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 $X(\omega)=3\omega$, $\omega\in\Omega$.


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

Any insight would be really appreciated, thank you!

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

1 Answer 1


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 $\mathbb E[X\mid \mathcal H] = X$.


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