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I already read the question Sample Standard Deviation vs. Population Standard Deviation about the difference between sample standard deviation $$S_n^2 = \dfrac{\sum\limits_{i = 1}^n \left(X_i-\bar{X}\right)^2}{n-1}$$ and population standard deviation $$S_N^2 = \dfrac{\sum\limits_{i = 1}^N \left(X_i-\bar{X}\right)^2}{N}$$

One thing stays unclear for me. Why doesn't sample standard deviation converge to population standard deviation? Other words, if we assume that $n=N$, formula (1) doesn't equal formula (2).

illuminato
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1 Answers1

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Assuming you're sampling without replacement from a population of size N, that formula only works when $n \ll N$. The formula is derived assuming the samples all come from the same distribution and are all independent. This is close enough to true if $n$ is a tiny fraction of $N$, since removing $n$ members of the population doesn't change the distribution of the remainder much. On the other hand, if $n$ is a significant fraction of $N$, then the distribution of the remaining population members will be different from the whole population, so the selection of each member of the sample can no longer be approximated as independent and identically distributed.

eyeballfrog
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