Convergence of Cauchy sample mean If $X_i$'s are iid Cauchy random variables, does the sample mean $\frac{\sum^{n}_{i=1}X_i}{n}$ converge in distribution to any limit? If yes, can someone explain to me how the limit can be computed step by step? I don't think it would, since the variance of a Cauchy random variable is infinity, but I am not so sure about this.
For every $n$ $\frac 1 n \sum\limits_{i=1}^{n} X_i$ also has the Cauchy distribution (!) so the Cauchy means certainly converge in distribution to the Cauchy distribution. The easiest way to prove this is to use characteristic functions: $Ee^{it \frac 1 n \sum\limits_{i=1}^{n} X_i}=(Ee^{itX_1/n})^{n}=(e^{-|t/n|})^{n}=e^{-|t|}$.
For the additional question in the comment below I will give a hint: If $S_n =\frac 1 n \sum\limits_{i=1}^{n} X_i$ converges in probability then $S_{2n}-S_n \to 0$ in probability, hence in distribution. So its characteristic function must tend to $1$ at every point. But the characteristic function converges to $e^{-|t|}$.
For every $n$ $\frac 1 n \sum\limits_{i=1}^{n} X_i$ also has the Cauchy distribution (!) so the Cauchy means certainly converge in distribution to the Cauchy distribution. The easiest way to prove this is to use characteristic functions: $Ee^{it \frac 1 n \sum\limits_{i=1}^{n} X_i}=(Ee^{itX_1/n})^{n}=(e^{-|t/n|})^{n}=e^{-|t|}$.
For the additional question in the comment below I will give a hint: If $S_n =\frac 1 n \sum\limits_{i=1}^{n} X_i$ converges in probability then $S_{2n}-S_n \to 0$ in probability, hence in distribution. So its characteristic function must tend to $1$ at every point. But the characteristic function converges to $e^{-|t|}$.