortunately this sample variance is biased, so the "unbiased sample
> variance" is used instead:
>
> s^2 = 1/(n-1) * ?(x-m)^2
>
> What makes this unbiased is that the expected value of the sample
> variances equals the true population variance. E.g. see
>
> http://en.wikipedia.org/wiki/Bessel's_correction
>
> The algebra convinces me -- I'm sure it's correct. But I'd like an easy
> example I can show people, but it's not working for me!
>
> Let's start with a population of: [1, 2, 3, 4]. The true mean is 2.5 and  
> the true (population) variance is 1.25.
>
> All possible samples for each sample size > 1, and their exact sample
> variances, are:
>
> n = 2
> 1,2 : 1/2
> 1,3 : 2
> 1,4 : 9/2
> 2,3 : 1/2
> 2,4 : 2
> 3,4 : 1/2
> Expectation for n=2: 5/3
>
> n=3
> 1,2,3 : 1
> 1,3,4 : 7/3
> 2,3,4 : 1
> Expectation for n=3: 13/9
>
> n=4
> 1,2,3,4 : 5/3
> Expectation for n=4: 5/3
>
> As you can see, none of the expectations for a particular sample size are
> equal to the population variance. If I instead add up all ten possible
> sample variances, and divide by ten, I get 1.6 which is still not equal
> to 1.25.
>
> What am I misunderstanding?


The formulae are correct only for a population with
a Gaussian distribution. The distribution of your
test population [1, 2, 3, 4] is not Gaussian, and
its difference from normality is enough to give
those differences in the sample variances.
--




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