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. For the instance where our population is number of heads from a coin toss, and a sample size of 2, the expected value calculations are all 0.25. However, that last expression $(1/n-1)( \sum \sigma^2(X_i)- n\sigma^2(\bar X) )$ would have four possibilities in this instance: X_0 X_1 \bar X 0 0 0 0 1 0.5 1 0 0.5 1 1 1 If I interpret the \sum in that expression as applying to everything to its right, my corresponding answers are never greater than 0, because X_i can never be greater than n*\bar X. Alternatively, if I interpret the \sum as applying only to \sigma^2(X_i) then my corresponding answers are always 0, because \sum X_i and n*\bar X must be equal. Am I missing something? Thanks, -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
