Sample median for a Lorentz (Cauchy) distribution - part 2

In a previous post I derived the distribution function for the median of a sample of size \(n = 2k +1\) (with \(k \geq 0\) integer) drawn from a Lorentz (or Cauchy) distribution:

\begin{equation}
\label{eq:result}
g(x) = \frac{n!}{(k!)^2 \, \pi ^{n} \, \gamma} \left [ \frac{\pi ^2}{4} - \arctan ^2 \left (\frac{x - x_0}{\gamma} \right )\right ]^k \frac{1}{1+ \left (\frac{x - x_0}{\gamma} \right )^2 }
\end{equation}

I will now consider some of its properties.

Sample median for a Lorentz (Cauchy) distribution

The Lorentz (or Cauchy) distribution
\begin{equation}
\label{eq:def}
f(x) = \frac{1}{\pi \gamma}\frac{1}{1+ \left (\frac{x - x_0}{\gamma} \right )^2 }
\end{equation}is pathological, in that it has no finite moments apart from the zero-order one (which ensures proper normalization of the density function.) Many results we usually take for granted (e.g. the central limit theorem) do not apply and, when sampled, the sample mean and sample variance are not good predictors for the position parameters \(x_0\) and \(\gamma\). This should not come as a surprise, since the distribution mean and variance do not even exist. How can we then estimate the position parameters?