On 6/28/07, John Randall <[EMAIL PROTECTED]> wrote:
Suppose we take a sample of size 2, so X_1 and X_2 are random
variables having the above distribution, \bar X=(X_1+X_2)/2.
Then \sum (X_i-\bar X)^2=((X_1-X_2)^2)/2, and by direct computation,
we get E(\sum (X_i-\bar X)^2)=1/4.
Ok thanks!
For some [lack of] reason, I was using mu in place of \bar X.
This, I believe, also answers the question you posed for me:
Why do you assign equal probabilities? If the die is fair, large values of
(X_i-\bar X)^2 are less likely than small values.
(And, in fact, I did not notice that I had made this mistake until
I tried to answer this question.)
Thanks again,
--
Raul
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