Let $X_n\Rightarrow X$ and $Y_n\Rightarrow Y$ as $n\to\infty$, where all these $X$ and $Y$ are well defined random variables and $\Rightarrow$ is the convergence in distribution.
Could you give me an example in which $X_n$ and $Y_n$ are independent, but $X$ and $Y$ are not independent?
Thank you!
On
The key here is the whole concept of convergence in distribution. The fact that $X_n\stackrel{d}{\to}X$ and $Y_n\stackrel{d}{\to}Y$ but for each $n$, $X_n$ and $Y_n$ are independent tells us nothing about the joint distribution of $X$ and $Y$.
The use of characteristic functions deals with all real-valued random variables. Let each pair of $Z_n=(X_n,Y_n)$ be independent (i.e. $X_n$ is independent of $Y_n$). Then for all $\xi=(\xi_1,\xi_2)\in \mathbb{R}^2$ we have $$E[e^{i\xi \cdot Z_n}]=E[e^{i\xi_1 X_n}]E[e^{i\xi_2Y_n}]\to E[e^{i\xi_1 X}]E[e^{i\xi_2Y}]$$ Let $Z^{(1)} \sim X,Z^{(2)}\sim Y$ be s.t. $Z^{(1)}$ is independent of $Z^{(2)}$. Then define $Z=(Z^{(1)},Z^{(2)})$. We have $$E[e^{i\xi \cdot Z}]=E[e^{i\xi_1 Z^{(1)}}]E[e^{i\xi_2 Z^{(2)}}]=E[e^{i\xi_1 X}]E[e^{i\xi_2Y}]$$ So $Z_n$ converges in distribution to a pair of independent rvs each with respective distribution $\sim X,\sim Y$. However, $(X,Y)$ might have any distribution.
Let $a$ be such that $F_X(x)$ is continuous at $a$, and $b$ be such that $F_Y(x)$ is continuous at $b$. Then $F_{X_n}(a)\to F_X(a)$, $F_{Y_n}(b)\to F_Y(b)$. On the other hand, $$ \mathbb{P}(X_n\le a,Y_n\le b)=\mathbb{P}(X_n\le a)\ \mathbb{P}(Y_n\le b) $$ by independence. As a result, the LHS of the above $$ F_{X_n,Y_n}(a,b)\to F_X(a) F_Y(b) $$ so if $F_{X,Y}(a,b):=F_X(a)F_Y(b)$ then $$ \mathbb{P}(X_n\le a,Y_n\le b)\to F_{X,Y}(a,b)\quad\text{as }n\to \infty $$ and thus the pair $(X_n,Y_n)$ converges in distribution to a random variable whose CDF is given by $F_{X,Y}(a,b)$, and given its product form, $X$ and $Y$ are independent.
The only thing which remains to consider is the case when $F_X(x)$ is not continuous at $a$ (and similar for $Y$).