Raul Miller wrote: > Sure -- that's why I called that numerical model a way of "checking > my work" instead of a "proof". But, if the math is valid, then the math > should remain valid when I plug in the numbers.
This is where the problem lies. Let S^2=(1/n)\sum (X_i-\mu)^2. Then E(S^2)=\sigma^2, the population variance. However, if you replace \mu by \bar X, it is not true that E((1/n)\sum (X_i-\bar X)^2=\sigma^2. You are assuming that E((X-Y)^2)=E(X-E(Y))^2, which is false. Assuming this is equivalent to saying E(Y^2)=E(Y)^2. >> If you are trying to calculate an expected value by averaging it over >> samples taken from the population, you will get an estimate, but what >> does it mean? This is precisely what estimation is about. > > In this case, my "samples" precisely represent the entire population. > > For example, let's consider your hypothetical case of number of > heads from a coin toss. > > With a fair coin, the population, with distribution is: > 0: 50% > 1: 50% > > The possible samples for a sample size of 2 are then > 0 0: 25% > 0 1: 25% > 1 0: 25% > 1 1: 25% > > I don't actually need to enumerate probabilities for this > case, since it's evenly distributed. However, if I had an > unfair coin I could deal with that as well, using basically > the same approach: > 0: 25% > 1: 75% > > with possibilities: > 0 0: 0.0625% > 0 1: 0.1875% > 1 0: 0.1875% > 1 1: 0.5625% > > Thus, for E(\sum (X_i-\bar X)^2)=\sum (x_i-\bar x)^2 I still don't get this. I am using the notation x_i for an actual sample value. The left hand side is a number. The right hand side will vary based on the actual sample chosen. > > 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. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
