Suppose $X_1, \ldots, X_n$ are iid standard Cauchy random variables, does $\frac{1}{n}\sum_{i=1}^{n}X_i$ converge in probability or almost surely? Suppose $X_1, \ldots, X_n$ are iid standard Cauchy random variables, does $\frac{1}{n}\sum_{i=1}^{n}X_i$ converge in probability or almost surely?
I know that by stable law, $\frac{1}{n}\sum_{i=1}^{n}X_i \sim Cauchy(0,1)$, but how can I find if it converges in probability or not by the definition of convergence in probability? 
 A: Let $Z_n = \frac{1}{n} \sum_{i=1}^n X_i$. 
Now $$Z_{2n}  - Z_{n} = \underset{A_n}{\underbrace{-\frac{1}{2n} \sum_{i=1}^n X_i}} +\underset{B_n}{\underbrace{\frac{1}{2n} \sum_{i=n+1}^{2n} X_i}}.$$
Then $A_n$ and $B_n$ are independent identically distributed Cauchy $(0,\frac 12)$. Hence, $A_n + B_n $ is Cauchy $(0,1)$. 
In particular, 
$$P(|Z_n -Z_{2n}|>\epsilon )=c,$$ 
where $c$ is the probability that Cauchy $(0,1)$ has absolute value larger than $\epsilon$ (explicitly, $c= \frac{2}{\pi} \arctan\epsilon$).  This is a constant independent of $n$. 
From this we conclude that the sequence $(Z_n)$ does not converge in probability and therefore also not a.s. It is Cauchy distributed, but unfortunately, it is not a Cauchy sequence is probability... 
Indeed, for any random variable $Z$ we have 
$$|Z_n - Z_{2n}|= |Z_n -Z+ Z-Z_{2n}|\le |Z_n - Z| + |Z_{2n}-Z|.$$ 
Therefore, the event  $\{|Z_n-Z_{2n}|>\epsilon\}$ is contained in the event  $\{|Z_n -z|+|Z_{2n}-Z|>\epsilon\}$. But the latter event is contained in the event $\{|Z_n-Z|>\epsilon/2\}\cup \{|Z_{2n}-Z|>\epsilon/2\}$. Or: 
$$c = P(|Z_n-Z_{2n}|>\epsilon) \le P(|Z_n -Z|>\epsilon/2)+P(|Z_{2n}-Z|>\epsilon/2).$$ 
Since the righthand side is bounded below by $c$, it follows that $Z$ is not the limit in probability of $(Z_n)$ (if it were, both summands on RHS would go to zero as $n\to\infty$). 
A: It's pretty clear that it can't converge (in either sense) to a number, since it always has the Cauchy(0,1) distribution. To see that it can't converge to anything, observe that $$ \frac{S_{2n}}{2n}-\frac{S_n}{n} = \frac{1}{2}\frac{S_{2n}-S_n}{n} -\frac{1}{2}\frac{S_n}{n}.$$ where $S_n = \sum_{k=1}^n X_k.$ $(S_{2n}-S_{n})/n$ and $S_n/n$ are independent Cauchys, so their difference is stable and the distance between $S_{2n}/2n$ and $S_n/n$ cannot go to zero as $n$ gets large.
