# convergence of Cauchy distribution

It is known that the Large Number Theorem does not apply to Cauchy distribution since it does not have an expectation value. That said, $S_n / n$ does not converge in any sense (almost sure, in probability, in distribution) where $S_n$ is the sum of $n$ i.i.d. Cauchy variables.

My question is: whether $S_n / n^2$ or $S_n / n^3$ will converge in any form?

The motivation of this question is that $S_n / n$ converges (both a.s. and in probability) according to Large Number Theorem, $S_n / \sqrt{n}$ converges in distribution to $\mathcal N(0,1)$ according to Central Limit Theorem. After learning the Law of Iterated Logarithm, I have a guess that we can make some similar expressions converge even for Cauchy variables.

• $S_n/n$ does converge in distribution in the Cauchy case. In fact, it does in a very strong sense: The sequence of distributions is constant! – cardinal Jan 12 '16 at 23:27

$$\frac{\sum_{i=1}^n X_i}{n^p}$$
for $p>1$, will converge in probability to zero as $n \to \infty$.
• +1. What if anything can we say about the almost sure properties of $S_n / n^p$? – dsaxton Jan 13 '16 at 0:43
• @Sheldon I suppose that if you can show that the second term of $A_n B_n$ is bounded in probability - and we know that convergence in distribution implies boundedness in probability- while the first term goes to zero, then the result follows. – JohnK Jan 13 '16 at 13:06
• I think you are correct. For the numerator, we can take its maximum value by setting every $X_i$ to the maximum value $X_{\rm{max}}$ in the support range. Then after dividing the numerator by $n$, we get $X_{\rm{max}}$. It is easy to show that $\frac {X_{\rm{max}}} {n^{p-1}}$ converge in probability to zero as $n$ goes to infinity. Thanks! – Sheldon Jan 13 '16 at 15:09