2
$\begingroup$

let me just ask one simple question, I am not sure if I understand this concept of conditioning w.r.t. sub-$\sigma$-algebras.

Let $(\Omega,\mathcal{A},\mathbb{P})$ be probability space and $X,Y:\Omega\rightarrow \mathbb{R}$ random variables. If someone writes $$E(X|Y=y), $$ does it mean $$E(X|\sigma(Y))(Y^{-1}(y))?$$

Generally conditional expectation is random variable on $(\Omega,\mathcal{F})$ where $\mathcal{F}\subset\mathcal{A}$ such that $\int_F E(X|\mathcal{F})\;d\mathbb{P}=\int_F X\;d\mathbb{P}$.

I am not sure if notation $E(X|Y=y)$ means conditioning on $\sigma(Y)$, while it is a function of $\omega$'s as preimages of values of $Y$.

I hope the meaning of this question is clear.

$\endgroup$
2
  • 3
    $\begingroup$ In all cases I can remember seeing such an expression, either the writer is being sloppy or is invoking the elementary concept of conditional expectation for discrete variables. One way to understand the notation, then, is that $E(X\mid Y)$ can be considered a partial function of real numbers $y$ that is supposed to be evaluated at any $\omega\in\Omega$ for which $Y(\omega)=y:$ that is, it is the push forward of the conditional expectation from $(\Omega,\mathcal F)$ to $(\mathbb R, \mathcal B(\mathbb R)).$ $\endgroup$
    – whuber
    Jul 8, 2022 at 19:40
  • $\begingroup$ $\mathbb{E}[X|Y]=e(Y)$ is a $\sigma(Y)$-measurable transform of $Y$ satisfying $$\mathbb E[e(Y)]=\mathbb E[X]$$ $\endgroup$
    – Xi'an
    Jul 8, 2022 at 19:42

0

Your Answer

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