Suppose I have a large number of positive vectors of different lengths that are drawn from a heavy tailed distribution. I'd like to summarize them in a way that ranks the ones with a large sum and a large average over ones that have only a large sum or only a large average. For instance, say I have

A = [10, .1]
B = [0.6, 0.6, 0.6, 0.6, ..., 0.6] (length 20)
C = [.1, .1, .1, ..., .1] (length 1000)

If I rank each by their average, I have A > B > C. If I rank each by its sum, I have C > B > A. But really, B is the most interesting to me, because it has the most big numbers.

I know that in this example, taking the median would give me what I want, but I don't think that's exactly what I'm looking for in general—I can easily construct examples where the median wouldn't give me what I want.

Is this a standard problem? Are there standard statistics that would help me here?

  • $\begingroup$ $(mean)^2+(sum)^2?$ $\endgroup$
    – Dave
    Aug 8, 2020 at 23:36
  • $\begingroup$ Note that in fact A has a larger sum (as well as a larger mean) than B so it might be difficult to justify anything which selected B. Also note that the sum is the mean multiplied by the number of terms $\endgroup$
    – Henry
    Aug 9, 2020 at 0:06
  • $\begingroup$ @Henry well, imagine B had length 20 $\endgroup$
    – crf
    Aug 9, 2020 at 0:25

1 Answer 1


Because sum and mean goes together by $mean = \dfrac{sum}{n}$, I suppose you can use either sum or mean together with $n$. Intuitively you'll want higher sum or mean with lower $n$.

Say that there are cases where two vectors have same sum or mean and same $n$, and you are interested in which vector has larger values. Median helps here.

If you say the vectors can somehow behave that median can't deal well, use five number summary instead of only the median. That way you'll also know the 75th percentile and the max instead of only median.


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