Random variable defined on the IID I understand that the mean and variance of the Cauchy distribution is undefined. I also understand that if we try to take independent and identically distributed random variable from the Cauchy distribution and attempt to use the Central Limit Theorem, it doesn't work. But what about the following case:
Let $X_1,X_2,X_3$ be independent and identical random variables from the Cauchy distribution ($x_0,\gamma$). Define
$$Y=median(X_1,X_2,X_3)$$
Is it possible to find the expectation and variance of $Y$? If so, how? If not, is it because of the fundamental fact that the central limit theorem cannot be applied to the Cauchy distribution?
Thanks for reading.
 A: This answer expands my comment and validates Henry's first comment. Let $X_1, X_2, X_3$ be i.i.d. Cauchy with symmetry about 0 (so $x_0=0$). Define $Y$ as the median.
Fix $t\in \mathbb{R}$.  The median is bigger than $t$ if and only if at least two of the random variables are larger than $t$:
$$ \{Y>t\} = \{X_1>t, X_2>t\} \cup \{X_1>t, X_3>t\} \cup \{X_2>t, X_3>t\}$$
So by the union bound and symmetry:
$$ P[Y>t] \leq 3P[X_1>t, X_2>t] = 3P[X_1>t]^2$$
where the final equality holds by independence of $X_1$ and $X_2$.
Write $Y = Y^+ - Y^-$ where $Y^+=\max[Y,0]$ and $Y^-=\max[-Y,0]$. We get
\begin{align}
Y^+ &= \int_0^{\infty} 1\{Y>t\}dt \\
\implies E[Y^+] &= \int_0^{\infty} P[Y>t]dt \\
&\leq 3 \int_0^{\infty} P[X_1>t]^2dt \\
&< \infty
\end{align}
So $E[Y^+]$ is finite.  By symmetry we know $E[Y^-]=-E[Y^+]$ and so $E[Y]$ exists and is 0.
To show infinite variance we notice:
$$ \{X_1>t, X_2>t\} \subseteq \{Y>t\}$$
So $$P[X_1>t]^2 \leq P[Y>t]$$
Now
$$ (Y^+)^2 = 2\int_0^{\infty} t 1\{Y>t\}dt $$
So
\begin{align}
E[(Y^+)^2] &= 2\int_0^{\infty} t P[Y>t]dt\\
&\geq 2 \int_0^{\infty} t P[X_1>t]^2dt = \infty
\end{align}
So $E[(Y^+)^2]= \infty$.  Thus, $Y$ has mean 0 but infinite variance.
If we consider shifting $Y$ (so we have Cauchy with $x_0\neq 0$) then the mean shifts to $x_0$ but the variance stays infinite.    You can validate Henry's last comment on $Z$ being the median of 5 i.i.d. Cauchy using a similar technique.
