# Unbiased estimator of variance for a sample drawn from a finite population without replacement

Previously, I do believe $$S^2$$ is an unbiased estimator of $$\sigma^2$$

$$S^2 = \frac{1}{n-1}\sum_{i=1}^n{\left(X_i-\bar{X}\right)^2}$$

is a correct conclusion.

However, I found the following statement:

Considering the sample variance:

$$s^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left(y_i -\bar{y}\right)^2$$

it can be shown (see Appendix A, Derivations) that

$$E(s^2) = \frac{N}{N-1}\sigma^{2}$$

This is an example based on simple random sample without replacement. It says $$S^2$$ is a biased estimator of $$\sigma^2$$.

So I am wondering "$$S^2$$ is an unbiased estimator of $$\sigma^2$$" can only be applied to some specific cases? How to understand this result based on simple random sample?

• Others should be aware that $n$ is the sample size, $N$ is the population size, and the sample is drawn from the finite population without replacement. Sep 17, 2016 at 23:22
• Mar 10 at 1:36

When sampling from a finite population without replacement, the observations are negatively correlated with each other, and the sample variance $s^2 = \frac{1}{n-1} \sum_i \left( x_i - \bar{x} \right)^2$ is a slightly biased estimate of the population variance $\sigma^2$.

The derivation in this link from Robert Serfling provides a clear explanation of what's going on. The author first proves that if the observations in a sample have constant covariance (i.e. $\mathrm{Cov}\left(x_i, x_j \right) = \gamma$ for all $i\neq j$) that: $$E[s^2] = \sigma^2 - \gamma$$

For independent draws (hence $\gamma = 0$), you have $E[s^2] = \sigma^2$ and the sample variance is an unbiased estimate of the population variance. But the issue you have with sampling without replacement from a finite population is that your draws are negatively correlated with each other!

In the case of sampling without replacement from a population of size $N$: $$\text{For i\neq j }\quad \mathrm{Cov}\left(x_i, x_j \right) = \frac{-\sigma^2}{N-1}$$ Hence: $$E\left[s^2\right] = \frac{N}{N-1}\sigma^2$$

The sample variance is indeed biased for a finite population with simple random sampling without replacement. And the solution to get an unbiased result is to multiply the sample variance by $$\frac{N-1}{N}$$, where $$N$$ is the population size.

I’m an engineer, not a mathematician. So my proof was to build a complete sampling distribution in Excel from a finite population and assuming sampling without replacement. I found that the mean of the sampling distribution sample variances ($$s^2$$) did not equal the population variance. $$s^2$$ is biased in this case. I don’t know why the literature so often ignores this fact. But if I multiply the mean $$s^2$$ by $$\frac{N-1}{N}$$, where $$N$$ is the population size, then lo and behold the product is exactly equal to the population variance.

Intuitively, as my sample size n increases and approaches and eventually equals the population size $$N$$ ($$n=N$$), I should expect the sample variance to approach the population variance if the sample variance is unbiased. That does not happen since the sample is divided by $$n-1$$ and the population by $$N$$. Multiplying the sample variance by $$\frac{N-1}{N}$$ solves this dilemma.

I don't know where your statements come from, but it the way you present them they are false. Taking directly the variance of the sample (that is, dividing by $$n$$) we get a biased estimator, but using sample variance (dividing by $$n-1$$) we get an unbiased estimator.

I think your statement comes from different conflicting sources or your source uses different notations in different parts. Maybe "$$s^2$$" means variance ($$n$$) in one page and sample variance ($$n-1$$) in the other. The fact that one formula uses "$$n$$" with the same meaning the other uses "$$N$$" makes me suspect that they aren't consistent.

• Sorry I forget to mention, as Gunns said: "that n is the sample size, N is the population size, and the sample is drawn from the finite population without replacement. " Sep 18, 2016 at 4:25