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I explore centrality measures for graphs and discretized regions.
The data array (for given point) is all distances from this point to the boundary of region.
At the beginning I used standard deviation of this array as measure of centrality of point
(the smaller deviation, the more centrality).
But recently I've found simple criterion (for strictly positive arrays): $$\kappa=\frac{\min(arr)}{\max(arr)}$$ This measure is very clear and illustrative:

enter image description here

I wonder if this measure is always used in statistics, what are its properties?

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  • $\begingroup$ Your example includes two regions with the same linear scale. You should see if this metric continues to make sense if the regions radially differ in scale. In particular, if the size of the region goes to zero, then the centrality by this metric will diverge. $\endgroup$
    – Dave
    Commented Jan 2 at 17:48
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    $\begingroup$ Though as far as I know not known/studied, this metric has some similarities to several cluster quality metrics like the Davies-Boulding index and the Dunn index $\endgroup$
    – Dave
    Commented Jan 2 at 18:00

1 Answer 1

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  • It's definitely not the most popular. You would not find it discussed among the commonly used statistics used for measuring dispersion.
  • The closest thing to it, that is quite popular is range, which just uses a different arithmetic operation, i.e. $\max(arr) - \min(arr)$.
  • The biggest problem with the measure is that both minimum and maximum are very volatile and susceptible to outliers. One outlier in either direction could change the measure a lot.
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  • $\begingroup$ Thank you! The problem of range is non-zero value for points of boundary. What about outliers, in this case we have smooth set of values (possible regions are not fractals etc) $\endgroup$
    – lesobrod
    Commented Jan 2 at 13:12
  • $\begingroup$ @lesobrod I don't understand what you mean by "non-zero value for points of boundary" but the range and the metric in your question would have the same problems with boundaries, as they are simple functions of the same values. $\endgroup$
    – Tim
    Commented Jan 2 at 14:58

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