The exercise you shared involves random variables $X$ and $Y$, and the goal is to prove that $Y_2$, the difference between $Y$ and its conditional expectation $E[Y | X]$, is uncorrelated with every $\sigma(X)$-measurable random variable.

Here is a detailed step-by-step explanation of the solution:

Problem Restatement

Let $X$ and $Y$ be integrable random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Define:

• $Y = Y_1 + Y_2$, where
  • $Y_1 = E[Y | X]$, the conditional expectation of $Y$ given $X$, which is $\sigma(X)$-measurable,
  • $Y_2 = Y - E[Y | X] = Y - Y_1$, the difference between $Y$ and its conditional expectation.

The task is to show that:

1. $Y_2$ and $X$ are uncorrelated.
2. More generally, $Y_2$ is uncorrelated with every $\sigma(X)$-measurable random variable.

Key Definitions and Setup

• Uncorrelated means that the covariance between two random variables $Z_1$ and $Z_2$ is zero, i.e., $\text{Cov}(Z_1, Z_2) = E[(Z_1 - E[Z_1])(Z_2 - E[Z_2])] = 0$.
• The conditional expectation $E[Y | X]$ is $\sigma(X)$-measurable, which means it is a function of $X$ and depends only on $X$.

Proof (More General Case)

We are asked to show that $Y_2$ is uncorrelated with every $\sigma(X)$-measurable random variable $\xi$.

By definition, $Y_2 = Y - E[Y | X]$. To show uncorrelation with $\xi$, we need to show that: $E[Y_2 \xi] = 0$

Now, proceed as follows:

1. Expectation of the product:
   [ E[Y_2 \xi] = E\left[(Y - E[Y | X])\xi\right] = E[Y \xi] - E[E[Y | X] \xi] ]

2. Simplification of terms:
   By the property of conditional expectation: [ E[E[Y | X] \xi] = E[\xi E[Y | X]] ] Since $\xi$ is $\sigma(X)$-measurable, it can be factored out of the conditional expectation: [ E[\xi E[Y | X]] = E[\xi \xi] = E[\xi] ] Which simplifies it to zero.

This solves the