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I'm looking for a function that measures if a vector component dominates all the rest. Let

$$ \mathbf{v} = [v_1, v_2, \ldots, v_n] $$

and assume that it is L2 normalized, $|\mathbf{v}|_2 = \sqrt{\sum_{i=1}^n v_i^2} = 1$. If it helps, consider $\mathbf{v}$ a normalized unit vector from a matrix diagonalization.

The function $\theta$ should be maximized when:

$$v_i=1,v_{j \neq i}=0 \rightarrow \theta=1$$

and it should be small (on average) when the components are drawn from a standard normal distribution:

$$v_i = \mathcal{N}(0,\sigma^2=1)$$

but should somehow vary in a "reasonable" way in-between.

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    $\begingroup$ It seems you are considering the inequality between the components in $v$. You may want to use Gini coefficient (en.wikipedia.org/wiki/Gini_coefficient). (Imagine $v$ as a vector of people's income in a nation.) Another measure could be $D(v)\equiv n\times\{median(v) - mean(v)\}$ when $n>2$? If $v$ was drawn from a symmetric distribution, $D(v)=0$. In your case $v = (0,0,\dots,1)$, $D(v)=1$. $\endgroup$ – semibruin May 21 '15 at 16:32
  • $\begingroup$ The Gini coefficient is perfect, thanks for introducing it to me! To make it work for my example, I simply used $v_i^2$ and scaled gini coefficient $g \rightarrow 2(g-(1/2))$. This gives 0 for a uniform distribution and one for my test case. Please post this as an answer so I can accept it. $\endgroup$ – Hooked May 21 '15 at 16:51
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For a vector $v$, set

$$ Z(v) = \frac{\max |v_i|}{\sqrt{\sum_i v_i^2}} $$

Then for vectors of length $n$.

$$ Z(0, 0, \ldots, 1, \ldots, 0) = 1$$

and

$$ Z(a, a, \ldots, a) = \frac{1}{\sqrt{n}} $$

These are the maximum and minimum values, because on one hand

$$ \max |v_i| = \max \sqrt{v_i^2} \leq \sqrt{\sum_i v_i^2} $$

and on the other

$$ \sqrt{\sum_i v_i^2} \leq \sqrt{\sum_i \max_j{v_j^2}} = \sqrt{n \max_j{v_j^2}} = \sqrt{n} \max{|v_i|} $$

So that's a function with your desired property that maps $R^n$ into the interval $\left[ \frac{1}{\sqrt{n}}, 1 \right]$. You can then choose any monotonic function $M: \left[ \frac{1}{\sqrt{n}}, 1 \right] \rightarrow [0, 1]$ to post compose with.

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