Suppose that $X_i$'s are iid Cauchy RV's with pdf $f_u (x) = \frac{1}{\pi} \frac{u}{u^2+x^2}$.
I am aware that the RV $Y:=\frac{1}{N}\sum_{k=1}^N X_k$ has the same density as the $X_i$'s. I am trying to show that $\lim_{N\rightarrow \infty} Y_N$ does not approach a constant value in probability.
In other words, I want to show that for a constant $\gamma$, $P(\lim_{N\rightarrow \infty} Y_N = \gamma) \neq 1.$
I am not sure if there's a better approach but I intend to use the definition of a limit, i.e. set $\epsilon >0$ then there exists $m$ such that for $N>m, |Y_N-\gamma| < \epsilon.$ I then intend to plug this expression into the probability above and compute the probability using the density function of $Y_N.$
Any thoughts on this approach/ is there a better alternative?
Insights very much appreciated
In fact $\lim_N Y_N$ exists almost surely is equivalent to $E(|X_1|) < +\infty$, which is not the case for Cauchy variable.
One direction of above assertion is the famous law of large number.
To prove the other direction, note that if $\lim_N Y_N$ exsits alomst surely, then $\lim_N\frac{X_N}{N} = 0$ alomst surely, which we will see is impossible if $E|X_1| = +\infty$
By $E(|X_1|) \le \sum_{k=1}^{\infty} k P(k-1\leq X_{1} < k) = \sum_{k=0}^{\infty}P(|X_1| \geq k) = \sum_{k=0}^{\infty}P(|X_k| \geq k)$, we see that if $E(|X_1|) = +\infty$, then $\sum_{k=0}^{\infty}P(|X_k| \geq k) = \infty$.
Since $X_k$ are indepedent, by Borel-Cantelli lemma, almost surely there exist infinitely many $k$, such that $|X_k| \geq k$.
So $\lim_N\frac{X_N}{N} = 0$ alomst surely is impoosible