I've $X_i$ random variable iid such that $E(X_i) = \mu_X$ and $V(X_i) = \sigma_X^2$ both finite, with $\mu_X \ne 0$. Also $Y_i$ iid with $E(Y_i) = \mu_Y$ and $V(Y_i) = \sigma_Y^2$. By definition $\bar X_n = \frac{1}{n}\sum_{i=1}^n X_i$.

I've to analyze the convergence in distribution of $\sqrt{n}(\frac{\bar Y_n}{\bar X_n} - \frac{\mu_Y}{\mu_X})$.

I know by CLT that $\sqrt{n}\frac{(\bar Y_n - \mu_Y)}{\sigma_Y}$ converge in distribution to the normal distribution, ie $N(0, 1)$.

Also by the weak law of great numbers $\bar X_n$ converge in probability to $\mu_X$, and since $\mu_X \ne 0$, I also have $\frac{1}{\bar X_n}$ converge in probability to $\frac{1}{\mu_X}$.

So if I can "factor out" $\frac{1}{\bar X_n}$ then I should be able to use Slutzky's theorem $\sqrt{n}(\bar Y_n - \mu_Y) \times \frac{1}{\bar X_n}$. First part converge in distribution, the second part converge in probability, so the product will converge in distribution.

But I'm unable to remove either $\bar X_n$ nor $\mu_X$. Any idea how to proceed?


An important detail I've left out is that $X_i, Y_j$ are independent of each other for every choice of $i,j$.


The difference between what you want, $\sqrt n ( \bar Y_n / \bar X_n - \mu_Y/\mu_X),$ and what Slutzky allows you to have, $\sqrt n (\bar Y_n/\bar X_n - \mu_Y/\bar X_n)$, is $\Delta = \sqrt n \mu_Y ( 1/\bar X_n - 1/ \mu_X)$. I think you want $\Delta \to 0$ in probability. By the delta method and the moment hypotheses on $X$, it looks like $\Delta$ has a non-trivial limiting gaussian distribution of its own, instead.

You should have started with the 2 dimensional CLT for $(X,Y)$, and applied the 2 dimensional delta method. What you had gotten was the $\partial / \partial y\, (y/x)$ part, but you left off the $\partial/\partial x \,(y/x)$ term.

  • $\begingroup$ Sorry, but I've missed an important detail $X_i$ and $Y_j$ are independant of each other for every choice of $i,j$. $\endgroup$ – Ismael Jul 18 '17 at 0:59

One possible solution was start with $$\sqrt{n}\left(\frac{\bar{Y}_n}{\bar{X}_n} - \frac{\mu_Y}{\mu_X}\right) = \sqrt{n}\times\frac{\bar{Y}_n \mu_X - \bar{X_n} \mu_Y}{\bar X_n \mu_X}$$

Now consider $W_n = Y_n \mu_X - X_n \mu_Y$ you can prove that $W_n$ has all the conditions to apply the C.L.T. So you have that $\sqrt{n}\bar{W}_n$ converge in distribution to a normal.

The remaining part is $\frac{1}{\bar{X}_n \mu_X}$. We knowe $\bar{X}_n \rightarrow \mu_X$ in probability by the Weak Law of Large Numbers, then also $\frac{1}{\bar{X}_n} \rightarrow \frac{1}{\mu_X}$ in probability.

With both results and applying Slutzky theorem we have the final result.


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