Sample variance expectation

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It is defined that the population variance is [tex]S^{2}= \frac{1}{N-1}\sum^{N}_{1}\left(y_{i} - \bar{y}_{N}\right)^{2}[/tex] or [tex]\sigma^{2}= \frac{1}{N}\sum^{N}_{1}\left(y_{i} - \bar{y}_{N}\right)^{2}[/tex].

Also that the [tex]V\left[\bar{y}_{n}\right] = \frac{N-n}{N}\frac{S^{2}}{n} = \left(\frac{1}{n} - \frac{1}{N}\right)S^{2}[/tex] and its unbiased estimator is [tex]\hat{V}\left[\bar{y}_{n}\right] = \frac{N-n}{N}\frac{s^{2}}{n} = \left(\frac{1}{n} - \frac{1}{N}\right)s^{2}[/tex] where [tex]s^{2}= \frac{1}{n-1}\sum^{n}_{1}\left(y_{i} - \bar{y}_{n}\right)^{2}[/tex]

To show that [tex]\hat{V}\left[\bar{y}_{n}\right][/tex] is unbiased, I understand that we only need to show that [tex]E\left[s^{2}\right] \right]= S^{2}[/tex]


I have done like this [tex]E\left[s^{2}\right] \right][/tex] = [tex]E\left[\frac{1}{n-1}\sum^{n}_{1}\left(y_{i} - \bar{y}_{n}\right)^{2}\right][/tex] and I have arrived at [tex]\left(n-1\right)E\left[s^{2}\right] = nE\left[y^{2}_{i}\right] - nE\left[\bar{y}^{2}_{n}\right][/tex] and come up at [tex]E\left[s^{2}\right] \right]= \sigma^{2}[/tex]

My question is why I did not come up at [tex]S^{2}[/tex] as the [tex]E\left[s^{2}\right][/tex]?
 

Answers and Replies

  • #2
mathman
Science Advisor
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For σ you use the true mean. For S you use the sample mean. That is why you need to divide by N-1 rather than N.
 

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