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.

  • $\begingroup$ It is a follow-up of this question: mathoverflow.net/q/349010/110691 $\endgroup$ Dec 26 '19 at 13:12
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    $\begingroup$ 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. $\endgroup$ Dec 26 '19 at 19:19
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    $\begingroup$ Of course, this suggests that the full question has the same answer, with just some technical difficulties added. $\endgroup$ 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 \begin{equation} E\Big(\sum_{i\in[n]}W_i\Big)^2=\sum_{i\in[n]}EW_i^2. \tag{1} \end{equation} 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 $[n]$.)


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