I noticed something strange or seemingly paradoxical in ArcMap 10.2.2. The median center is less influenced by spatial outliers than mean center, this is a fact. The thing is, I created 6 dispersed points and gave them similar attribute values. After this I calculated the mean and median center using those values as a weight field. The result was as I expected. The mean center point was influenced highly by the point with the highest value.

After this, I changed the value of a point and made it a high outlier. I recalculated the mean and median center and the result was very strange. The median seemed to be influenced more by that high outlier point. Why did this happen?

Thereafter, I tried the same thing but this time with an additional spatial outlier. At first I gave it a value similar to the others, and after this a very high value which is considered a high outlier value. The result was as I expected. The median center was less influenced by that outlier point (both with very high or similar to the others' value) than the mean center.

So the question is, does the spatial pattern of points play a role on weights? Does the nature of the outlier (spatial or value) point plays a role on the weights? Which of these exerts more influence to the position of the mean and median center points?

I'm uploading a screenshot which shows the first "experiment" I did.

The blue points are the means and the purple points are the medians. The heavily outlined points are unweighted. The smaller blue and purple points are the weighted with the high outlier value being under the selected point. The other two blue and purple points are weighted with the values of all the other points being similar

enter image description here

The reason for asking this question is because I do have empirical data uniformly spatially distributed with some values being outliers, and I'm trying to depict central tendency of several attributes. I noticed that "paradox" and didn't know which parameter to use. Mean center or median center?

  • It may help to interpret the weight as a multiplier of points. If one point is assigned a weight of four, the software may interpret this as implying that four points occur at that same location. If this is the case, then it makes sense that the median would be more heavily influenced by a change in weight as opposed to distance. – MannyG Nov 2 '15 at 18:57
  • Yes, but why median is influenced more than the mean regarding values (and not location alone) ? – KonstaNtie Nov 3 '15 at 17:26

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