I have a question regarding convergence in distribution, which arose during one of my research works. First, let me describe the setup a bit:
$\textbf{Description of the Problem}$:
Suppose that $X_n$ and $Y_n$ are two $\textbf{independent}$ sequences of random variables, such that $\sqrt{n} X_n \xrightarrow{d} F$ and $\sqrt{n}Y_n \xrightarrow{d} F$, where $F$ is a continuous probability distribution. As a consequence, $\frac{X_n}{Y_n} \xrightarrow{d} \frac{X}{Y}$, where $X$ and $Y$ are independent observations drawn from $F$.
Now, suppose that $W_n$ and $Z_n$ are two sequences of random variables such that the joint sequences $(X_n,W_n)$ and $(Y_n,Z_n)$ are independent with each other ($X_n$ and $W_n$ may be dependent, and similarly, $Y_n$ and $Z_n$ may be dependent). Also, assume that $X_n - W_n \xrightarrow{d} 0$ and $Y_n - Z_n \xrightarrow{d} 0$.
$\textbf{My Question}$:
Does the above setting imply that: $$\frac{W_n}{Z_n}\xrightarrow{d} \frac{X}{Y}~?$$
In my research, $F$ is the standard normal distribution $N(0,1)$, so I would also be happy if you can prove the claim with this additional assumption. And in case the claim is false, are there other technical assumptions which make it true?
Many thanks.
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $\sqrt{n}(W_n-X_n) \to_d 0$ and $\sqrt{n}(Z_n-Y_n) \to_d 0$. Then one can show that (by e.g. Portmanteau lemma) $$ \sqrt{n}(X_n - W_n) \to_d 0 \Leftrightarrow \sqrt{n}(X_n - W_n) \to_p 0. $$ Thus, one can conclude that $$ \sqrt{n}W_n = \sqrt{n}X_n +\sqrt{n} (W_n - X_n) \to_d F. $$ This is because $A_n \to_d A$ and $B_n \to_p c$ implies $(A_n,B_n) \to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n \to_d A+c$. In the same spirit, we have $Z_n \to_d F$, and hence $(W_n, Z_n) \to_d (X,Y)$ where $X,Y\sim F$ and $X\perp \!\!\!\perp Y.$ Continuous mapping theorem gives $$ \frac{W_n}{Z_n}\to_d \frac{X}{Y} $$as wanted.