I have a set of purchases with $ values like this: {1, 1, 1, 2, 7, 8, 10, 10, 20, 20}. Sum: 80. Mean: 8.

I had thought that if you split this into equal sums (of 40), this would occur at the mean (8). In fact you have to split at a different point in the set: {1, 1, 1, 2, 7, 8, 10, 10}, {20, 20}.

Clearly I was incorrect that this was a property of means, and this point isn't the median either, but I feel this is an average of some basic kind, as I'm splitting the data into two equal halves.

Is there a name for this? Is this one example of something analogous to, but not quite, percentiles?

  • $\begingroup$ Not exactly what you have here, but one concept that is "analogous to, but not quite, percentiles" in a similar way are expectiles. $\endgroup$ – Chris Haug May 23 '20 at 12:21
  • $\begingroup$ I am not sure how this quantity is defined exactly. What is the value of this statistic on $\{1,3,3,4\}$ for instance? $\endgroup$ – Federico Poloni May 24 '20 at 12:53

This looks like the weighted median

> x <- c(1, 1, 1, 2, 7, 8, 10, 10, 20, 20)   
> median(rep(x, times = x))   
[1] 15

If you do not use R rep(x, times = x) generates a new vector with each element repeated its own number of times. Note that this gives a value half way between 10 and 20 for your example as that is how it defines the median rather than 20 which you quote.

In a comment on the original question Federico Poloni poses the data-set {1, 3, 4, 4}. If we take the definition of the median as being that value such that no more than half the observations lie above it and no more than half lie below it then the analogous procedure here would be to take the value such that the sum of the values above it were no more than half the total sum and vice versa. So the answer would be 3 since any other value either has more than half the total sum (11) above or below it.

In another answer Carlo makes the valid point that this is an unusual choice for a measure of central tendency which is a very fair point.

  • $\begingroup$ This seems right - thank you! $\endgroup$ – Clive May 23 '20 at 18:38
  • 2
    $\begingroup$ Describing this statistic as "the weighted median" is misleading. It's a weighted median, for weights equal to the element values. This is usually not a sensible way to weight elements. $\endgroup$ – user2357112 supports Monica May 24 '20 at 21:50

I never heard of such statistic, but it occurs to me that it has terrible properties as a central tendency estimator: supposing that your distribution is always positive (negative values can actually worsen this) it is also probably quite skewed, meaning that your statistic will result in a value very high, next to the highest observed values.

Let's suppose for instance that every value in your sample is two times the preceding one, your statistic will pick the last value of them all.

This is also quite problematic from a probabilistic prospective, while mean and median have simple probabilistic meanings.


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