Convergence in distribution for correlated random variables

Question

What can be said about the convergence in distribution for the following expression

$\sqrt{n}\left(\frac{\sum_{i=1}^{n} X_{i}}{\sum_{i=1}^{n} Y_{i}}\right)$

Here the sequence of random variable ${X_i, Y_i}$ have finite means, variances, and correlated as defined by $\rho$

Does above expression converge to some Normal distribution? How can I show this if that is the case?

Answer 1

Denote $(Z_n,T_n)' = (\frac{1}{n}\sum_{i=1}^nX_i,\frac{1}{n}\sum_{i=1}^nY_i)'$, according to the central limit theorem:

$$ \sqrt{n}\left(\pmatrix{Z_n\\T_n}-\pmatrix{\mu_X\\\mu_Y}\right)\xrightarrow{\mathcal{D}} \mathcal{N}_2\left(\pmatrix{0\\0},\Sigma \right) $$ where $\Sigma$ the covariance matrix of $(X_i,Y_i)'$

Applying the multivariate delta method to the function $h(z,t)=z/t$ and suppose that $\mu_Y \ne 0$, we have then: $$ \sqrt{n}\left(\frac{\sum_{i=1}^{n} X_{i}}{\sum_{i=1}^{n} Y_{i}}-\frac{\mu_X}{\mu_Y}\right) = \sqrt{n}\left(h(Z_n,T_n)-\frac{\mu_X}{\mu_Y}\right)\xrightarrow{\mathcal{D}} \mathcal{N}_2\left(\pmatrix{0\\0},\Gamma'\cdot \Sigma \cdot \Gamma \right) $$ With $\Gamma=\left(\frac{\partial h}{\partial z}(\mu_X,\mu_Y),\frac{\partial h}{\partial t}(\mu_X,\mu_Y) \right)' = \left(\frac{1}{\mu_Y},-\frac{\mu_X}{\mu_Y^2}\right)'$

Answer 2

Assuming $(X_i, Y_i)$ are i.i.d., by the central limit theorem $$\sqrt{n}((\bar{X}_n, \bar{Y}_n) - (E(X_1), E(Y_1))) \to_d N(0, \Sigma),$$ where $\Sigma = Cov((X_1, Y_1), (X_1, Y_1)^T)$. For function $f$, by the delta method, $$\sqrt{n}(f(\bar{X}_n, \bar{Y}_n) - f(E(X_1), E(Y_1))) \to_d Df(E(X_1),E(Y_1))N(0,\Sigma) = N(0, Df(E(X_1),E(Y_1))\Sigma Df(E(X_1),E(Y_1))^T),$$ provided $f$ is differentiable at $(E(X_1), E(Y_1))$. For example, if $E(Y_1) \neq 0$, then taking $f(x, y) = \frac{x}{y}$ yields a result for $\sqrt{n}(\frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i} - \frac{E(X_1)}{E(Y_1)})$.