Ok, i have a question but i start with a definition first so that one can get the context.
(All variables in question have the same variance and under $H_0$ which we are considering - they have the same expected value, they are also independent $\sim$ $N(\mu,\sigma^2)$ )
error sum of squares SSE = $\sum\limits_{i}\sum\limits_{j} (X_{ij} - \bar{X})^2 = \sum\limits_{j} (X_{1j} - \bar{X} )+ ...+ \sum\limits_{j} (X_{Ij} - X_{I})^2$
apply this theorem to SSE get a $\chi^2$ distribution: if $X_1,...X_n$ are independent $N(0,1)$ then $\sum\limits_{i=1}^{n} (X_i -\bar{X})^2 \sim \chi^2(n-1)$
So SSE/$\sigma^2$ $\sim \chi^2$ with $I(J-1)$ degrees of freedom
Now, we know that the expected value of a chi square variable with $v$ degrees of freedom is just $v$ , thus (the implication is what i don't get)
$E\big(SSE/\sigma^2 \big) = I(J-1) \implies E(SSE/I(J-1)) = \sigma^2 $
So my question is how is it true $ E(SSE/I(J-1)) = \sigma^2$ there is no further explanation in my book or any proof, i just don't get this implication ..
Note that expectation is linear. So, what we really have is $$ \Bbb E(SSE / \sigma^2) = I(J-1) \implies\\ \frac 1{\sigma^2}\Bbb E(SSE) = I(J-1) \implies\\ \frac 1{I(J-1)}\Bbb E(SSE) = \sigma^2 \implies\\ \Bbb E(SSE / I(J-1)) = \sigma^2 $$