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The common approach for estimating the parameters of a normal distribution is to use the mean and the sample standard deviation / variance.

However, if there are some outliers, the median and the median deviation from the median should be much more robust, right?

On some data sets I tried, the normal distribution estimated by $\mathcal{N}(\text{median}(x), \text{median}|x - \text{median}(x)|)$ seems to produce a much better fit than the classic $\mathcal{N}(\hat\mu, \hat\sigma)$ using mean and RMS deviation.

Is there any reason to not use the median if you assume there are some outliers in the data set? Do you know some reference for this approach? A quick search on Google didn't find me useful results that discuss the benefits of using medians here (but obviously, "normal distribution parameter estimation median" is not a very specific set of search terms).

The median deviation, is it biased? Should I multiply it with $\frac{n-1}{n}$ to reduce bias?

Do you know similar robust parameter estimation approaches for other distributions such as Gamma distribution or the Exponentially modified Gaussian distribution (which needs Skewness in parameter estimation, and outliers really mess up this value)?

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    $\begingroup$ If you have outliers, it might be that your distribution is not really Gaussian normal. This does not answer your question, of course, but, IMO, this is a possibility one should always entertain. $\endgroup$ – sds Jan 30 '13 at 15:28
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    $\begingroup$ I don't have a simple, clean, mathematical distribution. I have real data, which is messy by nature. No distribution whatsoever will be a perfect fit, because you cannot handle the situation analytically anymore. And the outliers are actually my interest. :-) $\endgroup$ – Erich Schubert Jan 30 '13 at 16:01
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The observation that in an example involving data drawn from a contaminated Gaussian distribution, you'd get better estimates of the parameters describing the bulk of the data by using the $\text{mad}$ instead of $\text{med}|x-\text{med}(x)|$ where $\text{mad}(x)$ is:

$$\text{mad}=1.4826\times\text{med}|x-\text{med}(x)|$$

--where, $(\Phi^{-1}(0.75))^{-1}=1.4826$ is a consistency factor designed to ensure that $$\text{E}(\text{mad}(x)^2)=\text{Var}(x)$$ when $x$ is uncontaminated-- was originally made by Gauss (Walker, H. (1931)).

I cannot think of any reason not to use the $\text{med}$ instead of the sample mean in this case. The lower efficiency (at the Gaussian!) of the $\text{mad}$ can be a reason not to use the $\text{mad}$ in your example. However, there exist equally robust and highly-efficient alternatives to the $\text{mad}$. One of them is the $Q_n$. This estimator has many other advantages beside. It is also very insensitive to outliers (in fact nearly as insensitive as the mad). Contrary to the mad, it is not built around an estimate of location and does not assume that the distribution of the uncontaminated part of the data is symmetric. Like the mad, It is based on order statistics, so that it is always well defined even when the underlying distribution of your sample has no moments. Like the mad, It has a simple explicit form. Even more than for the mad, I see no reasons to use the sample standard deviation instead of the $Q_n$ in the example you describe (see Rousseeuw and Croux 1993 for more info about the $Q_n$).

As for your last question, about the specific case where $x\sim\Gamma(\nu,\lambda)$, then

$$\text{med}(x)\approx\lambda(\nu-1/3)$$

and

$$\text{mad}(x)\approx\lambda\sqrt{\nu}$$

(in both cases the approximations become good when $\nu>1.5$) so that

$$\hat{\nu}=\left(\frac{\text{med}(x)}{\text{mad}(x)}\right)^2$$

and

$$\hat{\lambda}=\frac{\text{mad}(x)^2}{\text{med}(x)}$$

See Chen and Rubin (1986) for a complete derivation.

  • J. Chen and H. Rubin, 1986. Bounds for the difference between median and mean of Gamma and Poisson distributions, Statist. Probab. Lett., 4 , 281–283.
  • P. J. Rousseeuw and C. Croux, 1993. Alternatives to the Median Absolute Deviation Journal of the American Statistical Association , Vol. 88, No. 424, pp. 1273-1283
  • Walker, H. (1931). Studies in the History of the Statistical Method. Baltimore, MD: Williams & Wilkins Co. pp. 24–25.
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    $\begingroup$ $\Phi^{-1}(0.75)^{-1} \approx 1.4826$ - is this the value to use, or is one of the two inversions extra? $\endgroup$ – Erich Schubert Jan 30 '13 at 11:47
  • $\begingroup$ @ErichSchubert: you are right: i forgot the second inverse..corrected. $\endgroup$ – user603 Jan 30 '13 at 12:32
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    $\begingroup$ +1. But I think you mis-characterize the "efficiency factor": it is not analogous to the $n/(n-1)$ factor for the variance because the latter is universal whereas your factor is specific to normal distributions only: with a different distribution in mind, you would have to change your factor. This difference is one crucial reason why variances and SDs have seen so many more applications than the MAD. $\endgroup$ – whuber Jan 30 '13 at 16:07
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    $\begingroup$ @whuber: thanks for this, i now realize my sentence 'this is similar in spirit' can easily be misunderstood. I removed it. $\endgroup$ – user603 Jan 30 '13 at 16:09
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    $\begingroup$ I have made the ExNormal part a separate question: stats.stackexchange.com/questions/48907/… But I have one more for you: LogNormal distribution - handle by applying log, then proceed as with normal distribution? $\endgroup$ – Erich Schubert Jan 30 '13 at 18:53
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If as you assert, the data are normal apart from some small proportion of outliers, the median and median absolute deviation will be robust to gross errors but won't make very efficient use of the information in the non-outlying data.

If you knew some a priori bound on the proportion of outliers you could trim that proportion for the mean and Winsorize the standard deviation. An alternative that doesn't require such knowledge would be to use M-estimators for the location and related quantities for the variance. The gain in efficiency if your assumptions are correct (such as the data really being normal apart from a small percentage of outliers) may in some circumstances be substantial.

The median deviation is biased as an estimate of the standard deviation - but not like the $\frac{n}{n-1}$ adjustment; the unadjusted sample mean square is asymptotically going to the variance, but the sample median absolute deviation is not asymptotically going to the population standard deviation; you need to multiply it by a constant simply to get consistency. After you have done that it's still small-sample biased in the same sense as the unadjusted mean square.

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