# Show the equivalence of these definitions of independence of random variables

Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space. There are two different definitions of the independence of random variables $X_1, X_2, ...$ on $(\Omega, \mathscr{F}, \mathbb{P})$:

1. The $\sigma$-algebras $\sigma(X_1), \sigma(X_2), ...$ are independent, i.e. for distinct indices $k_1, k_2, ..., k_m$, $\sigma(X_{k_1}), \sigma(X_{k_2}), ..., \sigma(X_{k_m})$ are independent.

2. Given Borel sets $B_1, B_2, ...$ and distinct indices $j_1, j_2, ..., j_p$,

$$\prod_{j=j_1}^{j_p} P(X_j \in B_j) = P(\bigcap_{j=j_1}^{j_p} (X_j \in B_j))$$

Why are these two definitions equivalent?

What I tried:

For distinct indices $k_1, k_2, ..., k_m$,

$\sigma(X_{k_1}), \sigma(X_{k_2}), ..., \sigma(X_{k_m})$ are independent.

$$\iff P(\bigcap_{k=k_1}^{k_m} A_k) = \prod_{k=k_1}^{k_m} P(A_k)$$

where

$$A_k \in \sigma(X_k) = \{ X_k^{-1}(B_k) | B_k \in \mathscr{B} \}$$

It looks like

$$P(\bigcap_{k=k_1}^{k_m} A_k) = \prod_{k=k_1}^{k_m} P(A_k)$$

is the same as

$$\prod_{j=j_1}^{j_p} P(X_j \in B_j) = P(\bigcap_{j=j_1}^{j_p} (X_j \in B_j))$$

Does that end the proof? Doesn't seem very rigorous. Can I somehow let or conclude that $m = p$ and $(k_1, ..., k_m) = (j_1, ..., j_p)$?

## 1 Answer

From definition 1, we can fix $m \in \mathbb{N}$, $k_1, k_2, \dots, k_m \in \mathbb{N}$ and $B_{k_1}, B_{k_2}, \dots, B_{k_m} \in \mathscr{B}$ s.t. $$P(\bigcap_{k=k_1}^{k_m} A_k) = \prod_{k=k_1}^{k_m} P(A_k)$$

Since $$\prod_{j=j_1}^{j_p} P(X_j \in B_j) = P(\bigcap_{j=j_1}^{j_p} (X_j \in B_j))$$ holds for all $p \in \mathbb{N}$, all distinct indices $j_1, \dots, j_p \in \mathbb{N}$ and all $B_{j_1}, B_{j_2}, \dots, B_{j_p} \in \mathscr{B}$ and since $m \in \mathbb{N}$, $k_1, k_2, \dots, k_m \in \mathbb{N}$ and $B_{k_1}, B_{k_2}, \dots, B_{k_m} \in \mathscr{B}$ we can choose $p = m$, $$j_i = k_i$$ and $B_{j_i} = B_{k_i}$ for $i = 1, \dots p \ ( = m )$. QED