We know that for a sample of size $n$ the sample variance is
$\displaystyle S^{2} = \frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}$
Suppose I used the whole population of size $N$ as my sample then in this case the variance I compute is
$\displaystyle S^{2} = \frac{1}{N-1}\sum_{i=1}^{N}(X_{i}-\mu)^{2}$ where $\mu=\frac{\sum_{i=1}^{N}X_i}{N}$
However, the population variance is
$\displaystyle \sigma^{2} = \frac{1}{N}\sum_{i=1}^{N}(X_{i}-\mu)^{2}$ where $\mu=\frac{\sum_{i=1}^{N}X_i}{N}$
and this is counterintuitive. One would think that by using the unbiased estimator of the sample variance, convergence when the whole population is considered is guaranteed as the variance of an estimator approaches $0$ as $N \rightarrow \infty$
Best Answer
Your $S^{2} = \dfrac{1}{n-1}\sum\limits_{i=1}^{n}(X_{i}-\overline{X})^{2}$ is an unbiased estimator of the population variance $\sigma^2$ using Bessel's correction and assumes that each sample point is independent of the others.
That independence assumption does not apply when you are sampling without replacement from a finite population size $N$. So your expression is not unbiased. Instead, in such a case, an unbiased estimator of $\sigma^2 = \dfrac{1}{N}\sum\limits_{i=1}^{N}(x_{i}-\overline{x})^{2}$ is $$S_{\text{w/o rep}}^{2} = \dfrac{N-1}{N(n-1)}\sum\limits_{i=1}^{n}(X_{i}-\overline{X})^{2}.$$
This has the desired properties: not only is $\mathbb E\left[S_{\text{w/o rep}}^{2}\right]=\sigma^2$, but when $n=N$ you have $S_{\text{w/o rep}}^{2}=\sigma^2$. You also have $S_{\text{w/o rep}}^{2}< S^2$, an intuitively credible result as you actually know the values of a positive fraction of the population.