# Do we have a formula (which maybe related to integral or probability) of $\mathbb{E}[X | \mathcal{G}] (\omega)$?

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!

• 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. – GEdgar Jan 27 '20 at 22:15
• 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? – LE Anh Dung Jan 27 '20 at 22:45

If $$\mu(\cdot,\cdot)$$ is the regular conditional distribution of $$X$$ given $$\mathcal{G}$$, then for an integrable function $$g$$, $$\mathsf{E}[g(X)\mid \mathcal{G}](\omega)=\int_{\mathbb{R}} g(x)\mu(\omega,dx) \quad\text{a.s.}$$
• 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? – LE Anh Dung Jan 27 '20 at 22:46