Convergence in distribution of independent random variables

Question

Let $(X_n)_{n\in \mathbb N}$ and $(Y_n)_{n\in \mathbb N}$ be random sequences independent of each other. Further suppose both $X_n$ and $Y_n$ converge in distribution (to some random variable).

I know that if $X'$ and $Y'$ are weak limits of $X_n$ and $Y_n$, then they need not be independent.

My question is, can I (always) find "independent" random variables $X$ and $Y$ that are weak limits of $X_n$ and $Y_n$ respectively?

Thanks and regards.

Answer 1

Yes, you can.

Suppose you are given two limit distributions $X'$ and $Y'$ which are dependent. What you can always do is to take an $X$ (random variable) which is equal in distribution to $X'$ but which is independent of $Y'$. You can view this $X$ as a random sample of size $1$ from the distribution of $X'$.

I think this explanation is a bit vague, so don't hesitate to comment. And I hope this helps!

Answer 2

Under the stated conditions, the pair $(X_n,Y_n)$ converges in distribution, say to a pair $(X',Y')$, and then $X'$ and $Y'$ are necessarily independent.

Answer 3

Let $X'$ & $X$ be $S_1$-valued r.v.s, and $Y'$ & $Y$ be $S_2$-valued ones. We denote the distribution that r.v. $X$ follows as $P_X(\cdot)$ ($= P(X \in \cdot)$, and similarly for other r.v.s.

We can construct a product measure $P_{X'} \otimes P_{Y'}$ on $(S_1 \times S_2, \sigma (\mathcal{P}) )$ that satisfies

$$ \forall A \in \mathcal{B}(S_1), B \in \mathcal{B}(S_2) \quad P_{X'} \otimes P_{Y'}(A \times B) = P_{X'} (A) P_{Y'} (B). $$

where $\mathcal{B}(\cdot)$ describes the Borel family and $\mathcal{P}$ is the family of "Borel rectangles", i.e.

$$ \mathcal{P} = \{ E \times F \mid E \in \mathcal{B}(S_1), \ F \in \mathcal{B}(S_2) \}. $$

When you think of a distribution of r.v.s $X$ & $Y$ as

$$ P_{(X, Y)}(A,B) = P_{X'} \otimes P_{Y'}(A,B) $$

then

\begin{eqnarray*} && P_X(A) = P_{(X, Y)}(A,\mathbb{R}) = P_{X'} \otimes P_{Y'}(A \times \mathbb{R}) = P_{X'}(A), \\ && P_Y(B) = P_{(X, Y)}(\mathbb{R},B) = P_{X'} \otimes P_{Y'}(\mathbb{R} \times B) = P_{Y'}(B). \end{eqnarray*}

(We used the last equal of the each equation the definition of $P_{X'} \otimes P_{Y'}$.)

We can say $X$ & $Y$ are independent, as

$$ P_{(X, Y)}(A,B) = P_{X'} \otimes P_{Y'}(A \times B) = P_{X'} (A) P_{Y'} (B) = P_X(A) P_Y(B). $$

I am not sure if this answer is clear enough...