Geometric average versus arithmatic average

[Niles]

Homework Statement

I have a range of numbers numbers $n_i$, each with a different weight $w_i$ 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 $(100^{0.5})*(30^{0.2})*(20^{0.3})=48.4991$. The arithmetic average of the numbers is $100*0.5 + 30*0.2 + 20*0.3=62$, so it is larger than the geometric average.

How can I find a new set of normalized weights $w_i'$ 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 $w_i'$ such that

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

The weights are all nonzero.My best attempt at the moment is

$$\sum_i (\text{GA} \frac{w_i}{n_i}) n_i$$

where GA is the geometric average. This sum yields GA as expected, but the weights are larger than 1.

 

[John Park]

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.

 

[haruspex]

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.