# Durrett Theorem 2.1.8

This is literally the theorem from the textbook:

In order for [random variables] $$X_1, \ldots, X_n$$ to be independent, it is sufficient that for all $$x_1, \ldots, x_n \in (-\infty, \infty]$$:

\begin{align*} P(X_1 \le x_1, \ldots, X_n \le x_n) = \prod\limits_{i=1}^n P(X_i \le x_i) \\ \end{align*}

Proof: Let $$\mathcal{A}_i =$$ the sets of the form $$\{ X_i \le x_i \}$$. Since

\begin{align*} \{X_i \le x\} \cap \{X_i \le y\} = \{X_i \le x \land y \} \end{align*} where $$(x \land y)_i = x_i \land y_i = \min\{x_i, y_i\}$$. $$\mathcal{A}_i$$ is a $$\pi$$-system. Since we have allowed $$x_i = \infty$$, $$\Omega \in \mathcal{A}_i$$. Exercise 1.3.1. implies $$\sigma(\mathcal{A}_i) = \sigma(X_i)$$, so the result follows from Theorem 2.1.7.

Is $$\mathcal{A}_i = \{X_i \le x_i\} = X_i^{-1}((-\infty, x_i))$$? That seems like a single set rather than a collection of sets. Also, this interpretation doesn't seem to be closed under intersection.

I understand $$\{X_i \le x\} \cap \{X_i \le y\} = \{X_i \le x \land y \}$$, but how about $$\{X_1 \le x_1\} \cap \{X_2 \le x_2\}$$. It does't seem that intersection would equal some $$\{X_i \le x_i\}$$ for some $$1 \le i \le n$$.

It seems my reading of $$\mathcal{A}_i$$ is not correct, but I'm not sure how else to read that.

The definition of $$\mathcal A_i$$ is $$\mathcal A_i:= \{ \{X_i \leq x_i\}: x_i \in \mathbb R \cup\{\infty\} \}.$$ It isn't just one set, and the proof goes on without intersecting sets of the form $$\{X_i \leq x_i\} \cap \{X_j \leq x_j\}$$, they just fix $$i$$ and prove that $$\sigma(A_i) = \sigma(X_i)$$.
• The expression $\{X_i \le x_i\}$ can be $\{X_1 \le x_1\}$ for $i=1$ or $\{X_2 \le x_2\}$ for $i=2$. $\{X_1 \le x_2\}$ would not be allowed as the indices are different. Next there are exactly $n$ random variables $X_1, \ldots, X_n$, and $n$ constant numbers $x_1, \ldots, x_n$. Is this right?
• For $i=2$, your expression would read $\mathcal{A}_2 := \{\{ X_2 \le x_2 \} : x_2 \in \mathbb{R} \cup \{ \infty \} \}$. That looks like a set of one and only one set.
• If I define $\{a : a \in \mathbb R\}$, the set isn't a singleton, in fact, is $\mathbb R$. $x_2$ is a variable as $a$ in the set i mentioned, it isn't a fixed value. If you like, you are intersecting sets of the form $\{X_i \leq a \}$ with $a \in \mathbb R \cup \{\infty\}$. Oct 7, 2021 at 19:52