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There are several summary statistics. When you want to describe the spread of a distribution you can use for example the standard deviation or Gini coefficient.

I know that the standard deviation is based on central tendency, i.e. deviation from the mean, and the Gini coefficient a general measurement of dispersion. I also know that the Gini coefficient has a lower and upper bound [0 1], and the standard deviation does not. These properties are good to know but what insights can the standard deviation give that the Gini cannot and vise versa? If I had to choose to use one of the two, what are the advantages of using one compared to other when it comes to being informative and insightfulness.

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4 Answers 4

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Two things to consider

The Gini is scale independent whereas the SD is in the original units

Suppose we have a measure bounded above and below. SD takes on its maximum value if half measurements are at each bound whereas Gini takes on the maximum is one is at one bound and all the rest at the other.

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  • $\begingroup$ Do you think we could we expand the use of Gini to account for concentration/heterogeneity in meta-analysis? It could be interesting as a means to quantify the concentration in heterogeneity... $\endgroup$ Commented May 9, 2016 at 12:59
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    $\begingroup$ Since the assumption is that the effects are normal, then no. But I think a fuller discussion is off-topic in this thread $\endgroup$
    – mdewey
    Commented May 9, 2016 at 20:52
  • $\begingroup$ @mdewey That last sentence was insightful and helped me the most. Thx! $\endgroup$ Commented May 19, 2016 at 14:47
  • $\begingroup$ @mdewey I tested this myself with some code, but is there a publication somewhere discussing this? Or a proof? (I'm referring to the last sentence) $\endgroup$ Commented May 19, 2016 at 14:59
  • $\begingroup$ @Ojtwist the Wikipedia article en.wikipedia.org/wiki/Gini_coefficient is helpful. $\endgroup$
    – mdewey
    Commented May 19, 2016 at 15:36
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The Gini coefficient is invariant to scale and is bounded, the standard deviation invariant to a shift, and unbounded, so they are difficult to compare directly. Now you can define a scale-invariant version of the standard deviation, by dividing by the mean (coefficient of variation).

However, the Gini index is still based on values, the second on squared values, so you can expect the second one will to more influenced by outliers (excessively low or high values). This can be found in Income inequality measures, F De Maio, ‎2007:

This measure of income inequality is calculated by the dividing the standard deviation of the income distribution by its mean. More equal income distributions will have smaller standard deviations; as such, the CV will be smaller in more equal societies. Despite being one of the simplest measures of inequality, use of the CV has been fairly limited in the public health literature and it has not featured in research on the income inequality hypothesis. This may be attributed to important limitations of the CV measure: (1) it does not have an upper bound, unlike the Gini coefficient,18 making interpretation and comparison somewhat more difficult; and (2) the two components of the CV (the mean and the standard deviation) may be exceedingly influenced by anomalously low or high income values. In other words, the CV would not be an appropriate choice of income inequality measure if a study's income data did not approach a normal distribution.

So the coefficient of variation is less robust, and still unbounded. To take a further step, you can remove the mean, and divide by the absolute deviation instead ($\ell_1(x-m)=\sum |x_n -m|$). Up to a factor, you end up with a $\ell_1/\ell_2$ norm ratio, which can be bounded, since, for an $N$-point vector, $\ell_2(x)\le \ell_1(x)\le \sqrt{N}\ell_2(x) $.

Now you have, with the Gini index and the $\ell_1/\ell_2$ norm ratio, two interesting measures of distribution sparsity, scale-invariant and bounded.

They are compared in Comparing Measures of Sparsity, 2009. Tested against different natural sparsity properties (Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies), the Gini index stands out as the best. But its shape makes it difficult to use as a loss function, and regularized versions of the $\ell_1/\ell_2$ can be used in this context.

So unless you want to characterize a nearly Gaussian distribution, if you want to measure a sparsity, use the Gini index, if you want to promote sparsity among different models, you can try such a norm ratio.

Additional lecture: The GMD (Gini’s Mean difference): A Superior Measure of Variability for Non-Normal Distributions, Shlomo Yitzhaki, 2002, whose abstract might appear of interest:

Of all measures of variability, the variance is by far the most popular. This paper argues that Gini’s Mean Difference (GMD), an alternative index of variability, shares many properties with the variance, but can be more informative about the properties of distributions that depart from normality

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The standard deviation has a scale (say, °K, meters, mmHg,...). Usually, this influences our judgement of its magnitude. So we tend to prefer the coefficient of variation or even better (on finite samples) the standard error.

The Gini coefficient is constructed on (scaleless) percentage values and thus has no scale on its own unit (like e.g. the Mach number). Use the Gini coefficient if you want to compare the equality of shares on something common (shares of 100%). Note that for this application the standard deviation could also be used, so I think your question to compare advantages and disadvantages only applies to this kind of application. In this case, the standard deviation would also be bounded to $[0,1]$. Both indicators would depend on the number of (non-negative) shares but in an opposite direction: Gini increases as the number increases, standard deviation decreases.

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I guess I am a bit late in answering, but the Gini coefficient is proportional to something very similar to the standard deviation divided by the mean.

To be precise, the "standard deviation" should be replaced by "mean absolute difference", which is the mean of $|x_i -x_j|$ for all pairs of points $x_i, x_j$.

The standard deviation measures how far the points are from the mean, and the mean absolute difference measures how far the points are from each other. Of course, the standard deviation uses squares and square roots instead of the absolute value. But we can say those are minor implementation differences. The idea is fundamentally the same.

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