Calculate limiting distribution of $\frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}$ [closed]

Let $$X_1,X_2,\ldots,X_n$$ be a random sample of Bernoulli distribution with parameter $$\frac{\theta_1}{\theta_1+\theta_2}$$ and $$Y_1,Y_2,\ldots,Y_n$$ be a random sample of geometric distribution with parameter $$\theta_1+\theta_2$$. If the two samples are independent of each other, how can I calculate the limiting distribution of $$\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}$$ ?

closed as unclear what you're asking by Did, Alex Provost, Namaste, Juniven, projectilemotionMar 26 '17 at 1:32

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By the law of large numbers, the ratios converge almost surely to $E(X_1)/E(Y_1)$ hence they converge in distribution to the same value. This uses that $(X_n)$ is i.i.d. and that $(Y_n)$ is i.i.d. but not the independence of $(X_n)$ and $(Y_n)$.
For all $i$, the expectation of $X_i$ is $\mu_X = \frac{\theta_1}{\theta_1 + \theta_2}$ and its variance is $\sigma_X^2 = \frac{\theta_1 \theta_2}{(\theta_1 + \theta_2)^2}$ . Similarly, the expectation of $Y_i$ is $\mu_Y = \frac{1}{\theta_1 + \theta_2}$ and its variance is $\sigma_Y^2 = \frac{1 - \theta_1 - \theta_2}{(\theta_1 + \theta_2)^2}$ .
Now, let us look for an asymptotic distribution. By virtue of the central limit theorem, we know the limiting distributions of the sum of i.i.d. variables: $\sum_{i=1}^n X_i \underset{n\rightarrow\infty}{\sim} \mathcal{N}(n\mu_X,n\sigma_X^2)$ and $\sum_{i=1}^n Y_i \underset{n\rightarrow\infty}{\sim} \mathcal{N}(n\mu_Y,n\sigma_Y^2)$ . The ratio of two independent unstandardized normal variables has a complex distribution, which can be expressed by Hermite polynomials (see Pham-Gia, 2006).
Note: alternatively, we can provide a limiting distribution based on the standardized version of the central limit theorem, under the form $\frac{\overline{X}_n -\mu_X}{\sigma_X/\sqrt{n}} \underset{n\rightarrow\infty}{\sim} \mathcal{N}(0,1)$ and $\frac{\overline{Y}_n -\mu_Y}{\sigma_Y/\sqrt{n}} \underset{n\rightarrow\infty}{\sim} \mathcal{N}(0,1)$ , where $\overline{X}_n$ and $\overline{Y}_n$ denote the sample averages. Since the ratio of two independent standardized normal variables is a Cauchy variable with parameters $0$ and $1$, the transformation properties of Cauchy variables yield \begin{equation} \frac{n\mu_X - \sum_{i=1}^n X_i}{n\mu_Y - \sum_{i=1}^n Y_i} \underset{n\rightarrow\infty}{\sim} \text{Cauchy}(0,\sigma_X/\sigma_Y) \, , \end{equation} which does not completely answer the question.