Suppose I have i.i.d. samples $X_1, \cdots, X_n$ from some unknown distribution $F$ and I wish to estimate the mean $\mu=\mu(F)$ of that distribution and I insist that the estimator be unbiased - i.e., $\mathbb{E}[T(X_1, \cdots, X_n)] = \mu$.

The canonical estimator is the sample mean $\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$. This is always unbiased and for many families of distributions, such as Gaussians, it is optimal or near-optimal in terms of variance.

However, the sample mean is not robust. In particular, the sample mean can change arbitrarily if a single $X_i$ is changed. This means it has a breakdown point of 0.

A more robust estimator is the sample median. Changing a few data points will not, for most samples, significantly change the median. This has a breakdown point of 0.5, which is the highest possible.

For Gaussian data, the sample median has higher variance than the sample mean (by a factor of $\pi/2$). However, for other distributions, such as the Laplace distribution or Student's $t$-distribution, the median actually has lower variance than the mean.

Furthermore, the median is always unbiased if the distribution is symmetric (about its mean). Many natural distributions are symmetric, but many are not, such as the following examples.

  1. Binomial
  2. Poisson
  3. log-Normal
  4. Gamma
  5. F-distribution
  6. Geometric distribution

My question is: Are there robust and unbiased estimators for the means of natural asymmetric distributions? By robust I simply mean a non-zero breakdown point and by natural I mean something from the above list or similar (just not a concocted example). I can't find any examples. I would be particularly interested in the Binomial case.

  • 3
    $\begingroup$ (+1) Because robust estimation is primarily of interest in nonparametric settings, likely little has been done along these lines. However, under your parametric models it's straightforward to concoct robust unbiased estimators out of order statistics: that will give you plenty of examples to consider. $\endgroup$ – whuber Mar 23 at 18:16
  • $\begingroup$ There is probably an asymmetric trimmed mean that would work in every one of those cases. However, I suspect the (asymmetric) trimming proportions would need to be derived on a case-by-case basis (and no, I've never seen it done). Of incidental interest, I asked a question about this in the non-parametric case, but there probably isn't an answer in that scenario. $\endgroup$ – Colin T Bowers May 30 at 5:04

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