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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?

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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

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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.

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$$\require{cancel} \begin{aligned} \operatorname{Var}(X) &= \operatorname{E}(X - \operatorname{E}X)^2 \\ &= \operatorname{E}\left(\operatorname{E}\left[(X - \operatorname{E}X)^2\mid Y\right]\right) \\ &\ne \operatorname E\left( \operatorname E\left[ (X-\operatorname E(X\mid Y))^2 \right] \mid Y \right) \\ &= \operatorname{E}(\operatorname{Var}(X\mid Y)) \end{aligned} $$

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