Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\Omega=\{a,b,c\}$ and $\mathcal{F}_1=\sigma(\{a\})$. Suppose $X$ is a random variable such that $X(a)=0$, $X(b)=1$ and $X(c)=0$. Define $Y=E(X|\mathcal{F}_1)$. How do we find $Y(a)$, $Y(b)$ and $Y(c)$?

By definition of conditional expectation, $Y \in \mathcal{F}_1$ and $\forall A \in \mathcal{F}_1, \sum_A XdP=\sum_A YdP$. Assuming $P$ assigns equal measure on $a,b$ and $c$ and using $\mathcal{F}_1=\{\emptyset, \{a\}, \{b,c\}, \Omega\}$, I have $Y(a)=0$ and $Y(b)+Y(c)=1$.

How do I find a way to say $Y(b)=Y(c)=1/2$?

share|cite|improve this question
up vote 4 down vote accepted

How do I find a way to say $Y(b)=Y(c)=1/2$?

You add to your considerations the condition that $Y$ is measurable with respect to $\mathcal F_1$. In discrete spaces, this means that $Y$ is constant on every atom of $\mathcal F_1$. Here, $Y$ is constant on $\{b,c\}$... et voilà!

share|cite|improve this answer
Thanks did! That's has a nice intuition to have on discrete spaces. – Bravo Feb 9 '13 at 16:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.