I should add that as near as I can tell, I must use a 1/n concept of variance,
rather than a 1/(n-1) concept of variance when dealing with population
distributions.
For example, consider
$E(S^2)=E(((1/n-1)\sum (X_i-\bar X)^2)
=(1/n-1)( \sum \sigma^2(X_i)- n\sigma^2(\bar X) )$
and working with an example where we count the number of heads flipped
with a fair coin, and taking two samples:
Here, if I try using the 1/n formula for population variance, I get:
n=2
E(S^2)= 0.25 = E(((1/n-1)\sum (X_i-\bar X)^2)
\sigma(\bar X) = 0.353553
\sigma^2(\bar X)= 0.125
\sigma(X_i)= 0.5
\sigma^2(X_i)= 0.25
0.25 = (0.25+0.25)-2*0.125
The equality holds as I expect it to.
But if I use the 1/(n-1) formula for population variance, the above
does not hold:
n=2
E(S^2)= 0.25 = E(((1/n-1)\sum (X_i-\bar X)^2)
\sigma(\bar X) = 0.408248
\sigma^2(\bar X)= 0.166667
\sigma(X_i)= 0.707107
\sigma^2(X_i)= 0.5
0.25 ~: (0.5+0.5)-2*0.166667
The asserted equality does not hold.
That said, further down in my numerical model, I may have
made a mistake -- on my last calculation (t7). The assertion is
$E(S^2)=(1/n-1)(n\sigma^2 -n(\sigma^2/n))=\sigma^2$.
But I believe n(\sigma^2/n) is \sigma^2 and I cannot find any
interpretation of $E(S^2)=(1/n-1)(n\sigma^2 -(\sigma^2)) that
seems numerically valid.
I am still looking into this.
And, I also note that I did not produce any calculations corresponding
to the previous line, which has the corresponding assertion
$\sigma^2(\bar X)=\sigma^2/n$
which worries me.
It seems to me that \sigma^2(X_i) is 0.25 which means I should expect
that \sigma^2/n is 0.125. But from previous calculations, I expect that
\sigma^2(bar X) is 0.125 which means that either I have woefully
misunderstood some of the notation, or that the assertion
$\sigma^2(\bar X)=\sigma^2/n$
must be false.
[Note also that these are not exclusive possibilities.]
--
Raul
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