Show: $\text{Var}(Y) = \text{Var}(Y - E[Y | X]) + \text{Var}(E[Y | X])$.

Question

Given: Let $X$, $Y$ be random variables on a common probability space $(\Omega, \mathcal{A}, P)$.

To prove: $\text{Var}(Y) = \text{Var}(Y - E[Y | X]) + \text{Var}(E[Y | X])$.

Attempt:

• $\text{Var}(Y) = E((Y-E(Y))^2)$ (by definition)
• $\text{Var}(Y) = E(Y^2) - (E(Y))^2$
• $\text{Var}(E[X|Y]) = E((E[X|Y])^2) - (E(E[X|Y]))^2 = E((E[X|Y])^2) - (E(Y))^2$
• $\text{Var}(Y-E[Y|X]) = \text{Var}(Y) + \text{Var}(E[Y|X]) - 2 \cdot \text{Cov}(Y, E[X|Y])$
• $\text{Cov}(Y, E[X|Y]) = E[YE[Y|X]] - E(Y)E[E[Y|X]] = E[YE[Y|X]] - (E(Y))^2$

And now I'm stuck with $E[YE[Y|X]]$. How can I proceed or is there something I'm missing that could make it easier?

Answer 1

You're almost there. It's useful to note by iterated expectation

$$\mathbb{E}\left[ Y \mathbb{E}\left[ Y \mid X \right] \right] = \mathbb{E}\left[ \mathbb{E}\left[ Y \mid X \right]^2 \right]$$

so

$$ \text{Var}\left( Y - \mathbb{E}\left[ Y \mid X \right] \right) = \text{Var}\left( Y \right) - 2 \text{Cov}\left( Y, \mathbb{E}\left[ Y \mid X \right] \right) + \text{Var}\left( \mathbb{E}\left[ Y \mid X \right] \right) $$

where

$$ \text{Cov}\left( Y, \mathbb{E}\left[ Y \mid X \right] \right) = \mathbb{E}\left[ \mathbb{E}\left[ Y \mid X \right]^2 \right] - \mathbb{E}\left[ Y \right]^2 = \text{Var}\left( \mathbb{E}\left[ Y \mid X \right] \right) $$

Putting it together we get

$$\text{Var}\left( Y - \mathbb{E}\left[ Y \mid X \right] \right) = \text{Var}\left( Y \right) - \text{Var}\left( \mathbb{E}\left[ Y \mid X \right] \right)$$

and the result follows.