Show $E[Y | E[Y | X]] = E[Y | X]$. [closed]

Question

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.

Answer 1

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

Answer 2

You should be done using the definition of the conditional expectation.

To be more precise, recall that on a probability space $(\Omega, \cal F, \mathbb{P})$, given a random variable $X$ and a $\sigma-$algebra $\cal G\subset\cal F$, $\mathbb{E}\left[X|\cal G\right]$ is the unique random variable $Z$ that verifies :

• $Z$ is integrable,
• $Z$ is $\cal G-$measurable,
• $\mathbb{E}[Z\mathbb{1}_G] = \mathbb{E}[X\mathbb{1}_G]$ for any $G\in\cal G$.

Recall also that if $Y$ is another random variable, then by definition $\mathbb{E}[X|Y]=\mathbb{E}[X|\sigma(Y)]$.

Let me know if you need any more help.