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Suppose a sequence of random vectors $(X_n,Y_n)$ converges jointly to some $(X,Y)$ in the weak topology. Question: If $X_n$ and $Y_n$ are independent for all $n$, are also $X$ and $Y$ independent?

This question has been answered in 1 under some hypothesis on the state space. However, since weak convergence only concerns the laws, and since independence can be inferred from the joint law, I feel like it should hold without any hypothesis on the state space.

The following seems to establish the answer if the random variables take values in separable metric spaces with their Borel $\sigma$-algebra, which we denote $(\mathcal{S}_1,\mathcal{B}_1)$ and $(\mathcal{S}_2,\mathcal{B}_2)$. We remark that the Borel $\sigma$-algebra of $\mathcal{S}_1 \times \mathcal{S}_2$ is generated by the Cartesian product of Borel sets, by separability.

Let $\mu_n$ be the law of $(X_n,Y_n)$ and let $\mu_n^1$, $\mu_n^2$ be the marginals. Denote their weak limits by $\mu$ and $\mu^1$, $\mu^2$. Consider $$ \mathcal{C} = \{ A \times B \colon A \in \mathcal{B}_1 , B \in \mathcal{B}_2, \mu^1(\partial A) = 0 = \mu^2(\partial B) \}. $$ Note that $$ \mu(\partial (A \times B)) \leq \mu(\partial A \times B) + \mu(A \times \partial B) \leq \mu^1(\partial A) + \mu^2(\partial B) = 0 $$ for $A \times B \in \mathcal{C}$. Thus, sets in $\mathcal{C}$ are sets of continuity for the measures $\mu^1$, $\mu^2$ and $\mu$, so by the Portmanteau theorem \begin{equation} \begin{split} \mathbb{P}( X \in A, Y \in B ) &= \mathbb{P}( (X,Y) \in A \times B ) = \lim_n \mathbb{P}( (X_n,Y_n) \in A \times B ) \\ &= \lim_n \mathbb{P}( X_n \in A ) \mathbb{P}( Y_n \in B ) = \mathbb{P}( X \in A ) \mathbb{P}( Y \in B ). \end{split} \end{equation} Thus, the probability measure $\mu$ and $\mu_1 \otimes \mu_2$ coincide on $\mathcal{C}$ and, hence, on the $\sigma$-algebra generated by $\mathcal{C}$ since $\mathcal{C}$ is closed under finite intersections.

Finally, it was shown in 2 that the $\sigma$-algebra generated by all sets of continuity is the entire Borel $\sigma$-algebra for separable metric spaces.

Is this proof correct? Can it be extended to more general settings or is there another proof that covers more general settings?

Cheers

Christophe Leuridan
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Florian R
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  • I reformulated a bit the first argument, and added that $\mathcal{C}$ is closed under finite intersections. But the last argument requires a proof, since $\mathcal{C}$ is only contained in the collection of all continuity sets for $\mu$. – Christophe Leuridan Nov 29 '23 at 20:10
  • I stumbled upon this post, which includes my question as a special case when considering separable metric spaces. https://math.stackexchange.com/questions/1606793/properties-of-weak-convergence-of-probability-measures-on-product-spaces – Florian R Dec 19 '23 at 09:28

1 Answers1

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The last argument should be completed a bit.

Since $\mathcal{C}$ contains all Cartesian products $A \times S_2$ where $A$ is a continuity set for $\mu_1$, $\sigma(\mathcal{C})$ contains all Cartesian products $A \times S_2$ with $A \in \mathcal{B}_1$. In the same way, $\sigma(\mathcal{C})$ contains all Cartesian products $S_1 \times B$ with $B \in \mathcal{B}_2$. Hence (by intersection) $\sigma(\mathcal{C})$ contains all Cartesian products $A \times B$ with $A \in \mathcal{B}_1$ and $B \in \mathcal{B}_2$. Thus $\sigma(\mathcal{C})$ contains $\mathcal{B}_1 \otimes \mathcal{B}_2$.

Christophe Leuridan
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    Thanks for filling in some details! I will leave the question and bounty open for now since I am still interested if someone can cook up a method that covers more general settings. I was hoping my approach for separable metric spaces could serve as inspiration for generalizations. But I will pick your answer if nothing shows within a week. – Florian R Nov 30 '23 at 05:27