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Geometric average versus arithmatic average

  1. Mar 23, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a range of numbers numbers [itex]n_i[/itex], each with a different weight [itex]w_i[/itex] that sum up to 1. To keep things simple, let's take the case where we have three numbers with the following weights:

    n_i w_i
    ------------------------------
    100 0.5
    30 0.2
    20 0.3

    Their geometric average is [itex](100^{0.5})*(30^{0.2})*(20^{0.3})=48.4991[/itex]. The arithmetic average of the numbers is [itex]100*0.5 + 30*0.2 + 20*0.3=62[/itex], so it is larger than the geometric average.

    How can I find a new set of normalized weights [itex]w_i'[/itex] that sum to 1 that can be used to find the arithmetic average of the numbers such that it is equal to the geometric average? In other words, I would like to find a new set [itex]w_i'[/itex] such that

    [itex]100*w_1' + 30*w_2' + 20*w_3' = (100^{0.5})*(30^{0.2})*(20^{0.3})[/itex] given that [itex]w_1'+w_2'+w_3'=1[/itex].

    The weights are all nonzero.


    My best attempt at the moment is

    [tex]
    \sum_i (\text{GA} \frac{w_i}{n_i}) n_i
    [/tex]

    where GA is the geometric average. This sum yields GA as expected, but the weights are larger than 1.
     
  2. jcsd
  3. Mar 23, 2017 #2
    Unweighted, the arithmetic mean is always >= geometric mean, so I suspect your weights may have to have a sum > 1. Think what happens if one of the numbers you're averaging is zero.
     
  4. Mar 23, 2017 #3

    haruspex

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    To get equality between AM and GM, you need all terms equal. What does that suggest for the weights? You can always get the weights to add to 1 by normalising: divide by the sum of the weights.
     
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