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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 $Y(b)=Y(c)=1/2$?

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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!

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

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