0
$\begingroup$

My lecture note defines conditional expectation as follows:

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathcal{G}$ be a sub-sigma field of $\mathcal{F}$. If $X$ is an integrable random variable, then the conditional expectation of $X$ given $\mathcal{G}$ is any integrable random variable $Z$ which satisfies the following two properties:

(CE1) $Z$ is $\mathcal{G}$-measurable.

(CE2) $$\forall \Lambda \in \mathcal{G}: \int_{\Lambda} Z \, d \mathbb{P}=\int_{\Lambda} X \, d \mathbb{P}$$

We denote $Z$ by $\mathbb{E}[X | \mathcal{G}]$.

Let $X$ be an integrable random variable and $\mathcal G$ a sub-sigma field.

I would like to ask if we can deduce a formula (which maybe related to integral or probability) of $\mathbb{E}[X | \mathcal{G}] (\omega)$ from (CE1) and (CE2).

Thank you so much!

Improve this question
$\endgroup$
  • 1
    $\begingroup$ In general, $\mathbb{E}[X | \mathcal{G}]$ is defined only up to sets of measure zero. If singleton $\{\omega\}$ has measure zero, then changing $\mathbb{E}[X | \mathcal{G}]$ on $\omega$ makes no difference. If you have studied measure theory, you may like to think of $\mathbb{E}[X | \mathcal{G}]$ as an example of a Radon-Nikodym derivative. $\endgroup$ – GEdgar Jan 27 '20 at 22:15
  • $\begingroup$ Hi @GEdgar, if $X$ is the indicator function $\mathbb{1}_\Lambda$, can we have a formula for $\mathbb{E}[\mathbb{1}_\Lambda | \mathcal{G}] (\omega)$ without Radon-Nikodym derivative? $\endgroup$ – LE Anh Dung Jan 27 '20 at 22:45
3
$\begingroup$

If $\mu(\cdot,\cdot)$ is the regular conditional distribution of $X$ given $\mathcal{G}$, then for an integrable function $g$, $$ \mathsf{E}[g(X)\mid \mathcal{G}](\omega)=\int_{\mathbb{R}} g(x)\mu(\omega,dx) \quad\text{a.s.} $$

Improve this answer
$\endgroup$
  • $\begingroup$ Hi @d.k.o, if $X$ is the indicator function $\mathbb{1}_\Lambda$, can we have a formula for $\mathbb{E}[\mathbb{1}_\Lambda | \mathcal{G}] (\omega)$ without regular conditional distribution? $\endgroup$ – LE Anh Dung Jan 27 '20 at 22:46
  • $\begingroup$ What kind of formula are you looking for? $\endgroup$ – d.k.o. Jan 27 '20 at 22:46

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.