# Population Variance vs Sample Variance [duplicate]

I'm particularly interested why N appears in the denominator for the formula of population variance, but (n-1) appears in the denominator for the sample variance, where N is the total number of elements in the population and n is the total number of elements in the simple random sample?

Basically, the (n-1) seems less intuitive to me.

I'd love to know the reasoning behind this, as well as seeing any mathematical derivations that back up that reasoning.

## marked as duplicate by Glen_b♦Apr 5 '17 at 17:53

• – Glen_b Apr 5 '17 at 16:58
• It is just because it is common to use the unbiased estimate of variance for the sample estimate. – Michael Chernick Apr 5 '17 at 17:24
• @Glen_b Thanks, that was exactly what I was looking for, especially the pdf that was linked to: maxwell.ict.griffith.edu.au/sso/biased_variance.pdf Something that was unclear to me in that paper is that I was not familiar with the "iid assumption" and why that means E{xi*xj} = (mu_x)^2? – EthanT Apr 5 '17 at 17:43
• "iid" means "independent and identically distributed". The expectation result is a direct consequence of that independence. See en.wikipedia.org/wiki/… – Glen_b Apr 5 '17 at 17:47
• Well, a couple more questions. The pdf starts by assuming a bias exists. How generally true is this assumption? Are all samples guaranteed to be biased in such a fashion that the (n=1) is a good corrector in ALL cases? From what I know in engineering, biases come in all "shapes and sizes". Could the same be said about biases found within simple random sampling? Or, regardless of origin, do all biases resulting from random sampling simply result in a shift of the mean away from the population mean, thereby making this simple correction of (n-1) sufficient? – EthanT Apr 5 '17 at 17:51