Is a weighted average, assuming the weights sum up to one, always smaller than the unweighted arithmetic average? As if the weights are smaller than one the weighted mean is a convex combination of the individual points. Is there any condition on the individual observation for the statement to hold true?

Noelanijd 2022-07-22 Answered
Is a weighted average, assuming the weights sum up to one, always smaller than the unweighted arithmetic average? As if the weights are smaller than one the weighted mean is a convex combination of the individual points. Is there any condition on the individual observation for the statement to hold true?
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Answers (2)

Anaya Gregory
Answered 2022-07-23 Author has 14 answers
This is not normally true, as examples in the comments show. However, there is a natural condition which will make it true: the condition that the weights are in the opposite order to the numbers, i.e. smaller numbers get larger weights.
More generally, if weights and numbers are negatively correlated then the weighted mean will be less than the arithmetic mean. However, this is not a very deep statement, since that is basically how "negatively correlated" is defined.
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Raynor2i
Answered 2022-07-24 Author has 6 answers
You can work it out:
WM     AM w 1 x 1 + w 2 x 2 w 1 + w 2     x 1 + x 2 2 w 1 x 1 + ( 1 w 1 ) x 2 1     x 1 + x 2 2 2 w 1 x 1 + 2 ( 1 w 1 ) x 2     x 1 + x 2 ( 2 w 1 1 ) ( x 1 x 2 )       0
Now, you have various options. For example, if w 1 > 1 2 , x 1 > x 2 , then W M > A M.
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