Probability Theory – Show E[Y | E[Y | X]] = E[Y | X]

conditional-expectationprobability theoryrandom variables

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

To show: $E[Y | E[Y | X]] = E[Y | X]$

For this problem, I'm unsure how to rewrite the left-hand side of the statement since I don't know how to deal with a conditional expectation as a condition in another conditional expectation; any initial hints would be appreciated.

Best Answer

@DagunDagun gave great hints. I will write it out more explicitly for my own benefit.

Let $\sigma(X)$ denote the $\sigma$-algebra generated by the random variable $X$. It would suffice to show $\mathbb{E}\left[ Y \mid X \right]$ is $\sigma\left( \mathbb{E}\left[ Y \mid X \right] \right)$-measurable and

$$\int_A \mathbb{E}\left[ Y \mid X \right] \ \text{d}\mathbb{P} = \int_A Y \ \text{d}\mathbb{P}$$

for all $A \in \sigma\left( \mathbb{E}\left[ Y \mid X \right] \right)$. We know $\mathbb{E}\left[ Y \mid X \right]$ is $\sigma\left( X \right)$-measurable and

$$\int_A \mathbb{E}\left[ Y \mid X \right] \ \text{d}\mathbb{P} = \int_A Y \ \text{d}\mathbb{P}$$

for all $A \in \sigma\left( X \right)$. Therefore it would suffice to show

$$\sigma\left( \mathbb{E}\left[ Y \mid X \right] \right) \subseteq \sigma\left( X \right)$$

Recall $\mathbb{E}\left[ Y \mid X \right]$ is $\sigma(X)$-measurable and $\sigma\left( \mathbb{E}\left[ Y \mid X \right] \right)$ is the smallest $\sigma$-algebra that makes $\mathbb{E}\left[ Y \mid X \right]$ measurable. We conclude $\sigma\left( \mathbb{E}\left[ Y \mid X \right] \right) \subseteq \sigma\left( X \right)$.

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