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Conditional expectation proof question

oyth94

Member
Jun 2, 2013
33
Here is a proof question: For two random variables X and Y, we can define E(X|Y) to be the function of Y that satisfies E(Xg(X)) = E(E(X|Y)g(Y)) for any function g. Using this definition show that E(X1 + X2|Y) = E(X1|Y) + E(X2|Y)

So what I did was I plugged in to X = X1 + X2
E(E(X1 + X2)|Y)g(Y))
= E(X1g(Y)) + E(X2g(Y))
= E(E(X1|y)g(Y) + E(X2|Y)g(Y))
= E(g(Y) [E(X1|Y) + E(X2|Y)]

am I on the right track? what do I do after that?
 

girdav

Member
Feb 1, 2012
96
Are the $y$ supposed to be $Y$?