# Series of independent random variables are independent again

Let $\{X^i_j : i=1,..n, j\in\mathbb{N}\}$ be an independent set of random variables on a probability space $(\Omega, A, \mathbb{P})$, $$X^k_l: \Omega \to \mathbb{R^+} := \{x \in \mathbb{R} : x \ge 0\}$$ $$Y^m := \sum_{i=1}^\infty X^m_i$$

Show that

$$Y^1, ..., Y^n$$

is an independent set of random variables

Defintion of independence for a set of random variables: A set of random variables is mutually independent if and only if for any finite subset $X_1, \ldots, X_n$ and any finite sequence of numbers $a_1, \ldots, a_n$, the events $\{X_1 \le a_1\}, \ldots, \{X_n \le a_n\}$ are mutually independent events.

Edit: The random variables are now non-negative

• So the set of $X_j^1$ that are in the sum of $Y^1$ is completely disjoint from the set of $X_j^2$ that are in the sum of $Y^2$, correct? – David K Apr 5 '15 at 15:26
• @DavidK yes exactly. For $(i, j) \ne (l, k): X^i_j \ne X^l_k$ – rasbi Apr 5 '15 at 15:28
• Consider the random variable $X^i=(X^i_j)_{j \in \mathbb{N}}$, which is $\mathbb{R}^{\mathbb{N}}$-valued. Then $X^1,...,X^n$ are independent as $\mathbb{R}^{\mathbb{N}}$-valued random variables. Define $f: \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ as $f= \limsup_n \sum_{i=1}^n \pi_i$, where the $\pi_i$ are the canonical projections. Then $Y^i=f(X^i)$, so the $Y^i$ are independent. – Shalop Apr 6 '15 at 15:59
• @Shalop Is there a way of proofing it without using random variables which have values in a room with infinite dimensions? – rasbi Apr 8 '15 at 5:56
• I don't know any, but there probably is. Maybe @saz knows? – Shalop Apr 8 '15 at 15:14

For each $1 \leq i \leq n$, define $X^i:=(X^i_j)_{j\in\mathbb{N}}$. More explicitly, $X^i$ is the mapping from $\Omega \to \mathbb{R}^{\mathbb{N}}$ defined by $\omega \mapsto (j \mapsto X^i_j(\omega))$. It is measurable with respect to the $\sigma$-algebra $\mathcal{F}$ on $\mathbb{R}^{\mathbb{N}}$ generated by the canonical projections $\pi_j:\mathbb{R}^{\mathbb{N}}\to \mathbb{R}$.
I claim that $X^1,...,X^n$ are independent. To show this, we'd need to show that $P(X^1 \in A^1,...,X^n\in A^n)=P(X^1\in A^1)\cdots P(X^n \in A^n)$ for all $A^i \subset \mathbb{R}^{\mathbb{N}}$ with $A^i \in \mathcal{F}$. It suffices to prove this when each of the $A^i$ has the form $\bigcap_{j=1}^{n_i} \pi_j^{-1}(B^i_j)$, where the $B^i_j \subset \mathbb{R}$ are Borel sets and $n_i \in \mathbb{N}$ (because such sets form a $\pi$-system generating $\mathcal{F}$). But this follows directly from the independence of the entire collection $\{ X^i_j \}_{i,j}$.
Define $f:= \sup_n \sum_{j=1}^n \pi_j: \mathbb{R}^{\mathbb{N}}\to \mathbb{R}$. Then $f(X^i)=Y^i$ for each $1 \leq i \leq n$ (because the $X^i$ are non-negative), so since the $X^i$ are independent, it follows that the $Y^i$ are independent.