1
$\begingroup$

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?

Edit:

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.