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I have been trying this for an hour now, but I can't seem to find out how this conclusion is drawn. This is from the book by Greene (Econometric analysis p 53).

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What may be confusing me is that I don't know what the difference between $E_{x,y}$ and $E_yE_x$ is. I have interpreted the first as expectation over both variables, and the second as $E(E(\cdot|Y))$ which by the law of iterated expectations is equal to simply the expectation. Therefore, I have interpreted them to be equivalent.

Is this correct? And So how do we derive this result?

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This notation is really imprecise, it would be better to write $E_y E_{x|y}$ for it does not make sense to integrate over $p(x)$ and then $p(y)$ unless $x\perp y$.

With this in mind, you are correct. Assuming everything is well behaved and that Greene's notation is simply telling you the order of integration (the correct distribution being implied by the context):

$$ E_{y, x}[g] = \int_{y, x}gp(y,x) d(y,x) = \int_y \int_x gp(y,x)dxdy = E_yE_x[g] $$

Also, as you correctly surmised:

$$ \begin{align} E_yE_x[g] &=\int_y \int_x gp(y,x)dxdy \\ &=\int_y \int_x gp(x|y)p(y)dxdy \\ &= \int_y \left( \int_x gp(x|y)dx \right) p(y)dy\\ &= E[E[g|y]] \end{align} $$

Finally, to prove the mean squared error decomposition, take $(y - \gamma x)^2$ and sum and subtract $E[y|x]$: $(y - \gamma x)^2 =(y - E[y|x] + E[y|x]- \gamma x)^2$. Expanding this square gives you: $(y - E[y|x])^2 + (E[y|x]- \gamma x)^2 + 2(y - E[y|x])(E[y|x]- \gamma x)$. Then just take the expectation.

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