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The tower law states: $$\mathbb{E}(X)=\mathbb{E}[\mathbb{E}(X|Y)]$$

But, from my understanding, we have the following:

Suppose we let $Z:=X|Y$ and $a,b\in\mathbb{R}$. Then:
$\mathbb{E}(X)=a$ is the expectation of a random variable, so it is a number
$\mathbb{E}(X|Y)=\mathbb{E}(Z)=b$ is also the expectation of a random variable, so it is a number
$\mathbb{E}[\mathbb{E}(X|Y)]=\mathbb{E}[\mathbb{E}(Z)]=\mathbb{E}(b)=b$ is the expectation of a number, so it is equal to that number itself
But, by this argument, the tower law doesn't hold

I'm guessing my understanding of $\mathbb{E}(X|Y)$ is wrong, but I don't know why. Please explain

YinWai Tse
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1 Answers1

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$W := E[X \mid Y]$ is a random variable that depends on $Y$. Then $E[E[X \mid Y]]] = E[W]$. The notation $Z := X \mid Y$ unfortunately does not make sense.

angryavian
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  • The first paragraph of your post makes for a correct and full answer. Re the second paragraph, one should note that the idea to define random variables $Z_y$ is often formulated but never adopted since it implies to change, for each $y$, the probability space $\Omega$ to $\Omega_y={Y=y}$. And this has tons of unintended consequences, for example, which probability is $E$ in $E(Z_y)$ based upon, how to define $Z_{y_1}+Z_{y_2}$ for $y_1\ne y_2$, and so on. – Did Feb 11 '18 at 19:11
  • @Did Fair point, thank you for pointing this out. I'll remove the second paragraph. – angryavian Feb 11 '18 at 19:17