What‘s wrong with my proof of the Law of Total Variance? According to the Law of Total Variance,
$$\operatorname{Var}(X)=\operatorname{E}(\operatorname{Var}(X\mid Y)) + \operatorname{Var}(\operatorname{E}(X\mid Y))$$
When trying to prove it, I write
$$
\begin{equation}
\begin{aligned}
\operatorname{Var}(X) &= \operatorname{E}(X - \operatorname{E}X)^2 \\
&= \operatorname{E}\left\{\operatorname{E}\left[(X - \operatorname{E}X)^2\mid Y\right]\right\} \\
&= \operatorname{E}(\operatorname{Var}(X\mid Y))
\end{aligned}
\end{equation}
$$
What's wrong with it?
 A: The third line is wrong, because you don't have $\text{E}[X|Y]$ in the second line. For example, if $Y$ is Bernoulli(1/2) and $X$ is 1 if $Y$ is 1 and -1 if $Y$ is 0, then $\text{E}[(X-\text{E}[X|Y])^2|Y] = 0$ (this is what you want) because $Y$ is totally informative of $X$, but what you have will give you $\text{E}[(X-\text{E}[X])^2|Y] = \text{E}[(X-0)^2|Y] = \text{E}[X^2|Y] = 1 \ne 0$.
Not gonna lie, you had me questioning myself and I had to stare at this for a bit before it hit me even though I've had to prove LOTV to myself a billion times :P
A: $$\require{cancel}
\begin{aligned}
\operatorname{Var}(X) &= \operatorname{E}(X - \operatorname{E}X)^2 \\
&= \operatorname{E}\left(\operatorname{E}\left[(X - \operatorname{E}X)^2\mid Y\right]\right) \\
&\ne \operatorname E\left( \operatorname E\left[ (X-\operatorname E(X\mid Y))^2 \right] \mid Y \right) \\
&= \operatorname{E}(\operatorname{Var}(X\mid Y))
\end{aligned}
$$
A: The transition from the second to the third line does not follow.  Since $\mathbb{E}(X) \neq \mathbb{E}(X|Y)$ you have:
$$\mathbb{E}[ (X - \mathbb{E}(X))^2 | Y ] 
\neq \mathbb{E}[ (X - \mathbb{E}(X|Y))^2 | Y ]  
= \mathbb{E}[ \mathbb{V}(X|Y) ].$$
In the special case where $\mathbb{E}(X) = \mathbb{E}(X|Y=y)$ for all $y \in \mathbb{R}$ your working and result would hold, and would be a special case of the more general result.
