MHB Conditional expectation proof question

AI Thread Summary
The discussion centers on proving that E(X1 + X2|Y) equals E(X1|Y) + E(X2|Y) using the definition of conditional expectation. The user attempts to substitute X with X1 + X2 and applies the expectation operator, breaking it down into components for X1 and X2. They question whether their notation is correct, specifically regarding the use of lowercase 'y' versus uppercase 'Y'. The conversation highlights the importance of maintaining consistent notation in mathematical proofs. The proof process is on the right track, and clarification on notation will enhance understanding.
oyth94
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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 into 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?
 
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Are the $y$ supposed to be $Y$?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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