Page 318 - Becoming Metric Wise

310 Becoming Metric-Wise

sources have or produce 80% of all items. Of course, this is just a rule of
thumb. More generally one may have a 100y/100x rule, where y . x,
i.e., 100x% of the most productive sources have 100y% of all items. In
(Egghe, 1993b) it is shown that, if Lotka’s law is valid then
1
y 5 x μ (9.20)
where μ denotes the average number of items per source. Eq. (9.20)
shows that y depends on x but not only on x.
A consequence: Assume that we have two IPPs following Lotka’s law
(9.7).If μ , μ and y 1 5 y 2 then x 1 . x 2 .
1
2
ð1=μ 1 Þ ð1=μ 2 Þ
Proof. The equality y 1 5 y 2 implies: x 1 5 x 2
lnðx 1 Þ lnðx 2 Þ
Taking the logarithm of both sides yields: 5
μ μ 1 μ 2
lnðx 1 Þ 1
and thus: 5 , 1
lnðx 2 Þ μ
2
From this inequality we derive that ln(x 1 ) . ln(x 2 ), using the fact that
x 1 and x 2 are smaller than 1 and hence their logs are negative.
Consequently: x 1 . x 2 (as a logarithmic function is increasing).
This result shows that the higher the average production per source,
the smaller the fraction needed to obtain a fixed percentage of items.
We recall from Chapter 4, Statistics that Lorenz curves can be seen as
graphical representations of the generalized 80/20-rule: For each percent-
age 100x of most productive sources they show the corresponding per-
centage 100y of items. Lorenz curves are the basis of a concentration
theory and related acceptable measures of concentration. We explain this.
Assume that an array of N nonnegative numbers is given. Which
requirement should one impose for an acceptable measure of concentra-
tion? For the moment we keep N fixed. A first requirement is that the
measure should not depend on the order in which the numbers are
given.
As this function, denoted by F, must measure inequality another
requirement is that if all values are equal the resulting concentration is
zero:
Fx; x; .. .; xÞ 5 0 (9.21)
ð
If all values are zero except one then F must attain its highest value
(given N). Further, we consider concentration as a relative property,
hence the value of F for (x 1 ,.. .,x N ) must be the same as for (cx 1 ,.. .,cx N ),
with c . 0.