Raul Miller wrote: > On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote: >> =\sum ((X_i-\mu)^2) -n(\bar X-\mu)^2$, >> since $\sum (X_i-\mu)=(\sum X_i) - n\mu = n(\bar X-\mu)$. > > Ok, but these are all zero for the case where you're working with an > accurate model -- where \mu and \bar X are equal. > > Mathematically speaking, I don't see this as a valid approach. >
It does not matter how accurate your model is except in the degenerate case of a constant distribution: \bar X is random variable, while \mu is a number. It makes no sense to talk about \mu and \bar X being equal. It makes sense to talk about P(\bar X=\mu): however in the case of a continuous distribution this is always zero. What is important is its expectation, and E(\bar X)=\mu. Note that \bar x, the value you get on an actual sample, is not the same as the random variable \bar X. Similarly S^2 is a random variable such that E(S^2)=\sigma^2, the population variance. > Numerically speaking, even if they're "just close" instead of "identical", > we're heading into unstable territory. We're talking about basic definitions here. There are several computational techniques that can give accurate answers. > > Put differently: as far as I can see, I have to accept the definition > of standard deviation as an axiom, if I am going to work with it at > all. > You do not have to accept this as an axiom. You can come up with any statistic that is an unbiased estimator of \sigma^2. The sample variance with denominator n will not do. However, you then need to compare the variance of your statistic with that of S^2. If it is greater, your statistic is not as good. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
