Raul Miller wrote:

> For one thing, the reasoning for using N-1 in the denominator of
> standard deviation seems specious.
>
> For another, in most cases where it really matters (correlation, for
> example), it just cancels out.
>
As I indicate below, it is necessary for estimation.

> 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.


> 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.
>
The denominator of the sample mean really should be n, and for the
sample variance it really should be n-1 if you are estimating the
population mean from the sample.

Suppose you have a population with distribution f, mean mu and
standard deviation sigma.  Let X1,...Xn be independent random
variables with distribution f (these correspond to how to select a
sample of size n).  You want to use the sample to estimate mu and
sigma, that is, find statistics whose expected values are mu and
sigma.

Let M=(X1+...+Xn)/n be the sample mean.  Then E(M)=mu.

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:  However you will find

S^2=((X1-M)^2+...+(Xn-M)^2)/(n-1).  As before, E(S^2)=sigma^2.

Best wishes,

John







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