Independence depth of linearly dependent random variables

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.

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 $$\{X_1, ... , X_n\}$$?

I only know the following fact:

Independence depth of $$\{X_1, ... , X_n\}$$ is strictly less than $$n - 1$$.

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, 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 $$x \mapsto e^{cx}$$. 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.

• It is a follow-up of this question: mathoverflow.net/q/349010/110691 Dec 26 '19 at 13:12
• If the rv's have expectations, then $k>1$ is impossible, by the argument I already gave in your first question: take $E(\ldots|X_1)$ to see that $X_1$ is constant. Dec 26 '19 at 19:19
• Of course, this suggests that the full question has the same answer, with just some technical difficulties added. Dec 26 '19 at 19:20

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 $$0=\operatorname{Var}(\sum X_i)=\sum \operatorname{Var}(X_i)>0.$$

The largest possible independence depth of $$\{X_1,\dots,X_n\}$$ is $$1$$. That is, for any natural $$n\ge2$$, there are no pairwise independent random variables (r.v.'s) $$X_1,\dots,X_n$$ such that (i) $$X_1+\dots+X_n=0$$ almost surely (a.s.) and (ii) for all real $$c_1,\dots,c_n$$ and all $$i\in[n]:=\{1,\dots,n\}$$ we have $$P(X_i=c_i)\ne1$$.

Indeed, suppose the contrary: that $$n\ge2$$, $$X_1,\dots,X_n$$ are pairwise independent r.v.'s such that $$X_1+\dots+X_n=0$$ a.s., and for all real $$c_1,\dots,c_n$$ and all $$i\in[n]$$ we have $$P(X_i=c_i)\ne1$$.

Let $$Z=(Z_1,\dots,Z_n):=X-Y$$, where $$X:=(X_1,\dots,X_n)$$ and $$Y=(Y_1,\dots,Y_n)$$ is an independent copy of $$X$$. Then $$Z_1,\dots,Z_n$$ are symmetric pairwise independent r.v.'s such that $$Z_1+\dots+Z_n=0$$ a.s., and for all $$i\in[n]$$ we have $$P(Z_i=0)\ne1$$.

Take now any real $$a>0$$ and introduce $$W_i:=W_{i,a}:=Z_i\,I\{|Z_i|\le a\},$$ where $$I$$ denotes the indicator. Then the $$W_i$$'s are bounded symmetric pairwise independent r.v.'s, whence $$$$E\Big(\sum_{i\in[n]}W_i\Big)^2=\sum_{i\in[n]}EW_i^2. \tag{1}$$$$ On the other hand, recalling the condition $$Z_1+\dots+Z_n=0$$ a.s., introducing the random set $$\mathcal J_a:=\{j\in[n]\colon |Z_j|>a\}$$, and finally letting $$a\to\infty$$, we have \begin{align*} E\Big(\sum_{i\in[n]}W_i\Big)^2 &=\sum_{J\subseteq[n]}E\Big(\sum_{i\notin J}W_i\Big)^2\,I\{\mathcal J_a=J\}\\ &=\sum_{J\ne\emptyset}E\Big(\sum_{i\notin J}W_i\Big)^2\,I\{\mathcal J_a=J\}\\ &\le n\sum_{J\ne\emptyset}\sum_{i\notin J}EW_i^2\,I\{\mathcal J_a=J\}\\ &= n\sum_{J\ne\emptyset}\sum_{i\notin J}EZ_i^2\,I\{|Z_i|\le a\}\,I\{\mathcal J_a=J\}\\ &\ll\max_{i\ne j}EZ_i^2\,I\{|Z_i|\le a\}\,I\{|Z_j|>a\} \\ &=\max_{i\ne j}EZ_i^2\,I\{|Z_i|\le a\}\,P(|Z_j|>a) \\ &=\max_{i\ne j}EW_i^2\,P(|Z_j|>a) \\ &=o\Big(\max_{i\in[n]}EW_i^2\Big) =o\Big(\sum_{i\in[n]}EW_i^2\Big), \end{align*} which contradicts (1), as desired. (In the above multi-line display, $$\sum_{J\ne\emptyset}$$ denotes the summation over all non-empty $$J\subseteq[n]$$, $$A\ll B$$ means $$A\le CB$$ for some real $$C>0$$ not depending on $$a$$, and $$\max_{i\ne j}$$ denotes the maximum over all distinct $$i$$ and $$j$$ in $$[n]$$.)