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Given two $n-$dimensional vector, namely $U = [u_1,u_2 \cdots u_n]$ and $V=[v_1,v_2 \cdots v_n]$ such that $\sum_i u_i = \sum_i v_i = 1$ and $u_i \in [0,1]$ and $v_i \in [0,1]$, i would like to derive a $\text{Share_coefficient}$ and $\text{NoShare_coefficient}$ measure which satisfies the below constraint.

  1. $\text{Share_coefficient} + \text{NoShare_coefficient} = 1$
  2. $\text{Share_coefficient}\geq 0$ and $\text{NoShare_coefficient}\geq 0$
  3. When the vector $U$ and $V$ are identical then $\text{Share_coefficient}=1$
  4. When the vector $U$ and $V$ are disjoint (e.g., $U=[1\;\; 0\;\; 0]$ and $V = [ 0\;\; 0.5 \;\; 0.5]$) then $\text{NoShare_coefficient}=1$
  5. For other combination of vector $U$ and $V$, the coefficients are assigned accordingly; i.e. when the overlap between the vectors is more, then $\text{Share_coefficient}$ should be greater than $\text{NoShare_coefficient}$ and vice versa.

I derived the following measure.
$$\text{Share_coefficient} = 1- \frac{1}{2}\sum_i|u_i-v_i|\quad\text{ and }\quad \text{NoShare_coefficient} = \frac{1}{2}\sum_i|u_i-v_i|$$

But it is unable to distinguish the below two example

  1. $U=[1\;\; 0\;\; 0]$ and $V=[0.5\;\; 0.5\;\; 0]$ (only 2-dimension are involved out of 3-dimension)
  2. $U=[0.5\;\; 0.5\;\; 0]$ and $V=[0\;\; 0.5\;\; 0.5]$ (all the dimension are involved)

Both these example provides $\text{Share_coefficient} = \text{NoShare_coefficient} = 0.5$. However I would like to distinguish these two example by assigning different values.

Can anyone help me out in deriving a measure which distinguish the above example and satisfies the mentioned constraint?

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    Presumably, if you don't want these two cases to result in the same share coefficient, you also have some preference which of them should result in the higher one?2011-12-20
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    When you say you'd like to dinstinguish between these two cases, do you know which one you'd like to be greater? It seems like the most likely candidate for your share and no-share functions are the $cos^2$ and $sin^2$ of the angle between the two vectors.2011-12-20
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    I would like the $share\_coefficient$ of first example to be greater than the second example.2011-12-20
  • 0
    Have you tried the correlation coefficient $$\text{share-coeff} = \frac{\sum_i u_iv_i}{\sqrt{\sum_i u_i^2}\sqrt{\sum_i v_i^2}}, ~\text{no-share-coeff} = 1 - \text{share-coeff}$$2011-12-20
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    The cosine of the angle between two vectors can be computed as $\frac{u\cdot v}{|u||v|}$, so $cos^2$ can be written as $\frac{(u\cdot v)^2}{(u\cdot u)(v \cdot v)}$ which gives $0.5$ for your first example and $0.25$ for the second example.2011-12-20
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    @Learner: Great, then Thomas' suggestion is perfect. (Dilip's, which is the square root of Thomas', also works, but Thomas' avoids the square roots and also has a nice symmetry between share and no-share.)2011-12-20

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