Definition

$E (X) = E_{Y} [E_{X} (X ∣ Y)]$

Proof

By definition of conditional expectation $\forall G \in G, \int_{G} E (X ∣ G) d P = \int_{G} X d P$ Since $G$ is sub-Sigma-Field, $Ω \in G$ and $\int_{Ω} E (X ∣ G) d P = \int_{Ω} X d P$ hold.

Then, By definition of Expected Value $E (X) = \int_{Ω} X d P$ $E (E (X ∣ G)) = \int_{Ω} E (X ∣ G) d P = \int_{Ω} X d P = E (X)$

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