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I don’t see how this relates to Simpson’s paradox at all. If you split the set into two, you need to correct for number of elements in each split set. If you
by j7ake 4y ago
I don’t see how this relates to Simpson’s paradox at all.
If you split the set into two, you need to correct for number of elements in each split set.
If you have a set of N numbers, the mean is the weighted sum with weights 1/N.
If instead you split set into n1 and n2 where n1+n2=N, then you need weights n1/N and n2/N for first and second split.
Naively applying the mean formula without reweighting elements properly is not the Simpson’s paradox.
- graderjs 4y agoI haven't considered the question you raise, but could it have something to do with: when you want to compute a partial average you may not have the information of the total number of elements available to you?
- j7ake 4y agoIf you consider the definition of mean for a continuous variable you can see that you can indeed break up the operations into different partitions of the data then add them up. The weight in continuous case is explicit while for discrete case it is sometimes implicit (1/N). When I think Simpson’s paradox I think of regressions. It is indeed true that if you do regression on half your data and then regression on the other half, you won’t get the same result as doing regression on the full data.