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