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I saw two versions of the law of iterated expectations, this one: \begin{align} E(E(Y\vert X)) = E(Y) \end{align} and this one: \begin{align} E(E(Y\vert X_1, X_2)\vert X_1) = E(Y \vert X_1) \end{align}

I was wondering if this one is also correct or not: \begin{align} E(E(Y\vert Z) \vert X) = E(Y\vert X) \end{align}

Here $X,X_1,X_2,Y,Z$ are random variables.

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2 Answers 2

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The conjecture is false. If $X=Y$, $Y$ and $Z$ are independent, $E(Y)=0$ then $E(Y|X)=Y$, $E(Y|Z)=0$, and $E(E(Y|Z)|X)=0$.

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The reason the second version you state works but the third one does not is that "the smaller information set dominates", see this elaborate answer: A generalization of the Law of Iterated Expectations

Conditioning on $X_1,X_2$ means conditioning on more information than conditioning on $X_1$ alone, which implies that first conditioning on $X_1,X_2$ and then conditioning on $X_1$ is as good as the latter only.

In the third statement, we have no such relationship between $X$ and $Z$.

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