Convergence of $\sqrt{n}\left(\frac{S_n}{T_n}-\frac{\mu_X}{\mu_Y} \right)$ where $S_n=\sum\limits_{i=1}^nX_i$ and $T_n=\sum\limits_{i=1}^nY_i$

Question

Suppose $(X_i, Y_i),i\geq 1$ are i.i.d. bivariate r.v. wit $E(X_1)=\mu_x$, $E(Y_1)=\mu_y$ and $Var(X_1)=\sigma^2_x$, $Var(Y_1)=\sigma^2_y$ and $Corr(X_1,Y_1)=\rho_{xy}$. If $X_1$ and $Y_1$ are positive r.v.s, show that $$ Z_n=\sqrt{n}\left(\frac{\sum_{i=1}^{n}X_i}{\sum_{i=1}^{n}Y_i}-\frac{\mu_x}{\mu_y} \right) $$ converges in distribution to normal distribution with mean $0$ and variance $$ \frac{1}{\mu^4_y}(\mu^2_y\sigma^2_x+\mu^2_x\sigma^2_y-2\rho_{xy}\mu_x\mu_y\sigma_x\sigma_y) $$ One of my friend solved it by approximating the denominator, but if anyone can give a simple solution, that would be great. Right now I am trying to find a rather easy to calculate solution, any hint or trick would be appreciable. No need to give the full solution.

Answer 1

Solution of the problem using the help of Did and NCh:

$$Z_n=\frac{U_n}{\mu_y V_n}$$ $$U_n=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}W_i,\space W_i=\mu_yX_i-\mu_xY_i, \space \frac{1}{n}\sum_{i=1}^{n}Y_i$$ Using CLT, we have $U_n\Rightarrow N(0,\sigma^2)$, where $\sigma^2=\mu^2_y\sigma^2_x+\mu^2_x\sigma^2_y-2\rho_{xy}\mu_x\mu_y\sigma_x\sigma_y$

$V_n\rightarrow_p \mu_y$. Note that since $Y_1>0$ a.e., we have $\mu_y>0$.

Using Slutsky's theorem, we have the final result, $$Z_n\Rightarrow N\Big{(}0,\frac{\sigma^2}{\mu^4_y}\Big)$$ which is our desired result.