Raul Miller wrote: > >> > I can determine \sum(X_i-\bar X)^2 for each of those >> > potential sample cases (0, 0.5, 0.5, 0) and then >> > average them. For the fair coin, this average is >> > 0.25. For that unfair coin, I get 0.1875 for this >> > average. >> >> This precisely illustrates the point. The population variance is 3%8, >> not 3%16. You can see in this case that using denominator n=2 >> gives the wrong answer, using denominator n-1 gives the correct >> answer. > > I am not sure this is a valid point.
Why? This shows that the method you are using does not produce an unbiased estimator of variance, but mine does. > > See > http://en.wikipedia.org/wiki/Standard_deviation#Estimating_population_standard_deviation_from_sample_standard_deviation > as an example treatment which seems to conflict with the assertions > you have advanced in this last paragraph. > This gives a biased estimator. > As I understand the wiki write up, even if you consider denominator > n=1 as valid for samples which do not represent the population, > you must still use denominator n=2 when you are dealing with > the population distribution. What is a sample which does not represent the population? Bset wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
