Why are the random variables independent? This question is about independence in the ANOVA analysis.
Assume that you have $I\cdot J$ independent random variables $X_{i,j}$. We also assume that each is normally distributed with common mean $\mu$ and common variance $\sigma^2$. Define:
$\bar{X_{i.}}=\sum_{j=1}^JX_{i,j}/J$.
$\bar{X_{..}}=\sum_{i=1}^I\sum_{j=1}^JX_{i,j}/(IJ)$.
$S_i^2=\sum_{j=1}^J(X_{i,j}-\bar{X_{i.}})^2/(J-1)$.
Why are then $\sum_{i=1}^I(\bar{X_{i.}}-\bar{X_{..}})^2$ and $\sum_{i=1}^JS_i^2$ independent?
I know that there is a result that says that $\bar{X_{i.}}$ and $S_i^2$ are indpendent. But I do not see how we get independence when we introduce $\bar{X_{..}}$? My book doesn't explain why we have this independence in detail. It says that we have indpendence because the first expression is based on $\bar{X_{i.}}$ and the second expression is based on $S_i^2$. 
 A: The result is a consequence of Cochran's Theorem, one of the results of which also states that, if a $\chi^2$ variate $X$ with $n$ degrees of freedom can be written as the sum,
$X = X_1 + ... + X_k$ where each $X_i \sim \chi^2_{r_{i}}$ such that $r_1 + ... + r_k = n$, then $X_i$'s are independently distributed.
Notice that the total sum of square can be decomposed into following terms,
$\displaystyle \sum_{i=1}^I\sum_{j=1}^J\left(X_{i,j} - \bar{X}_{..}\right)^2 = \displaystyle \sum_{i=1}^I\sum_{j=1}^J\left(X_{i,j} - \bar{X}_{i.}\right)^2 + \sum_{i=1}^I\sum_{j=1}^J\left(\bar{X_{i.}} - \bar{X}_{..}\right)^2$
Dividing by $\sigma^2$ on both sides,
$\displaystyle \sum_{i=1}^I\sum_{j=1}^J\left(\dfrac{X_{i,j} - \bar{X}_{..}}{\sigma}\right)^2 = \displaystyle \sum_{i=1}^I\sum_{j=1}^J\left(\dfrac{X_{i,j} - \bar{X}_{i.}}{\sigma}\right)^2 + \sum_{i=1}^I\sum_{j=1}^J\left(\dfrac{\bar{X_{i.}} - \bar{X}_{..}}{\sigma}\right)^2$
Futher, since $X_{i,j}$ are normally distributed, the LHS is distributed according to a $\chi^2$ variate with $IJ-1$ degrees of freedom. Also, it can be verified that the term on the extreme right is a $\chi^2$ variate with $I - 1$ degrees of freedom and the term in the middle is the sum of $I$ independent $\chi^2$ variate each with $J-1$ degrees of freedom.
So, we have $IJ - 1 = I(J-1) + (I-1)$.
Hence, by Cochran's theorem, the independence is verified.
