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I have a set of numbers $S$, and for each $s_i\in S$, $0\lt s_i \lt 1$. I would like to transform them so that they sum up to $1$.

An obvious way to do it is to calculate $t_i=\frac{s_i}{\sum_i{s_i}}$, so that $\sum_i{t_i}=1$. I wonder if there is any other way to achieve the same effect, and what are the differences between them.

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A little generalization of this would be: $t_i=\frac{f(s_i)}{\sum f(s_i)}$ with $\sum f(s_i)$ not equal to zero. $f$ would allow you to "bend" $S$ (give different weights).

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  • $\begingroup$ I almost updated the same. I am deleting my post. $\endgroup$ – vinux Jan 10 '12 at 16:05
  • $\begingroup$ @Arthur, thanks for the generalization, could you please elaborate with an example for what you mean by "bending" the set $S$? Appreciated. $\endgroup$ – MLister Jan 10 '12 at 16:07
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    $\begingroup$ +1. But the "little generalization" is too general: given any target set of values $t_i$ (summing to unity), you can find such an $f$ for which each $s_i$ is transformed into $t_i$, effectively losing all information about the original $s$'s. Normally, then, we restrict $f$ to be positive, continuous, and strictly monotonically increasing so that at least the order of the original sequence is preserved. $\endgroup$ – whuber Jan 10 '12 at 16:57
  • $\begingroup$ MLister, some illustrated examples of such re-expressions appear at stats.stackexchange.com/a/10979. When $f$ is linear (with zero constant term), its effect is identical to your "obvious" formula. When $f$ is nonlinear, it expands or contracts the differences between the $t_i$ by different amounts: this is the "bending" Arthur refers to. $\endgroup$ – whuber Jan 11 '12 at 13:17
  • $\begingroup$ @whuber that's true. However it really depends on what you're looking for. You could imagine that $s_i$ is the proportion of women in profession $i$, and use $f(x)=|x-0.5|$ as the "distance from perfect diversity". Moreover, nothing in the question forbids choosing $f=1_{\mathbb{Q}}$, although I agree my generalization is not "little" at all. $\endgroup$ – Arthur Jan 11 '12 at 14:59

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