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If $G_1$ is a sub-$\sigma$-algebra of $G_2$, then

$$ E[E[X \mid G_1 ] \mid G_2 ] = E[X \mid G_1 ] = E[X \mid G_1 ] \mid G_2 ].$$

No prob for the 2nd equality, but I just cannot prove the 1st one...

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(1) You can use LaTeX here, see the FAQ. (2) The first and third terms are the same? (3) Unbalanced brackets. –  Nate Eldredge Oct 26 '12 at 3:42
first because $\mathbb E(X \vert G_1)$ is $G_2$ measuralble –  mike Oct 26 '12 at 11:19

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