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Help with proof of sample variance

by fadecomic
Tags: proof, sample, variance
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fadecomic
#1
Mar18-09, 02:47 PM
P: 10
Hi,

I'm trying to prove to myself that the formulation of sample variance as [tex]s^2=\frac{1}{n-1}\sum_{i=1}^{n}(y_i - \mu_y)^2[/tex] is an unbiased estimator of the population variance [tex]\sigma^2[/tex]. Of course, I proceed by checking the expected value of the sample variance, which flows smoothly until I get to [tex]\frac{n-1}{n}E[s^2] = E[y_i^2]-E[\mu_y^2][/tex]. Fine. Now, from the definition of variance: [tex]V[y_i^2]=E[y_i^2]-(E[y_i])^2[/tex]. Most of the sources I've checked insert the population variance here for [tex]V[y_i][/tex] (which is what you get to if you sub in the above). That make absolutely no sense to me. Shouldn't this be the sample variance since [tex]y_i[/tex] is the sample? Calling it the population variance seems circular to me. Can someone explain to me why this variance is the population variance and not the sample variance?

Thanks.
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fadecomic
#2
Mar18-09, 03:02 PM
P: 10
Oh, never mind. I figured it out. Above, y_i is a random draw from the population, not the sample, so by definition, its variance is the population variance, not the sample variance.


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