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 . Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
