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We have four random variables say W,X,Y,Z where W and X has the same distribution and Y, Z also has the same distribution. Bad news is EX and EY may not exist but E(W+Z) is zero. if we assume that E(X+Y) exists, could we conclude that E(X+Y) is zero ? ( I know if EX and EY where defined we used linearity and it is obvious, also we know nothing more about X, Y)

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The title seems rather misleading. If this is linearity not holding, then limits of real sequences aren't linear either, since the sum of two sequences can have a limit even though the individual sequences don't. Perhaps you mean something like "where linearity cannot be used"? –  joriki Oct 21 '12 at 21:41

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Correlation may kill the cat.

If the distribution is (forsimplicity) symmetric about the origin, we may have that $Z=-W$ whereas $X,Y$ are independent, even though all four variables have the same distribution. With $X,Y$ independent $E(X+Y)$ will not exist if $E(X),E(Y)$ don't exist. But $Z+W=0$ a.s., hence $E(Z+W)=0$.

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In my last version I said that E(X+Y) is defined, I mean assume that this exists so now your counterexample does not work –  peanut Oct 22 '12 at 6:14
    
(In fact it was already the original version that said $E(X+Y)$ is defined.) Hagen, peanut duplicated this question, presumably because the answers that don't answer the question were causing it to appear resolved to others. I voted to close the duplicate. May I suggest that you delete (or fix) this answer to make this question unanswered again? –  joriki Oct 22 '12 at 10:07

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