# Conditional Expectation of a Discrete Random Variable Given A Sigma Field

In my undergraduate Mathematics of Finance class, we are studying conditional expectation given a sigma field. I understand what a sigma field is by its definition, but I don't understand how to compute conditional expectation of a discrete random variable given a sigma field.

Can someone please give a very basic intuitive explanation of what it means to condition on a sigma field generated by some partition?

It would be extra helpful if someone could give the explanation in the context of the following problem I am working on:

Consider probability space $$(\Omega,\mathcal F,\mathbb P)$$ in which $$\Omega= \{1,2,3,4\}$$, $$\mathcal F$$ is the sigma-field of all subsets of $$\Omega$$, and $$\mathbb P(A)= |A|/|\Omega|$$ for each $$A \in\mathcal F$$. Each outcome is equally likely. Let $$\mathcal G$$ be the sigma-field generated by the partition $$\mathcal P=\{\{1,2\}, \{3,4\}\}$$ and $$\mathcal H$$ be the sigma-field generated by the partition $$\mathcal Q=\{\{1\}, \{2\}, \{3\}, \{4\}\}.$$ Observe that $$\mathcal G$$ is contained in $$\mathcal H$$. Let $$X$$ be the random variable defined by $$X(\omega)=3\omega$$, $$\omega\in\Omega$$.

Compute:

1. $$\mathbb E[X \mid \mathcal G]$$,
2. $$\mathbb E[X \mid \mathcal H]$$.

Any insight would be really appreciated, thank you!

• Please edit and use MathJax to properly format math expressions. Commented Feb 4, 2020 at 23:59

By definition, $$\mathbb E[X\mid\mathcal G]$$ is the unique (up to probability) $$\mathcal G$$-measurable random variable which satisfies $$\int_E \mathbb E[X\mid\mathcal G]\ \mathsf d\mathbb P = \int_E X\ \mathsf d\mathbb P$$ for all $$E\in\mathcal G$$. Now, $$\mathcal G = \{\varnothing,\{1,2\},\{3,4\},\Omega\},$$ so we must have $$\mathbb E[\mathsf1_{\{1,2\}}(\omega)\mathbb E[X\mid\mathcal G]] = \mathbb E[X\mathsf 1_{\{1,2\}}(\omega)] = 3\cdot\mathbb P(X=1) + 6\cdot\mathbb P(X=2) = \frac 92,$$ and similarly $$\mathbb E[\mathsf1_{\{3,4\}}(\omega)\mathbb E[X\mid\mathcal G]] = \mathbb E[X\mathsf 1_{\{1,2\}}(\omega)] = 9\cdot\mathbb P(X=3) + 12\cdot\mathbb P(X=4) = \frac{21}2.$$ This implies that $$\mathbb E[X\mid \mathcal G](\omega) = \begin{cases} \frac92,& \omega\in\{1,2\}\\ \frac{21}2,&\omega\in\{3,4\}. \end{cases}$$
As for question 2. - note that $$\mathcal H=\Omega$$, and so $$\mathbb E[X\mid \mathcal H] = X$$.