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I am trying to get $E(X|F)$. But I am not sure if I am right.

The only single variables available in $F$ are $\{1\}$ and $\{ 2\}$. So, the probability of 2 possible events out of 6 events in F is $\frac{2}{6}=1/3$.

Am I right? If not right, how do we get $ E(X|F)$ ? I would appreciate your help.

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Hint. $F=\sigma(\{1 \},\{2 \},\{3,4 \})$

$D=\{ \{1 \},\{2 \},\{3,4 \} \}$

$$E(X|F)=E(X|\{1\})1_{\{1 \}} +E(X|\{2\})1_{\{2 \}}+E(X|\{3,4\})1_{\{3,4 \}} $$

On the other hands

\begin{eqnarray} E(X|F)(\omega)= \left\{ \begin{array}{cc} E(X|\{1\}) & \omega \in \{1 \} \\ E(X|\{2\}) & \omega \in \{2 \}\\ E(X|\{3,4\}) & \omega \in \{3,4 \} \end{array} \right. \end{eqnarray} see conditional-expectations-when-conditioning-on-a-discrete-sigma-algebra and a finite or countable partition now

$$E(X|A)=\frac{E(X1_{A})}{P(A)}$$ Conditional_expectation_with_respect_to_an_event

for example

$$E(X|\{3,4\})=\frac{E(X1_{\{3,4\}})}{P(\{3,4\})}$$

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