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I am wondering how I can show for discrete random variables $Y, X, Z, W$ that:

$$ \mathbb{E}[Y\mid X,Z] = \sum_{w \in \mathcal{W}} \mathbb{E}[Y\mid X,Z,W=w]\mathbb{P}(W=w\mid X,Z) $$

I am trying to use the law of iterated expectations and total probability, but am unable to get the right form. Is it because I am abusing notation on the conditional? (which should be conditioned on a value)?

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The notation is fine. If you understand $E[Y] = \sum_{w \in \mathcal{W}} E[Y \mid W=w] P(W=w)$ holds for any [discrete] random variables $Y$ and $W$, then you are done.

In the original question, conditioning just changes the probability distribution of $Y$ and $W$, and is notated using $P(\cdot \mid X, Z)$ and $E[\cdot \mid X, Z]$. You can then apply the law of total expectation.

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  • $\begingroup$ Thanks! For the equation $E[Y] = \sum_{w \in \mathcal{W}} E[Y \mid W=w] P(W=w)$, is there a special name for it? Or is it something independently proved? $\endgroup$ – user321627 Nov 2 '18 at 1:44
  • $\begingroup$ Yes. It is the Law of Total Expectation: $$\mathsf E(Y) ~{=\mathsf E(\mathsf E(Y\mid W))\qquad\text{in general}\\= \sum_{w\in\mathcal W}\mathsf E(Y\mid W=w)\mathsf P(W=w)\qquad\text{for discrete random variables}}$$ $\endgroup$ – Graham Kemp Nov 2 '18 at 3:07

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