0
$\begingroup$

Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.

Let $Z_n$ = $X_n$ + $Y_n$

I want to study the law convergence of the sample mean of $Z_n$. That is:

$$ \overline{Z_n} = \frac{\sum X_i + \sum Y_i}{n} $$

So, first of all there is an hint: I cannot use the Law of large numbers. Why is that?

Anyway, I am really out of tools to attack this problem. Does someone have a hint?

EDIT: Managed to prove that sample mean of Cauchy is still Cauchy! Still not sure about the solution.

$\endgroup$
1
$\begingroup$

Hint: what does $\sum X_i / n$ converge to? As you note, the sum $\sum Y_i / n$ is equal in distribution to a standard Cauchy.

$\endgroup$
  • $\begingroup$ I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$ $\endgroup$ – qcc101 Jan 15 at 17:08
  • $\begingroup$ @qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $\overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't. $\endgroup$ – Marcus M Jan 15 at 17:10
  • $\begingroup$ I think I would try to use law of large numbers and then Slutsky to conclude, is that right? $\endgroup$ – qcc101 Jan 15 at 17:15
  • $\begingroup$ @qcc101, yep that'll do it! $\endgroup$ – Marcus M Jan 15 at 17:17
  • $\begingroup$ Nice, thank you for the follow-ups it was exactly how I wanted to solve it. $\endgroup$ – qcc101 Jan 15 at 17:19

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.