Why Does the Expected Value of Sample Variance Differ From Population Variance?

[safina]

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

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

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


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

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

 

[mathman]

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.