What is wrong with this computation? Law of Iterated Expectations

Question

Suppose that $X$ is a random variable and $U$ is some event. I'm interested in computing $E[XPr(U|X)]$. What is wrong with the following reasoning?

1. $E[XPr(U|X)]=E[E[XPr(U|X)|X]$ (Law of Iterated Expectations)
2. $E[E[XPr(U|X)|X]=E[X[E[Pr(U|X)|X]$
3. $E[X[E[Pr(U|X)|X]=E[X[E(Pr(U|X)]$
4. $E[X[E(Pr(U|X)]=E[X*Pr(U)]$
5. $E[XPr(U)]=Pr(U)E[X]$

Can someone spot a mistake? I'm pretty sure the equality $E[XPr(U|X)]=Pr(U)E[X]$ is not true (I tried some numerical examples), but I really don't know where is my mistake.

Thanks guys!

Answer 1

If you write $\Pr(U\mid X)=\mathsf E(\mathbf 1_U\mid X)$ then you have :

$$\begin{align}\mathsf E(X\Pr(U\mid X)) &= \mathsf E(X~\mathsf E(\mathbf 1_U\mid X)) \\ &= \mathsf E(\mathsf E(X~\mathbf 1_U\mid X)) \\ & = \mathsf E(X~\mathbf 1_U)\end{align}$$

However, the next proposed step (5), $\mathsf E(X\mathbf 1_U)=\mathsf E(X)\Pr(U)$ is only true if $X$ and $\mathbf 1_U$ are uncorrelated.   Is this warentted?

Answer 2

The mistake is the following. $\mathbb{P}(U\mid X)$ is actually a function of $X$, not a constant. Therefore, $\mathbb{E}(\mathbb{P}(U\mid X)\mid X)=\mathbb{P}(U\mid X)$. That conditional probability goes out of the conditional expectation as a constant.