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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. 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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    $\begingroup$ Has a certain homeworky aspect to it, voting to close. $\endgroup$
    – Igor Rivin
    Commented Oct 21, 2012 at 20:08

1 Answer 1

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Consider the following example. $W, X, Y, Z$ all have standard Cauchy distribution, but $X$ and $Y$ are independent while $Z = -W$. Then $X+Y$ also has a Cauchy distribution so its expected value does not exist, while $W+Z=0$ identically.

EDIT: With the additional assumption that $E[X+Y]$ exists, I believe you'll find that $$ E[X+Y] = \int_{-\infty}^\infty (F_{-X}(t) - F_{Y}(t))\ dt $$ (where $F_{-X}$ and $F_Y$ are the cumulative distribution functions of $-X$ and $Y$), so that this is the same as $E[W+Z]$.

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