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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.

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    $\begingroup$ Let $\Omega = \{\text{point}\}$ be a set with one element, endowed with its (unique) probability measure and $X, Y, Z: \Omega\to \mathbb{R}$ identically zero. Then their sum is zero and they are pairwise independent. $\endgroup$
    – Chris
    Dec 23 '19 at 20:13
  • $\begingroup$ @Chris, I forgot to see that they should be non-constant. Thank you for pointing that out to me. $\endgroup$ Dec 23 '19 at 20:36
  • $\begingroup$ The answer is also no if 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. $\endgroup$ Dec 23 '19 at 20:50
  • $\begingroup$ For what it worth, here is a construction of $X,Y,Z$ pairwise independent such that $X+Y+Z$ is the identity function. Let $\omega$ be a number uniformly drawn from the interval $\Omega:=[0,1)$. Write $\omega=0.d_1d_2d_3\ldots$ and define $X(\omega):=0.d_100d_400d_7\ldots$, $Y(\omega):=0.0d_200d_500d_8\ldots$, and $Z(\omega):=0.00d_300d_600d_9\ldots$. Then $X(\omega)+Y(\omega)+Z(\omega)=\omega$ and $X,Y,Z$ are pairwise independent. $\endgroup$
    – Seva
    Dec 23 '19 at 21:09
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    $\begingroup$ Of course you mean to say that $X,Y,Z$ are real variables. 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 $X+Y+Z=0$. $\endgroup$
    – Gro-Tsen
    Dec 23 '19 at 22:23
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Replace $Z$ by $-Z$, so that $Z=X+Y$. Let $f_X$ and $f_Y$ the characteristic functions of $X$ and $Y$, so that $f_X(s)=Ee^{isX}$ for real $s$. Suppose the pairwise independence. Then for all real $s$ and $u$ $$f_X(u)f_Y(u)f_X(s)=f_Z(u)f_X(s)=Ee^{iuZ+isX} \\ =Ee^{i(u+s)X+iuY}=f_X(u+s)f_Y(u). \tag{1} $$ Therefore and because $f_Y$ is continuous with $f_Y(0)=1\ne0$, we have $$f_X(u+s)=f_X(u)f_X(s) \tag{2} $$ for all real $u$ close enough to $0$ and all real $s$.

Note that (2) (together with the conditions that $f_X$ is continuous with $f_X(0)=1\ne0$) implies that $f_X$ is nowhere $0$. Similarly, $f_Y$ is nowhere $0$. So, (2) actually holds for all real $u$ and $s$. So, $f_X(s)=e^{isa}$ for some real $a$ and all real $s$.

So, $X$ is a constant almost surely. Similarly, $Y$ is a constant almost surely.

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  • $\begingroup$ What about $k$ random variables of which any $k-1$ are independent? $\endgroup$ Dec 24 '19 at 5:51
  • $\begingroup$ @BrendanMcKay : I guess that can be handled similarly, but I have not really thought about it. $\endgroup$ Dec 24 '19 at 6:24
  • $\begingroup$ @BrendanMcKay : Your question (as well as a more general one) has been answered at mathoverflow.net/questions/349156/… $\endgroup$ Jan 2 '20 at 2:26

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