On 6/28/07, John Randall <[EMAIL PROTECTED]> wrote:
Raul Miller wrote:
> I am stuck again. I should probably have waited to reach this point
> before sending my previous response... Here's where I am at:
>
> On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
>> Then $E(S^2)=E((1/n-1)\sum (X_i-\bar X)^2
>> =(1/n-1)( \sum E((X_i-\mu)^2)-nE((\bar X-\mu)^2 )
>>=(1/n-1)( \sum \sigma^2(X_i)- n\sigma^2(\bar X) )$
>
> But this seems trivially wrong.
Sorry, the notation is less than crystal clear, since I am using \sigma^2
to mean both the population variance and a variance in general.
In the last line quoted:
\sigma^2(X_i)=variance of X_i=\sigma^2 (population variance)
\sigma^2(\bar X)=variance of \bar X=\sigma^2/n .
I think by 'variance of X_i' you mean $((X_i - \bar X)/(n-1))^2$
but this gets me no closer to understanding what
you meant by \sigma^2(X_i) or \sigma^2(\bar X).
For example: in our simple example, $((X_i - \bar X)/(n-1))^2$ can be
zero in some cases (when both sampled values are the same), which
clearly is not the same as \sigma^2(population variance). So either
I do not understand what you mean by = in
variance of X_i=\sigma^2 (population variance)
or I do not understand what you mean by variance of X_i
or... something else.
I can see that E(\sum(variance of X_i))=\sigma^2(population variance)
for my understanding of variance of X_i, but that's not what you
seemed to have said.
My impression, at the moment, is that if I started by defining
that sigma corresponds to my "substandard deviation" for the
population, and if I also replace n-1 with n in your above expressions,
that the modified statements would remain valid. However, I am getting
mired down in notational ambiguities -- I can't even show that the
statements mean what you clearly intend them to mean, so I am not at
all prepared to declare that I understand your underlying logic.
--
Raul
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