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.

  • $\begingroup$ +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) $\endgroup$ – Balarka Sen May 19 '19 at 1:03
  • $\begingroup$ (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) $\endgroup$ – Balarka Sen May 19 '19 at 1:07
  • $\begingroup$ The class generating it in my case is a pi system so I think it works? (Recalling that if two measures on a sigma algebra are equal on a pi system that generates the sigma algebra and they are sigma finite on the pi system, then they are equal on the whole sigma algebra). $\endgroup$ – user661541 May 19 '19 at 6:37
  • $\begingroup$ Yes, that works. $\endgroup$ – Kavi Rama Murthy May 19 '19 at 7:05
  • $\begingroup$ Thanks for the help! $\endgroup$ – user661541 May 19 '19 at 7:19

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