# Relationship between population variance and sample variance as estimate of population variance

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.