# Do there exist three pairwise independent random variables, such that their sum is zero?

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.

• 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. Dec 23 '19 at 20:13
• @Chris, I forgot to see that they should be non-constant. Thank you for pointing that out to me. Dec 23 '19 at 20:36
• 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. Dec 23 '19 at 20:50
• 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.
– Seva
Dec 23 '19 at 21:09
• 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 $X+Y+Z=0$. Dec 23 '19 at 22:23

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.
• What about $k$ random variables of which any $k-1$ are independent? Dec 24 '19 at 5:51