Does correlation affect the expectation value of a sum?

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The discussion centers on the expectation value of the sum of two random variables, X and Y, and whether correlation affects this value. It is established that E(X+Y) can be expressed as the sum of their individual expectations, E(X) + E(Y), only if X and Y are uncorrelated. The participants clarify that the calculation of the expected value using joint distribution does not depend on the independence or correlation of the variables. The derivation holds true regardless of correlation, provided the expectations exist for the distributions involved. Overall, the correlation does not impact the expectation value of the sum of two random variables.
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We have for two random variables X and Y (one sum runs over j and one over k):

E(X+Y) = ƩƩ(xj+yk)P(X=xk,Y=yk)
= ƩƩxjP(X=xk,Y=yk) + ƩƩykP(X=xk,Y=yk)

Now this can be simplified to obtain E(X+Y)=E(X)+E(Y) if we use that:
P(X=xk,Y=yk) = P(X=xk)P(Y=yk), because then (and same goes the other way around):
ƩjP(Y=yk)P(X=xj)= P(Y=yk)

But all this requires X and Y to be uncorrelated. Does the derivation above also hold if generally:

P(xj,yk)=P(xj l Y=yk)P(Y=yk)?
 
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Which indexes are varying on each of your summation signs isn't clear.

I suggest you rewrite your question using the forums LaTex. (See https://www.physicsforums.com/showpost.php?p=3977517&postcount=3)

Calculating the expected value of the sum of two variables using their joint distribution does not depend on whether the two variables are independent or not. "Correlation" is also not relevant.
 
Work with your original double sum - you should find the result of interest is true in each case (as long as expectations exist for your distributions)
 
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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