Derivation of the Satterthwaite appproximation Using the method of moments, one can try to approximate the sum of $\chi_{r}^{2}$ variables as $\sum a_{i}Y_{i}$ by equating the $n$-th movements of the sample with the $n$-th movement of the population, and "solve" the parameters this way. However I am stuck in the derivation of Satterthwaite appproximation. The author (Berger&Casella) suggested the following(page 314):
"..to do this we must match second moments, and we need $$\mathrm{E}\left(\sum^{k}_{i=1}a_{i}Y_{i}\right)^{2}=\mathrm{E}\left(\frac{\chi_{v}^{2}}{v}\right)^{2}=\frac{2}{v}+1$$
Applying the method of moments, we can drop the first expectation and solve for $v$, yielding $$\hat{v}=\frac{2}{\left(\sum^{k}_{i=1}a_{i}Y_i\right)^{2}-1}$$"
My naive question is why we can drop the expectation at all? This is not clear from the author's description of the method of moments, in which one merely equate $$m_{j}=\frac{1}{n}\sum^{n}_{i=1}X_{i}^{j}\text{  with  } EX^{j}$$
And it seems clear to me that the expectation sign cannot be dropped. Similarly in the last step of the derivation of the approximation formula, the author suggested:
"..substituting this expression for the variance and removing the expectations, we obtain...."(page 315)
Can anyone give a hint? Sorry the question is really "low".
Edit:
A fellow at here suggested that the method of moments assume $E(Z)=Z$ because one equals the two moments. I do not think this follows straightaway from the definition. Even when $j=1$ one has to equate $\frac{1}{n}\sum^{n}_{i=1}X_{i}$ with $EX^{1}$. I do not think this implies $E(Z)=Z$ in general, such that one can use $Z=\sum a_{i}Y_{i}$.
 A: Background: Understanding the method of moments (MoM) in a basic way
Motivation for the method: The (strong) Law of Large Numbers (LLN) gives us reason to think that (at least for large samples), a sample expectation will be close to the population expectation (note that the LLN applies to higher moments by taking $Z=X^j$). Thus, if we have $iid$ $X_i, i=1,\ldots,n$ we have Casella & Berger's $m_j = \frac{1}{n} \sum_{i=1}^n X_i^j$ is set equal to $\text{E}(m_j) = \text{E}(X_i^j) = \mu_j$.
Why you only need consider first moments: Consider Casella & Berger's $m_j = \frac{1}{n} \sum_{i=1}^n X_i^j$ and note that (as we did in the motivating argument), for any $j$ we can just take $Z_i = X_i^j$ and be left with $m_1$ for a different random variable. That is, all MoM estimators can be thought of as first moment MoMs; we can simply make that substitution to get any other moment we need. So MoM is really just setting $m=\mu$ where $m = \frac{1}{n} \sum_{i=1}^n X_i$ for some set of $iid$ $X_i \sim f_X$.
Why you can think of MoM as 'drop expectations': (i) Take $Z = \frac{1}{n} \sum_{i=1}^n X_i$ and note that $\text{E}(Z)=\text{E}(X)$ by linearity of expectation, so MoM simply takes $Z=\text{E}(Z)$. Similarly, taking $Z^j = \text{E}(Z^j)$ follows immediately from the argument we already used - i.e. we can think of MoM as 'drop expectations', and it will be reasonable because we have some random variable which will be close to its expectation; (ii) more generally, we could reasonably do this ('drop expectations') for any $Z$ that we had reason to think would be 'close to' its expectation.
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Now for the expression in the section relating to Satterthwaite in Casella & Berger
Casella & Berger match first and second moments of $Z=\sum_{i=1}^k a_iY_i$ that is, they take
$\text{E}(Z) = Z$ and $\text{E}(Z^2)=Z^2$, the second of which gives an estimate of $\nu$.
Note that $Z=\sum_i a_iY_i$ is a constant times a sample expectation; there's a clear sense in which we might expect that $Z\approx \text{E}(Z)$ and $Z^2 \approx \text{E}(Z^2)$, but we don't actually have to justify it here, we're just following their argument about what happens when we do it.
A: As pointed out patiently by Glen_b above, since $X_{i}$ is an independent random sample, we have $$E(\sum a_{i}Y_{i})=\sum a_{i}E(Y_{i})=\sum a_{i} Y_{i}$$where the last equation $$E(Y_{i})=Y_{i}$$ follows from the fact that from the definition of method of moments we have $$E(X)=\frac{\sum X_{i}}{n}=\overline{X}=X \text{when $n$=1} $$So the author's proof is justified. 
