On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
Raul Miller wrote:
> For another, "standard deviation" jumps from a manageable quantity to
> an unknowable quantity when the result is determined, and it seems
> to me that it should be zero in that case.
I think here you are confusing the random variable sample standard
deviation, which has a distribution, with the value of the sample standard
deviation on a specific sample, which does not.
If I understand you correctly, you're saying that it's not meaningful
to talk about probabilities of 1? (For example: no cumulative
distribution across all sets.)
I guess you could say that I'm uncomfortable working with a
model of probability which imposes that limitation.
> To my mind, the reasoning for the N-1 factor in standard deviation
> may be validly applied to the mean -- if I include all the deviation
> terms AND the deviation of the mean itself from itself, I should
> exclude the count of the mean in the divisor. But no one bothers
> to express it that way, so this seems an exercise in futility.
...
Suppose you have a population with distribution f, mean mu and
standard deviation sigma. ...
Now let S^2=((X1-mu)^2+...+(Xn-mu)^2)/n. Then E(S^2)=sigma^2.
The problem in the latter is that you do not know mu: you have to
estimate it from the sample.
If you write (Xi-mu)^2=((Xi-M)+(M-mu))^2 and expand it out, you will
be able to eliminate mu from the sum
Can I?
I think I have to divide by M-mu, which would be bogus if it turned
out that M=mu (In other words, if I were working with an accurate
model).
That said, I've not actually proven that there's no other way to
work the math on this -- have I overlooked something? [Let's
limit this discussion to the cases where n=2, for now.]
--
Raul
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