I am trying to make the link between the population variance, $\sigma^{2}$ and the sample variance $s^{2}$ with a particular example.
Say I have a population of five elements $\{0,1,2,3,4\}$, so $N = 5$. I was initially asked to calculate the population variance, and $V(\bar{y})$ for the sample mean, using samples of size $n = 2$. I did all of this, I'll provide the necessary results after this paragraph. The second part of the question asked show numerically that
$$E(s^{2}) = \frac{N}{N-1} \sigma^{2} $$
So now my curiosity kicked in. I know that the above result is a bias estimator of the population variance, so in order to make it unbias, I would have to multiply $s^{2}$ by $\frac{N-1}{N}$. So I attempted to do this with one calculation of $s^{2}$ from the set of samples of size $n = 2$ hoping to get the population variance I calculated earlier, but the calculation was not near what the population variance was. Using the following explicit values:
$\sigma^{2} = 2$
the sample element I used was $\{0,1\}$,
which gave me the following estimates:
$\bar{y} = 0.5$
which gave me a value sample variance of $s^{2} = (0-0.5)^{2} + (1-0.5)^{2} = 0.5$
Thus: $\frac{N-1}{N}s^{2} = \frac{4}{5}(0.5) = 0.4$
I thought this would equal the "theoretical" variance regardless of what value of $s^{2}$ was obtained. Am I interpreting that wrong? or is it because since $s^{2}$ is only an "estimation" of what the population variance is, then this estimator may very well not be exactly the value of the population parameter. Other values I used did fall closer to the population variance, but I was under the idea that making the estimator unbias it would always be equal to the population parameter value.
Best Answer
Expected value of an estimator should be equal to the "theoretical" variance (in the case of unbiased estimator). Particular numerical result may differ. In fact, in many applications, "theoretical" variance is not known at all.