# An example of conditional expectation

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?

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.

• Sorry, I a bit miss-wrote my question. Nov 5 '20 at 16:23

$$X+Y=1$$ 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 $$\{5, 6\}$$.

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