# Estimation of variance: How to bring Bessel's correction together with degrees of freedom?

I have been considering multiple textbooks to find out the reason that the denominator of the estimation of the population variance is n-1 rather than n. Depending on the book, two reasons are given:

Some others refer to Bessel's correction. As far as I understood, the variance of the sample underestimates the population variance systematically in magnitude of the variance of the mean. If you subtract this variance and solve the equations, n-1 is left as denominator.

Other authors refer to degrees of freedom: the sample mean is used to estimate the population mean. This value is then fixed in the calculation of the variance and therefore you lose one degree of freedom, resulting in n-1.

I wonder how this two explanations go together. Is it a coincidence that both explanations fit? Is there one right one? Why are even two explanations spread?

English is not my native language, I hope I still got the terms right. Thank you for your answers!

• I have attempted to explain the distinction in an answer at stats.stackexchange.com/a/17148. In some respects it really is a coincidence, which is why there is so much confusion. However, for a common application--ordinary least squares--there is an intimate connection between the two explanations. You will find that pointed out in other answers in the same thread.
– whuber
Jul 15 '15 at 17:55
• So, if you were to teach students , would you go for Bessel's correction or degrees of freedom-explanation? Jul 15 '15 at 18:07
• It depends on the context, the students, and the teaching objective. I provided one answer at stats.stackexchange.com/a/3932 for an introductory course: just ignore the whole issue. If the purpose is to explain the Student t-statistic, Bessel's correction (using an unbiased variance estimator) is more meaningful (and consistent with Student's original explanation). In a regression course, one would want to make the connection with degrees of freedom. For teaching $\chi^2$ testing to advanced students, I would have to cover much of what's in the answer referenced in my first comment.
– whuber
Jul 15 '15 at 18:27