Suppose, $\Xi$ is a collection of random variables. We call $\Xi$ $k$-independent, iff any $k$ distinct elements of $\Xi$ are mutually independent. For example, $2$-independence is pairwise independence and $|\Xi|$-independence is the full mutual independence of random variables from $\Xi$.


Let's define independence depth of $\Xi$ as the maximal number $k$, such that $\Xi$ is $k$-independent.



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> Suppose $X\_1, ... , X\_n$ are non-constant real random variables, such that $X\_1 + ... + X\_n = 0$. What is the largest possible independence depth of $\{X\_1, ... , X\_n\}$? 
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I only know the following fact:



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> Independence depth of $\{X\_1, ... , X\_n\}$ is strictly less than $n - 1$.
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Suppose $\{X\_1, ... , X\_{n}\}$ is $(n-1)$-independent. Then suppose $Y = -X\_n$. Thus we have $Y = \sum{i = 1}^{n - 1} X\_k$.


Let's define $\chi\_X$ as a characteristic function of a random variable $X$. Then we have $\forall x, y \in \mathbb{R}, k \leq n -1 $:


$$\chi\_{X\_k}(x)\Pi\_{i = 1}^{n-1} \chi\_{X\_i}(y) = \chi\_{X\_k}(x)\chi\_{Y}(y) = Ee^{i(xX\_k + yY)} = Ee^{i((x + y)X\_k + \sum\_{i = 1}^{k - 1}yX\_i + \sum\_{i = k+1}^{n-1} yX\_i)} = (\Pi\_{i = 1}^{k - 1}\chi\_{X\_i}(y))\chi\_{X\_k}(\Pi\_{i = k + 1}^{n-1}\chi\_{X\_i}(y))$$


From that and the facts, that characteristic functions are continuous and $\chi\_{X\_1}(0) = ... = \chi\_{X\_{n-1}}(0) = 1$ it follows, that $\exists \epsilon > 0$, such that $\forall x \in \mathbb{R}, |y| < \epsilon, k < n - 1$ we have $\chi\_{X\_k}(x + y) = \chi\_{X\_k}(x)\chi\_{X\_k}(Y)$. From that and the fact, that $\mathbb{R}$ is an [Archimedean field](https://en.wikipedia.org/wiki/Archimedean_property), we can conclude, that $\forall x, y \in \mathbb{R}, k < n - 1$ we have $\chi\_{X\_k}(x + y) = \chi\_{X\_k}(x)\chi\_{X\_k}(Y)$. And we know, that [all non-zero functions with this property are of the form $x \mapsto e^{cx}$](https://math.stackexchange.com/q/293371/407165). Thus we can conclude, that $\forall k < n - 1$ we have $\chi\_{X\_k}(x) = e^{ic\_kx}$ and thus $X\_k = c\_k$ almost surely. Thus all $X\_k$ and $Y$ (as the sum of them) are constants.



Independence depth of linearly dependent random variables
Here's a different argument.
Pick $t>0$ such that $P[|X\_i|>t]\leq 1/n$ for all $i.$ The event $|X\_i|>tn$ is a subset of the union of the events $|X\_j|>t$ for $j\neq i,$ so


$$P[|X\_i|>tn] \leq \sum\_{j\neq i} P[|X\_j|>t \text{ and }|X\_i|>tn] \leq \frac{n-1}{n}P[|X\_i|>tn].$$


Since each $X\_i$ is essentially bounded you can use the argument in Christian Remling's comment, or $0=\operatorname{Var}(\sum X\_i)=\sum \operatorname{Var}(X\_i)>0.$


