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Let $(\Omega, F, P)$ be a probability space.

$P=1/6 \ (\omega \in \{1,2,3, 4, 5, 6\})$

$X=0 \ (\omega \in \{1,2,3\}), \ 1 \ (\omega \in \{4\}), \ 2 \ (\omega \in \{5, 6\})$
$Y=1 \ (\omega \in \{1,2,3\}), \ 0 \ (\omega \in \{4, 5, 6\})$

What is $E(X|\sigma(X+Y))$ ?

I think $X+Y=1 \ (\omega \in \{1,2,3,4\}), \ 2 \ (\omega \in \{5,6\})$ so,

$E(X|\sigma(X+Y))= 1/4 \ (X+Y=1), \ 2 \ (X+Y=2)$

Is this true?

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You calculation of X+ Y is wrong. If $\omega$ is 1, 2, or 3, X is 0 and Y is 1 so X+ Y= 1. That you have right. If $\omega$ is 4 or 5, X= 1 and Y= 0 so X+ Y= 1. X+ Y is NOT 2 for $\omega$= 5. It is only for $\omega$= 6 that X= 2 and Y is 0 so X+ Y= 2.

You should have X+ Y= 1 for $\omega\in \{1, 2, 3, 4, 5\}$, X+ Y= 2 for $\omega$= 6.

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  • $\begingroup$ Sorry, I a bit miss-wrote my question. $\endgroup$ Nov 5 '20 at 16:23
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$X+Y=1$ on $\Omega \backslash \{5,6\}$ and $X + Y = 2$ on $\{5,6\}$. Therefore, $E(X|X+Y)$ must be a constant on $\Omega \backslash \{5,6\}$, say $a$. We have, $\frac{1}{6} = E(X, X+Y = 1) = a P(X + Y = 1) = a \frac{4}{6}$. Thus, $a = \frac{1}{4}$. Obviously, $E(X|X+Y) = 2$ on $\{5,6\}$.

Therefore, $E(X|X+Y) = \frac{1}{4}$ on $\Omega \backslash \{5,6\}$ and $E(X|X+Y) = 2$ on $\{5, 6\}$.

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  • $\begingroup$ Sorry, I a bit miss-wrote my question. $\endgroup$ Nov 5 '20 at 16:22
  • $\begingroup$ Okay. Edited, but the idea is the same. $\endgroup$
    – user295959
    Nov 5 '20 at 16:27
  • $\begingroup$ Thanks, I understand. $\endgroup$ Nov 5 '20 at 16:30

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