# $X_n$ independent variables if equations hold

Let $$X_1,X_2,\dots ,X_n$$ be random variables.

Statement 1

If for any $$B_i \in \mathcal{B}^n, i=1,2,\dots,n$$,

$$P\left( \overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) \right) = \prod_{i=1}^{n} P\left(X_i^{-1} \left(B_i \right) \right)$$

then $$X_i$$ are independent.

Statement 2

If we set $$X=\left( X_1, X_2, \dots , X_n \right)$$

and since $$\overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) = X^{-1}\left(B_1 \times B_2 \times \dots X_n \right)$$

then the equation in Statement 1 is equivalent to

$$P_X \left( B_1 \times B_2 \times \dots \times B_n \right) = P_{X_1}(B_1) \times P_{X_2}(B_2) \times \dots \times P_{X_n}(B_n)$$

Does anyone have guidance on how to proceed proving these two statements?

EDIT

I want to show that if either Statement 1 OR Statement 2 is true for $$B_i \in \mathcal{P}^n, i=1,2,...,n$$, then $$X_i$$ are mutually independent.

$$\mathcal{P}^n = \{ (a_1,b_1]\times(a_2,b_2]\times \dots \times (a_n,b_n]\} \cup \{ \emptyset \}$$

Attempt

My idea is that if Statement 1 is true for $$B_i \in \mathcal{P}^n, i=1,2,...,n$$ then, then by setting some $$B_i = \mathbb{R}^n \in \mathcal{P}^n$$ we get that for every subset of indices $$I = \{i_1, \dots, i_k \} \subset \{1,2, \dots, n\}$$ we have

$$P\left( \overset{k}{\underset{m=1}{\bigcap}}X_{i_m}^{-1}\left(B_{i_m} \right) \right) = \prod_{m=1}^{k} P\left(X_{i_m}^{-1} \left(B_{i_m} \right) \right)$$

for any $$B_{i_m} \in \mathcal{P}^n$$.

But the above does not imply that $$X_i$$ are mutually independent since we need it to be true for any $$B_{i_m} \in \mathcal{B}^n$$. How can I proceed from here?

If Statement 2 is true I don't know how to proceed to prove independence.

• Is not what you wrote the definition of independence? What is your definition? Commented Jan 12, 2021 at 11:44
• @Shashi Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $X_i$ be a family of random variables, $i \in I$, $X:\Omega \rightarrow \mathbb{R}^n$. $X_i$ are independent iff for every finite set $\{i_1,i_2,\dots,i_k\} \subset I$ and for every $B_1, B_2,\dots,B_k \in \mathcal{B}^n$, $$P\left( \overset{k}{\underset{m=1}{\bigcap}}X_{i_m}^{-1}\left(B_{i_m} \right) \right) = \prod_{m=1}^{k} P\left(X_{i_m}^{-1} \left(B_{i_m} \right) \right)$$
– Fib
Commented Jan 13, 2021 at 10:49
• How the statement you want to prove any different than the definition? Commented Jan 13, 2021 at 11:21
• @Shashi It is different because Statement 1 states that this equation holds when using all $n$ of the variables $X_i$. Where the definition says it bust be true for every finite subset of indices for $X_i$. To make it more clear, as I understand it, It is true that $$\overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) \subset \overset{k}{\underset{m=1}{ \bigcap}}X_{i_m}^{-1}\left(B_{i_m} \right)$$ $$\overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right)$$ What is true for a set $B \subset C$ is not necessarily true for $C$ as well.
– Fib
Commented Jan 13, 2021 at 13:21
• @Shashi Disregard my last comment, I had some errors in my latex code and the limit for editing passed. Read my next comment.
– Fib
Commented Jan 13, 2021 at 13:27

Let $$(\Omega,\Sigma,\mathbb{P})$$ be a probability space and $$X:(\Omega,\Sigma,\mathbb{P})\to(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$$ be a random vector.
\begin{align*} \mathbb{P}_{X}(B_{1}\times B_{2}\times\ldots\times B_{n}) & = \mathbb{P}(X^{-1}(B_{1}\times B_{2}\times\ldots\times B_{n}))\\\\ & = \mathbb{P}(\left\{\omega\in\Omega\mid X(\omega)\in B_{1}\times B_{2}\times\ldots\times B_{n}\right\})\\\\ & = \mathbb{P}(\{\omega\in\Omega \mid (X_{1}(\omega)\in B_{1})\wedge(X_{2}(\omega)\in B_{2})\wedge\ldots\wedge(X_{n}(\omega)\in B_{n})\}\\\\ & = \mathbb{P}(\{\omega\in\Omega \mid X_{1}(\omega)\in B_{1}\}\cap\ldots\cap\{\omega\in\Omega \mid X_{n}(\omega)\in B_{n}\})\\\\ & = \mathbb{P}\left(X^{-1}_{1}(B_{1})\cap X^{-1}_{2}(B_{2})\cap\ldots\cap X^{-1}_{n}(B_{n})\right)\\\\ & = \mathbb{P}(X^{-1}_{1}(B_{1}))\times\mathbb{P}(X^{-1}_{2}(B_{2}))\times\ldots\times\mathbb{P}(X^{-1}_{n}(B_{n}))\\\\ & = \mathbb{P}_{X_{1}}(B_{1})\times\mathbb{P}_{X_{2}}(B_{2})\times\ldots\times\mathbb{P}_{X_{n}}(B_{n}) \end{align*}
• Thank you for the response. My aim is to prove independence of the random variables $X_i$ starting with the assumption that either Statement 1 or Statement 2 is true, not the opposite.