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I'm working on highly skewed data, so I'm using the median instead of the mean to summarise the central tendency. I'd like to have a measure of dispersion While I often see people reporting mean $\pm$ standard deviation or median$\pm$quartiles to summarise the central tendency, is it ok to report median $\pm$ median absolute dispersion (MAD)? Are there potential issues with this approach?

I would find this approach more compact and intuitive than reporting lower and upper quartiles, especially in large tables full of figures.

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    $\begingroup$ I think mean, median, lower and upper quartiles jointly would describe the data better. You can find some other descriptive statistics here. $\endgroup$
    – user10525
    Commented May 24, 2012 at 21:16
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    $\begingroup$ I want to be as concise as possible: is the median + 2 quartiles ok? $\endgroup$
    – Mulone
    Commented May 24, 2012 at 21:27
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    $\begingroup$ MAD is a fine statistic for expressing the dispersion of a batch of data--it is more resistant to outliers even than the interquartile range. But you might want to think about what median $\pm$ MAD would really mean and how your audience ought to interpret it. It does not enjoy the same asymptotic or Chebeyshev inequality-like properties of mean $\pm$ SD. That, perhaps, is why such expressions are rarely, if ever, used. $\endgroup$
    – whuber ♦
    Commented May 24, 2012 at 21:35
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    $\begingroup$ I always thought MAD stood for mean absolute deviation the analogue to mse which is mean squared error. it is the average of the absolute deviations from the mean not the median. Am I right or am I going MAD? $\endgroup$
    – Michael R. Chernick
    Commented May 24, 2012 at 23:05
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    $\begingroup$ picture is a thousand words, if possible showing histogram is very powerful. $\endgroup$
    – bdeonovic
    Commented Mar 15, 2017 at 18:53

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I don't think median $\pm$ mad is appropriate in general.

You can easily build distributions where 50% of the data are fractionally lower than the median, and 50% of the data are spread out much greater than the median - e.g. (4.9,4.9,4.9,4.9,5,1000000,1000000,100000,1000000). The 5 $\pm$ 0.10 notation seems to suggest that there's some mass around (median + mad ~= 5.10), and that's just not always the case, and you've got no idea that there's a big mass over near 1000000.

Quartiles/quantiles give a much better idea of the distribution at the cost of an extra number - (4.9,5.0,1000000.0). I doubt it's entirely a co-incidence that the skewness is the third moment and that I seem to need three numbers/dimensions to intuitively visualize a skewed distribution.

That said, there's nothing wrong with it per se - I'm just arguing intuitions and readability here. If you're using it for yourself or your team, go crazy. But I think it would confuse a broad audience.

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    $\begingroup$ (+1) I would like to add that the definition of skewness in terms of the third moment is not the most accepted nowadays because it can only be applied on distributions with light tails. More modern definitions of skewness are based on quantiles, some of them can be found here. $\endgroup$
    – user10525
    Commented May 25, 2012 at 8:08
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    $\begingroup$ @amoeba Is it? The Wikipedia page for MAD defines it as Median(|Xi - Median(X)|), which is 0.1 with the data given. $\endgroup$
    – Upper_Case
    Commented Mar 16, 2017 at 20:26
  • $\begingroup$ @Upper_Case Thank you. I was wrong (forgot about 5-5=0 term). I will delete my comment above not to confuse future readers! $\endgroup$
    – amoeba
    Commented Mar 16, 2017 at 20:29
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Using the MAD amounts to assuming that the underlying distribution is symmetric (deviations above the median and below the median are considered equally). If you're data is skewed this is clearly wrong: it will lead you to overestimating the true variability of your data.

Fortunately, you can choose one of the several alternative to the mad that are equally robust, almost as easy to compute and that do not assume symmetricity.

Have a look at Rousseeuw and Croux 1992. These concepts are well explained here and implemented here. These two estimators are members of the so-called class of U-statistics, for which there is a well developed theory.

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"In this paper a more accurate index of asymmetry is studied. Specifically, the use of the left and right variance is proposed and an index of asymmetry based on them is introduced. Several examples demonstrate its usefulness. The question of evaluating more accurately the dispersion of data about the average emerges in all non-symmetric probability distributions. When the population distribution is non-symmetric, the average and variance (or standard deviation) of a set of data do not provide a precise idea of the distribution of the data, especially shape and symmetry. It is argued that the average, the proposed left variance (or left standard deviation) and right variance (or right standard deviation) describe the set of data more accurately."

Link

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    $\begingroup$ You quoted the abstract of a paper and provided something that resembles a URL (I took the liberty of fixing the link). That's not really the type of answers that we are looking for here; I encourage you to edit your answer and to try to add some comments of your own about why this link helps answer the question. The answer would be much improved if you explained how this asymmetry index is related to the mean central tendency and MAD. $\endgroup$
    – MånsT
    Commented Aug 1, 2012 at 17:53

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