# Let $X,Y$ be independent r.v. on a probability space, then are the conditional expectations $E[X|\Sigma]$ and $E[Y|\Sigma]$ independent?

Fix a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $X,Y:\Omega\to\mathbb{R}$ be $L^1$ random variables. Let $\Sigma\subset\mathcal{F}$ be a sub-$\sigma$-algebra. Is it true that $$\mathbb{E}[XY|\Sigma]=\mathbb{E}[X|\Sigma]\,\mathbb{E}[Y|\Sigma]?$$ I'm totally unable to prove this, and I'm beginning to suspect that it's false.

## 1 Answer

Let $X$ and $Y$ be independent, identically distributed random variables on $(\Omega, \mathcal{F}, \mathbb{P})$ with $\mathbb{P}(X = 1) = 1 - \mathbb{P}(X = 0) = \frac{1}{2}$. Let $\Sigma = \sigma(X+Y)$.

Then note that $\mathbb{E}[XY 1_{\{X+Y = 1\}}] = 0$ since $X+Y = 1$ implies one of $X$ or $Y$ is $0$. But we have $\mathbb{E}[X | \Sigma] = \mathbb{E}[Y | \Sigma] = \frac{X+Y}{2}$ so that $$\mathbb{E} \big [ \mathbb{E}[X | \Sigma] \, \mathbb{E}[Y | \Sigma] 1_{\{X+Y = 1\}} \big ] = \frac{1}{4} \mathbb{E}[(X+Y)^2 1_{\{X+Y = 1\}}] = \frac{1}{4} \mathbb{P}(X+Y = 1) \neq 0$$

So, since $\{X+Y = 1\} \in \Sigma$, $\mathbb{E}[XY|\Sigma] \neq \mathbb{E}[X|\Sigma]\,\mathbb{E}[Y|\Sigma]$