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.

  • 3
    $\begingroup$ $S_n/n$ does converge in distribution in the Cauchy case. In fact, it does in a very strong sense: The sequence of distributions is constant! $\endgroup$
    – cardinal
    Jan 12, 2016 at 23:27

1 Answer 1


The problem with the Cauchy distribution is that the sample mean has itself the same Cauchy distribution. Thus, as far as I can see, terms such as

$$\frac{\sum_{i=1}^n X_i}{n^p}$$

for $p>1$, will converge in probability to zero as $n \to \infty$.

  • $\begingroup$ +1. What if anything can we say about the almost sure properties of $S_n / n^p$? $\endgroup$
    – dsaxton
    Jan 13, 2016 at 0:43
  • $\begingroup$ Can we generalize this result to any distribution (does not necessarily require the existence of mean and variance)? $\endgroup$
    – Sheldon
    Jan 13, 2016 at 3:56
  • $\begingroup$ @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. $\endgroup$
    – JohnK
    Jan 13, 2016 at 13:06
  • $\begingroup$ 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! $\endgroup$
    – Sheldon
    Jan 13, 2016 at 15:09

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