I have a question.

Let $X_n$ converge to $X$ in distribution, on the other hand, $Y_n$ converges to $Y$. What can we obtain about convergence of division of $X_n/Y_n$ in distribution? Does it converge to division of $X / Y$ in distribution? For more study, is there any references which explain my question in details?

Any help, hint will be greatly appreciated.

  • 2
    $\begingroup$ Unless you have very strong information about the joint distribution of $(X_n,Y_n)$ it's difficult to see how you could draw any conclusions at all about their ratios. Indeed, forget about convergence for a moment and contemplate what you might be able to say about $X/Y$ based only on information about the marginal distributions of $X$ and $Y:$ I hope it's obvious that the distribution of $X/Y$ is not determined unless at least one of them is almost surely constant. $\endgroup$
    – whuber
    Dec 26 '18 at 16:21
  • $\begingroup$ Might take a look at this: en.wikipedia.org/wiki/Slutsky%27s_theorem $\endgroup$
    – olooney
    Dec 26 '18 at 17:39
  • 1
    $\begingroup$ Well, Slutsky (avery good friend) assumes that one of the sequences converges in probability. Convergence in distributions to a constant implies Convergence in probability $\endgroup$ Dec 26 '18 at 18:32
  • 2
    $\begingroup$ Just to make absolutely sure, does $Y_n$ converge to $Y$ in distribution, or is $Y$ a constant, or... ? $\endgroup$
    – jbowman
    Dec 26 '18 at 20:56

In the comments there are some requests for clarification, you should really respond. As it stands there is not really enough information! Specifically, you did not specify the joint distribution of the $(X_n, Y_n)$. Without some knowledge about that, we cannot answer.

So, first let us assume independence. Then you should be able to prove that $(X,Y)$ also are independent. There could still be some problems. Observe that $x/y =f(x,y)$ and $f$ is not defined for $y=0$ and not continuous in the second argument $y$ in any punctured (at zero) interval around $y=0$. So think about what would happen if $Y$ has probability mass around there.

Without independence: Have a look at Slutsky's theorem, but then you need that either $X$ or $Y$ is a constant (almost surely).


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.