# Suppose $X_1, \dots, X_n, Y$ are independent random variables. Prove that $X = (X_1, \dots, X_n)$ and $Y$ are independent variables.

Suppose $$X_1, \dots, X_n, Y$$ are independent random variables. Prove that $$X = (X_1, \dots, X_n)$$ and $$Y$$ are independent variables.

My attempt:

Fix $$A \in \mathcal{R}$$ (a Borel subset of the real numbers). Define measures

$$\mu_1(B) = P(Y \in A, X \in B), B \in \mathcal{R}^n$$

$$\mu_2(B) = P(Y \in A)P(X \in B), B \in \mathcal{R}^n$$

We have to show that $$\mu_1 = \mu_2$$ and then we will have shown that $$P(Y \in A, X \in B) = P(Y \in A)P(X \in B)$$, which is what we needed to prove.

Observe that $$\mathcal{R}^n = \sigma(\mathcal{P})$$ where

$$\mathcal{P}:= \{A_1 \times\dots \times A_n \mid A_1 , \dots, A_n \in \mathcal{R}\}$$

By independence of $$X_1, \dots, X_n, Y$$, we see that $$\mu_1$$ and $$\mu_2$$ are equal on $$\mathcal{P}$$ and by the unicity theorem of measures they are equal on $$\mathcal{R}^n$$.

Is this proof correct? Is there a way to avoid the uniqueness theorem?

If two probability measures $$P$$ and $$Q$$ are equal on class of sets generating a sigma algebra you cannot immediately conclude that they are equal on the sigma algebra. This requires a proof unless the class you started with is an algebra. In this case sets of the form $$X_1^{-1}(A_1)\cap ..\cap X_n^{-1}(A_n)$$ do not form an algebra. This class of sets is closed under finite intersections and the standard argument for this is to apply Dynkin's $$\pi -\lambda$$ theorem: $$\{A:P(A)=Q(A)\}$$ is a $$\lambda$$ system containing the given $$\pi$$ system so it contains the generated sigma algebra.
• +1, one can also argue by noting $\mathcal{P}$ forms a semialgebra, and any $\Bbb R \cup \{\infty\}$-valued nonnegative function $\mu$ on $\mathcal{P}$ which satisfies $\mu(\emptyset) = 0$, $\mu(\sqcup_{i = 1}^n A_i) = \sum_{i = 1}^n \mu(A_i)$ for any finite disjoint collection $(A_i)_{i = 1}^n \in \mathcal{P}$ and $\mu(\cup_{i \geq 1} A_i) \leq \sum_{i \geq 1} \mu(A_i)$ for any collection $(A_i)_{i \geq 1} \in \mathcal{P}$, extends to a measure on $\sigma(\mathcal{P})$, and conclude by Caratheodory extension theorem (as the relevant measures are probability measures, they are $\sigma$-finite) – Balarka Sen May 19 '19 at 1:03
• (Although the proof of Caratheodory extension theorem itself factors through Dynkin's $\pi-\lambda$ theorem; I suppose I find the former conceptually easier to invoke than the latter) – Balarka Sen May 19 '19 at 1:07