Given Cauchy distribution with pdf $p(x) = \frac{1}{\pi ((x - \theta)^2 + 1)}$
how can I find a consistent unbiased estimator for $\theta$?
My reasoning so far
Tried MLE, but there seems to be no closed-form expression. See Maximum likelihood estimator of location parameter of Cauchy distribution
MoM does not work for Cauchy - the distribution does not have moments.
By googling the keyword 'estimation of location parameter of Cauchy distribution', I gather that this has been a pretty well-studied problem. Similar to the answer by Robin Ryder, I suggest taking a look at the paper A Note on Estimation from a Cauchy Sample (Rothenberg et al., 1964) for an unbiased estimator of $\theta$ for a $C(\theta,1)$ distribution based on a sample of size $n$.
Quoting the relevant result from Exercise 8.8.6 (page 237) of Order Statistics, third edition, by David and Nagaraja, which cites the above paper as reference:
For samples of $n=2m+1 \quad(m=0,1,\ldots)$ from the Cauchy distribution
$$f(x)=\frac{1}{\pi(1+(x-\theta)^2)}\qquad,\,-\infty<x<\infty$$
, the trimmed mean
$$\frac{1}{n-2[nk]}\sum_{i=m-[nk]}^{m+[nk]}X_{(i)}\qquad,\,0\le k\le \frac{1}{2}$$
is an unbiased estimator of $\theta$ with asymptotic variance
$$\frac{1}{nk}\left[\frac{1-k}{k}\tan^2\left(\frac{\pi k}{2}\right)+\frac{2}{\pi k}\tan\left(\frac{\pi k}{2}\right)-1\right]$$
This is a minimum for $k=0.24$, when the trimmed mean is almost the midmean.
Indeed, computing the MLE is complicated. The most straightforward choice is to use the sample median, which is consistent, but there are more efficient estimators available.
Zhang (2010) shows that a more efficient estimator is to take the mean of the central 24% of the data, i.e. use the order statistics $X_{(1)}\leq X_{(2)}\leq\ldots\leq X_{(n)}$, take $r=\lfloor 0.38n\rfloor$ and $$T_n = \frac{1}{n-2r}\sum_{i=r+1}^{n-r}X_{(i)}$$
The efficiency of the estimator $T_n$ is 88% of the Cramér-Rao bound, whereas the sample median has an efficiency of 81%.
The paper by Zhang also suggests other estimators, which are more complicated but can achieve a higher efficiency.
This is an answer I've only just come up with, so there might be a mistake somewhere. First I use a demonstration to show you how this works, and then give you a consistent estimator.
Assume for the sake of demonstraton that $EX$ exists. That is, $EX < \infty$, and take it as the mode of the distribution as with other symmetric bell-curve distributions.
It can be proven that under certain regularity conditions, order statistics are consistent estimators of their associated quantiles.
For example, let $\tau \in (0,1)$, let $f$ be a continuous density function such that $f(\tau) > 0$
Define $k = [n\tau]$ where $n$ is your sample size and $[-]$ denotes the integer part of $n\tau$.
Finally, define $\epsilon_\tau$ as the $\tau$th quantile of the density $f$. That is, $\epsilon_\tau$ is the smallest value such that $F(\epsilon_\tau) = \tau$
Then given the continuity of $f$ along with $f(\tau) > 0$ we can prove that
$$X_{(k)} \to \epsilon_\tau$$
almost surely, where $k$ is the $k$th order statistic of the sample $X_1, \dots, X_n$ (the $k$th largest value).
But note the following!
The Cauchy density is continuous and $f(0.5) > 0$.
The mean (if it existed!) of the Cauchy distribution is also the median $$P(X < m)=P(X>m)=0.5$$
Hence taking $\tau = 0.5$ and noting that since the Cauchy as equal mean and median, the above statement implies that
$$X_{(k)} \to E[X]$$
almost surely, where again $k = [\tau n] = [0.5 n]$.
Hence the $k$th order statistic is a consistent estimator of $E[X]$.
The above shows that if you can find a value $\tau_\theta$ such that $F(\theta) = \tau_\theta$, then $X_{(k)}$ will be a consistent estimator of $\theta$, where once again $k =[n \tau_\theta]$.
The CDF of the $Cauchy(\theta,1)$ distribution is
$$F(x) = \frac{1}{\pi} \tan^{-1} \left( x-\theta \right) + \frac{1}{2}$$
So to obtain $\tau_\theta$ we simply need to solve the following for $y$:
$$y = \frac{1}{\pi} \tan^{-1} \left( \theta-\theta \right) + \frac{1}{2}$$
But $\tan^{-1}(0) = 0$, hence
$$y = \frac{1}{2}$$.
Thus, finally, a consistent estimator of $\theta$ is $X_{(k)}$ where $k = [0.5 n]$ and $X_{(k)}$ is again the $k$th order statistic.
That is, the median is a consistent estimator of $\theta$.
