Several texts say that the mean of all values of an unbiased estimator is equal to the parameter it is estimating.

They also say that the sample variance using Bessel's correction is an unbiased estimator of the population variance.

Here's what I'm confused about:

Consider a population with N=3, let's say {3,5,13}.

The population mean is \(\frac{3+5+13}{3} = 7\)

The population variance is \(\frac{(3-7)^2+(5-7)^2+(13-7)^2}{3} = 18.6...\)

There are three possible samples of n=2: {3,5}, {5,13}, and {3,13}.

Sample mean of {3,5} is 4.

Sample variance using Bessel's correction is \(\frac{(3-4)^2+(5-4)^2}{2-1} = 2\)

Sample mean of {5,13} is 9.

Sample variance using Bessel's correction is \(\frac{(5-9)^2+(13-9)^2}{2-1} = 32\)

Sample mean of {3,13} is 8.

Sample variance using Bessel's correction is \(\frac{(3-8)^2+(13-8)^2}{2-1} = 50\)

But the mean of the sample variances is \(\frac{2+32+50}{3}=28\neq18.6...\)

Am I doing something wrong? If I'm calculating correctly, have I misunderstood what it means for an estimator to be unbiased? Any help would be appreciated.

They also say that the sample variance using Bessel's correction is an unbiased estimator of the population variance.

Here's what I'm confused about:

Consider a population with N=3, let's say {3,5,13}.

The population mean is \(\frac{3+5+13}{3} = 7\)

The population variance is \(\frac{(3-7)^2+(5-7)^2+(13-7)^2}{3} = 18.6...\)

There are three possible samples of n=2: {3,5}, {5,13}, and {3,13}.

Sample mean of {3,5} is 4.

Sample variance using Bessel's correction is \(\frac{(3-4)^2+(5-4)^2}{2-1} = 2\)

Sample mean of {5,13} is 9.

Sample variance using Bessel's correction is \(\frac{(5-9)^2+(13-9)^2}{2-1} = 32\)

Sample mean of {3,13} is 8.

Sample variance using Bessel's correction is \(\frac{(3-8)^2+(13-8)^2}{2-1} = 50\)

But the mean of the sample variances is \(\frac{2+32+50}{3}=28\neq18.6...\)

Am I doing something wrong? If I'm calculating correctly, have I misunderstood what it means for an estimator to be unbiased? Any help would be appreciated.

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