This is something that has come up in my research. I originally posted this question on CrossValidated but realized it might be better suited for this site. I have deleted the question there (in case it annoys someone).

Let $X_1, X_2, \dots, $ be a sequence of iid Cauchy$(x, 1)$ random variables for some $x \in \mathbb{R}$. Define, $$ S_n(x) = \frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n |X_i|},\quad n \in \mathbb{N}. $$ What, if it exists, is the limit of $S_n(x)$ as $n \to \infty$? I don't think an almost sure limit exists. Monte Carlo simulations suggest there should be a non-trivial limit. Perhaps, $S_n$ converges in distribution or in probability.

For fixed $n = 20000$, below are the density plots of $S_n(x)$ for different $x$. For each $x$, I generated $20000$ instances of $S_n(x)$ and used them to estimate the density. On the $x-$axis is the location parameter $x$.

enter image description here

A paper I found in an answer to another question studies the above limit for stable distributions. It conveniently ignores the Cauchy case. They cite Feller's second volume but I haven't been able to find a result there. I wonder if any references consider the above limit for Cauchy random variables.

Another excellent answer I found concerning the median of the sum of half-Cauchy random variables. I think it suggests that the half-Cauchy might somehow belong to the domain of attraction of an asymmetric $\alpha=1$ stable law. If this is so, it might help understand the limit of $S_n$. But I am not sure.

  • $\begingroup$ Noting the answer of mike: This is one example where simulation is unfortunately of no use. $\endgroup$ Feb 28 at 19:27

1 Answer 1


It converges to 0 in probability. $\frac{\Sigma_1^nX_i}n $ always has the same cauchy distribution while $\frac{\Sigma_1^n|X_i|}n \rightarrow \infty $ by the law of large numbers

  • $\begingroup$ It turns out the convergence is extremely slow. The simulations are so misleading as a result. $\endgroup$ Mar 14 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.