I was working on an explicit example for a student of mine, and I ran into a bump.
Assume that the population is $\{1,2,4\}$. I draw samples of size 2 (without replacement) from the population. From the samples, I compute the sample variance of each. This gives me the following:
{1,2}-->$s_1^2=\frac{(1-1.5)^2+(2-1.5)^2}{2-1}=0.5$.
{1,4}-->$s_2^2=\frac{(1-2.5)^2+(4-2.5)^2}{2-1}=4.5$.
{2,4}-->$s_3^2=\frac{(2-3)^2+(4-3)^2}{2-1}=2$.
These should be unbiased estimators of the population variance.
It seems as though the computation for the expectation of the sample variances should be $\mathbb{E}[s^2]=\frac{0.5+4.5+2}{3}=\frac{7}{3}$.
It also seems as though population variance should be $\frac{(1-7/3)^2+(2-7/3)^2+(4-7/3)^2}{3}=\frac{14}{9}$.
Why aren't these two values the same? I thought they would be because sample variance is an unbiased estimator of population variance.
Where am I going wrong?
I did notice that if I divide the sum in the population variance computation by 2 (instead of 3), I do get the value from the sample variance computation; however, I don't have a good explanation for why this should be so (besides the fact that the arithmetic works).
