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If we consider a set of random variables $X_1$ up to $X_n$ that are independent, then their sum cannot be a constant (given that at least one of them is non-constant). That is not too hard to prove.

Does this assertion still hold, if the random variables are just pairwise independent? The answer is yes, if the variances exist. Then the variance of the sum exists and is the sum of the individual variances, hence larger than zero and the sum cannot be constant.

However, is it possible to construct a counterexample if we do not require finite variance (or even expectation) and only pairwise independence?

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No, it can not. Suppose, $X_1, ... X_n$ are pairwise independent random variables, such that $\sum_{i = 1}^n X_i = c$ is a constant. Then we can without the loss of generality assume, that $c = 0$ (otherwise we simply replace all $X_i$ with $X_i - \frac{c}{n}$). Now, suppose $t \in \mathbb{R}$ satisfies $P(\max(X_1, ... , X_n) > t) \leq \frac{1}{n}$ (such $t$ exists because for any random variable $Y$ we have $\lim_{t \to \infty}P(y > t) = 0$). Also note, that as $\sum_{i = 1}^n X_i = c$, we can conclude, that $|X_i| > nt$ implies $\exists j \neq i$ such that $|X_j| > t$ (the follows from $|X_i| = |(\sum_{j = 1}^i X_j) + (\sum_{j = i}^n X_j)| \leq (\sum_{j = 1}^i |X_j|) + (\sum_{j = i}^n |X_j|)$) Thus

$$P(|X_i| > nt) = P((|X_i| > nt)\cap(\exists j \neq i|X_j| > t)) \leq \sum_{j < n,j \neq i}P((|X_i| > nt)\cap(|X_j|>t)) = \sum_{j < n,j \neq i}P(|X_i| > nt)P(|X_j|>t) \leq \frac{n-1}{n}P(|X_i| > nt)$$

And as $\frac{n-1}{n} < 1$, the above inequality is only possible when $|X_i| \leq nt$ almost surely. But we know, that if $|Y| \leq a$ almost surely then $Var[Y] \leq a^2$. Thus all $X_i$ have finite varieties. Then we can conclude, that $0 = Var[0] = Var[\sum_{i = 1}^n X_i] = \sum_{i = 1}^n Var[X_i]$, which results in all $X_1, ... , X_n$ being constants (as $\forall i$ we have $Var[X_i] = 0$)

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  • $\begingroup$ Thank you! Except for some typos, this looks really nice. I'll go through it once more as soon as I find the time and then accept as the answer. $\endgroup$ – floxbr Jan 15 at 12:27

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