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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.

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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.

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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.

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  • $\begingroup$ I think that the Pitman estimator is unbiased and consistent as well. And equivariant. $\endgroup$ – Xi'an Nov 5 '18 at 21:29
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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!

  1. The Cauchy density is continuous and $f(0.5) > 0$.

  2. 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 $E[X]$.

Returning to your problem

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 $\theta$.

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  • $\begingroup$ But is it unbiased? $\endgroup$ – Greenparker Oct 24 '18 at 14:36
  • $\begingroup$ Thanks Xiaomi! What about unbiasedness? Any thoughts? $\endgroup$ – Ignacio Oct 24 '18 at 14:46
  • $\begingroup$ Oh I did not see the unbiased part of your question. Im not sure if the estimator is unbiased, but it is certainly asymptotically unbiased. I will think about this a bit and give an answer if someone else hasn't. $\endgroup$ – Xiaomi Oct 24 '18 at 15:40

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