Probability question: proving independence.

Question

Let $X,Y,X_n,Y_n: \Omega \rightarrow \mathbb{R}, n \ge 1$ be random variables.

1. If for all $n\ge1$, $X_n$ and $Y_n$ are independent and if $(X_n,Y_n) \xrightarrow[]{d}(X,Y)$, then X and Y are independent.

I found that if two random variables are independent then for $(\xi_1,\xi_2) \in\mathbb{R}^2$ $$\mathbb{E}(e^{i\langle (\xi_1,\xi_2), (X_n,Y_n)\rangle})=\mathbb{E}(e^{i\langle\xi_1, X_n\rangle})\cdot\mathbb{E}(e^{i\langle\xi_2, Y_n\rangle})\\ \implies \lim_{n\rightarrow \infty}\mathbb{E}(e^{i\langle (\xi_1,\xi_2), (X_n,Y_n)\rangle})=\lim_{n\rightarrow \infty}\mathbb{E}(e^{i\langle\xi_1, X_n\rangle})\cdot\mathbb{E}(e^{i\langle\xi_2, Y_n\rangle})=\mathbb{E}(e^{i\langle (\xi_1,\xi_2), (X,Y)\rangle})$$ But since it isn't given that $X_n,Y_n$ are weakly convergent then how do I proceed?

It feels like I am missing something very trivial, I am trying to read Brownian Motion An Introduction to Stochastic Processes by Schilling. It feels like there is a lot that I already need to know before tackling this book. I have had a basic course in Stochastic Processes, can someone recommend a book to bridge the gap.

Answer 1

$(X_n,Y_n)$ converging to $(X,Y)$ in distribution implies that $X_n$ converges to X in distribution and $Y_n$ converges to Y in distribution so your argument using characteristic functions is correct; it is also the best way to prove the result.

Answer 2

I guess the idea here is that you go to the definitions and the elementary properties.

The hypothesis is that for each $n\in\mathbb N$ the r.v. $X_n$ and $Y_n$ are independent, and a theorem/property/whatever states that this happens if and only if (important that is an equivalence, not a one way implication!) $$F_{X_nY_n}(x,y)=F_{X_n}(x)F_{Y_n}(y)$$ (that is, the joint cumulative distribution is the product of both marginal cumulative distributions).

You also know that $(X_n,Y_n)\xrightarrow{\mathcal D}(X,Y)$ which means that $F_{X_nY_n}(x,y)\to F_{XY}(x,y)$ as $n\to \infty$ (except maybe for points $(x,y)$ where $F_{XY}$ is not continuous).

Imagine that you could prove, maybe as a consequence of the aforementioned, that $F_{XY}(x,y)=F_X(x)\cdot F_Y(y)$ (not evident at all!), what would you conclude?

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UPDATE: Resolution

Well, the answer is that $F_{XY}(x,y)=F_X(x)\cdot F_Y(y)$ would mean that $X$ and $Y$ are independent (because the property I mentioned above is an equivalence, an if and only if).

So this reduces to prove that $$F_{X_nY_n}(x,y)=F_{X_n}(x)F_{Y_n}(y)\to F_{X}(x)F_Y(y)\text{ as } n\to \infty.$$

But this will be true if we show that $F_{X_n}(x)\to F_X(x)$ and the same for $Y$.

Now, as $F_{X_n}(x)=\lim_{y\to+\infty}F_{X_nY_n}(x,y)$, we can say that $$\lim_{n\to\infty}F_{X_n}(x)=\lim_{n\to\infty}\lim_{y\to+\infty}F_{X_nY_n}(x,y)$$ and if limits are to be swaped, then $$\lim_{n\to\infty}F_{X_n}(x)=\lim_{y\to+\infty}\lim_{n\to\infty}F_{X_nY_n}(x,y)=\lim_{y\to+\infty}F_{XY}(x,y)=F_X(x).$$

In the same way, we can prove that $$\lim_{n\to\infty}F_{Y_n}(y)=F_Y(y),$$ and so $$F_{XY}(x,y)=\lim_{n\to\infty}F_{X_nY_n}(x,y)=\lim_{n\to\infty}F_{X_n}(x)F_{Y_n}(y)=F_X(x)F_Y(y),$$ which proves $X$ and $Y$ are independent.

Notes:

1) The swapping of the limits is justified by the fact that $\lim_n F_{X_n}(x,y)$ is finite for every $y\in\mathbb R$ (except maybe at discontinuities of $F_{XY}$ which are at most countably infinite) as well as $\lim_{y\to+\infty}F_{X_nY_n}(x,y)$ is finite for every $n\in \mathbb N$ and the convergence is uniform.

2) It's worth noting that incidentally it's also been proven that if $(X_n,Y_n)\xrightarrow{\mathcal{D}}(X,Y)$, then $X_n\xrightarrow{\mathcal{D}}X$ and $Y_n\xrightarrow{\mathcal{D}}Y$.