Unorthodox tricks for explaining that this mixed model $SSE$ is proportional to a chisquare I usually deal with problems like these easily, but this time, I am stuck. I've pulled out every trick for showing chisquare distribution that I can think of, but nothing worked. You will soon see the problem. We are in a linear mixed effect model setting. Suppose $Y_{ij} = \mu + A_i + \varepsilon_{ij}$, such that $A_i \sim N(0,\sigma_A^2)$ for all $i$, $i = 1,\cdots,I$ and $\varepsilon_{ij} \stackrel{\text{iid}}{\sim} N(0,\sigma^2)$, $1 \leq j \leq J$. $A_i, \varepsilon_{ij}$ are all independent for all combinations of $A_i, \varepsilon_{ij}$ and all $i,j$. I want to explain why
$$
\frac{1}{\sigma^2} \sum_{i=1}^I\sum_{j=1}^J (Y_{ij}- \overline Y_ {i \cdot})^2 \sim \chi^2_\nu,
$$
for some $\nu$ where $\overline Y_ {i \cdot}$ denotes the mean across all $j$ for fixed $i$. Clearly $Y_{ij}$s are not not independent for fixed $i$ (right?). This is the problem causing all my normal tricks to fail. Can you suggest some new tricks that I may need to add to my bag of tricks for dealing with this?
 A: Hint: Fix the numerical values of $i$ and $j$ and write out explicitly what $Y_{ij}- \overline Y_ {i \cdot}$ is, expanding out that second term as the sum of $J$ different terms etc. Then, figure out whether the
quantity $(Y_{ij}- \overline Y_ {i \cdot})$ depends in any way on either
$\mu$ or $A_i$.  If your answer is No, (as I hope t will be) , then
$\sum_{j=1}^J (Y_{ij}- \overline Y_ {i \cdot})^2$ is pretty much like
a sum $\sum_{j=1}^J (X_i - \bar{X})^2$ where the $X_i$ are independent
$N(0,\sigma^2)$, right? And we know that the
distribution of the sum $\sum_{j=1}^J (X_i - \bar{X})^2$ is a scaled
$\chi^2$ distribution with the appropriate degrees of freedom (figure out
how many), don't we?
Finally, that outer sum is the sum of $I$ independent scaled $\chi^2$ distributions
with the same scale factor in each case, and so the double sum is
yet another scaled $\chi^2$ random variable with the same scale
factor as everybody else, and even more degrees of freedom (once again,
figure out how many).
S, there are no unorthodox techniques involving normal distributions
that you need to learn: it is all standard stuff. About the only thing
that might require some work is if you need to prove that 
$\sum_{j=1}^J (X_i - \bar{X})^2$ is a scaled $\chi^2$ random variable
on the grounds that the definition of a (scaled) $\chi^2$ random variable is the sum of the
squares of independent identically distributed normal random variables whereas the $(X_i - \bar{X})$ are not independent (though they
are identically distributed). But this too is a standard result;
see, for example, Wikipedia.
