0
$\begingroup$

I'm investigating the behaviour of some random variables obtained from standard Cauchy random variables $X_n$. Suppose $Y_n=\textrm{sgn}(X_n)|X_n|^{\alpha}$ for $\alpha\in[0,1]$. Let $S_n=Y_1+\dots+Y_n$. It's fairly easy to see that $S_n/n$ converges in distribution to another Cauchy random variables for $\alpha=1$, by simply considering the characteristic function.

My question is, how would I compute the limit of $S_n/n$ for $\alpha<1$? The characteristic functions of the $Y_n$ seem a bit awkward here, and the Central Limit Theorem only tells me about $S_n/\sqrt{n}$. The only thing I've gleaned thus far is that the $Y_n$ have mean $0$ in the case $\alpha<1$, and are obviously iid since the $X_n$ are. Any help would be much appreciated!

$\endgroup$
3
  • 4
    $\begingroup$ When $\alpha\lt1$, the random variables $Y_n$ are integrable hence $S_n/n\to0$ almost surely, in $L^1$, in probability and in distribution. $\endgroup$
    – Did
    Commented May 28, 2012 at 16:54
  • $\begingroup$ Ah right so I use the strong law of large numbers and the fact the convergence a.s. implies convergence in distribution? $\endgroup$
    – Edward Hughes
    Commented May 28, 2012 at 16:58
  • $\begingroup$ Do you need to think of convergence in distribution if you know that something converges to a constant a.s.? $\endgroup$
    – SBF
    Commented May 29, 2012 at 5:53

1 Answer 1

Reset to default
0
$\begingroup$

For the sake of having an answer:

When $\alpha\lt1$, the random variables $Y_n$ are integrable hence $S_n/n\to0$ almost surely, in $L^1$, in probability and in distribution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .