Asymptotic distribution of $\frac{\overline x_n+ \overline y_n}{\overline x_n- \overline y_n}$

Question

Let $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ be two independent random samples from $X$ and $Y$. We have $\mu_X = E (X ) > 0, \mu_Y = E (Y ) > 0$ and $\sigma^2_X = Var (X )$ and $\sigma^2_Y = Var (Y )$. Derive the asymptotic distribution of $\frac{\overline x_n+ \overline y_n}{\overline x_n- \overline y_n}$, where ${\overline x_n}$ is the sample average of the $x_i$s.

So I looked at the following first:

It is known as the central limit theorem that the sample mean of any distribution has an asymptotic Gaussian distribution. So both Xn & Yn have a Gaussian distribution.

second: the sum of 2 Gaussian random variable is a gaussian random variable, the mean is just the sum of each Xn and Yn, the variance is also the sum of the individual variances since theres independence of each Xi third: the distribution of the ratio is more complicated since it involves doing an integral I guess.

So this would be the intuition could someone please help me formalize it especially step 3, I don’t have the mathematical capacity yet to do it.

Really need the help.

Answer 1

Assuming $(X_n)$ and $(Y_n)$ are two i.i.d sequences with $E|X_1|<\infty$ and $E|Y_1|<\infty$.

By the law of large numbers, we have the following convergence in probability:

$$T=\frac{\frac1n\sum\limits_{i=1}^n X_i+\frac1n\sum\limits_{i=1}^n Y_i}{\frac1n\sum\limits_{i=1}^n X_i-\frac1n\sum\limits_{i=1}^n Y_i}\stackrel{P}\longrightarrow\frac{E(X_1)+E(Y_1)}{E(X_1)-E(Y_1)}\,,$$

provided $E(X_1)\ne E(Y_1)$.

This only uses that $(X_n)$ and $(Y_n)$ are (separately) i.i.d, not the fact that they are independent.

As a result $T$ also converges in distribution to $\frac{E(X_1)+E(Y_1)}{E(X_1)-E(Y_1)}$.

I am not sure if we can say anything about a non-degenerate asymptotic distribution of $T$ in general.

Answer 2

Don't know the answer but by the CLM you know that:
$\sqrt{n}({\overline x_𝑛}-µ_x) \rightarrow_p N(0,σ^2_X)$

Maybe someone else can chime in, as to how to proceed.