3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.