Do there exist such three non-constant pairwise independent random variables $X, Y, Z$ such that $X + Y + Z = 0$?

I managed only to prove the following two facts:

If such $X, Y, Z$ exist, they are not independent.

*Proof:*

If they are, then $X$ and $-X = Y + Z$ are also independent, which is impossible.

If such $X, Y, Z$ exist, then at least two of them do not have finite second moment.

*Proof:*

$\DeclareMathOperator\Var{Var}$Suppose, they all have finite second moments. Then $\Var(X) + \Var(Y) + \Var(Z) = 0$, which implies that all $X$, $Y$ and $Z$ are constants. Now suppose that without the loss of generality $Y$ and $Z$ have finite second moment. Then $\Var(X) = \Var(-Y-Z) = \Var(Y) + \Var(Z) \leq \infty$ and we return to the previous case.

However, those facts are clearly insufficient to solve this problem.

noif the rv's have a first moment because then you can take the conditional expection $E(\ldots |X)$, say, to see that $X+EY+EZ=0$, so $X$ is constant. $\endgroup$realvariables. Because for variables in a finite abelian group $G$, we can let $X,Y$ be independent and uniformly distributed on $G$ and $Z=-(X+Y)$ and then $Z$ is independent from any one of $X,Y$ (and also uniformly distributed on $G$) and we have $X+Y+Z=0$. $\endgroup$