On 6/28/07, John Randall <[EMAIL PROTECTED]> wrote:
X_i has the population distribution, so the variance of X_i is \sigma^2. The parenthetical (population variance) is an explanatory note, not an argument for \sigma^2.
Ok, fair enough. That said, I believe I now understand your proof well enough to assert that your proof works just as well if I use the same calculation for sample standard deviation as I use for population standard deviation. (With no "n-1" adjustment for the variance when working with samples). In other words: Let $\mu$ and $\sigma$ be the population mean and standard deviation. Fix the sample size $n$. Let $\bar X$ be the sample mean, $S^2$ the sample variance given by $$ S^2=(1/n)\sum (X_i-\bar X)^2$$ It is elementary to show $E(\bar X)=\mu$. We now show $E(S^2)=\sigma^2$. $\sum (X_i-\bar X)^2 =\sum ((X_i-\mu)-(\bar X -\mu))^2 =\sum ((X_i-\mu)^2) -n(\bar X-\mu)^2$, since $\sum (X_i-\mu)=(\sum X_i) - n\mu = n(\bar X-\mu)$. Then $E(S^2)=E((1/n)\sum (X_i-\bar X)^2) =(1/n)( \sum E((X_i-\mu)^2)-nE((\bar X-\mu)^2 )) =(1/n)( \sum \sigma^2(X_i)- n\sigma^2(\bar X) )$ But $\sigma^2(X_i)=\sigma^2$, $\sigma^2(\bar X)=\sigma^2/n$, so $E(S^2)=(1/n)(n\sigma^2 -n(\sigma^2/n))=\sigma^2$ I've checked my work on this -- I built myself a numerical model representing most of the above statements and have tested both versions of it [yours and mine] against several different populations and sample sizes. -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
