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$\def\prop#1#2#3{#1:#2:#3}$Let's say I have a proportion, $\prop 135$

And then a set of other ones, $$ \begin{array}{l} \prop 235\\ \prop 145 \\ \prop 136 \end{array} $$

Which one is closer to $\prop 135$?

How to define it closer with mathematics?

(If you think it needs editing, edit away or comment)

martini
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Shane Hsu
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  • You can define a distance in many ways, which one depends on what you are trying to do. The sum of the absolute differences is a simple one. – copper.hat Sep 19 '13 at 07:13
  • @copper.hat sum of absolute difference, I will think about that. But how? For example 1 : 3 : 5 and 3 : 4 : 5, one can say their absolute difference is 2 + 1 + 0 = 3 but if we take 1 : 3 : 5 and multiply it by 4/3, we get 4/3 : 4 : 20/3, and their difference will become 5/3 + 0 + 5/3 = 10 / 3, a bit larger than 3. So there will be cases where there are different difference for two proportions. – Shane Hsu Sep 19 '13 at 07:43

1 Answers1

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This depends on which metric you define. Possible metrics would, for example, result in converting the $a:b:c$ to a vector $$\left( \begin{matrix}a\\b\\c\end{matrix} \right)$$ And using some standard $\mathbb R^3$ metrics, like $$d(x,y) := \Vert x-y \Vert_p$$ For $1\leq p\leq \infty$. The results vary with the choice of $p$. For $p=\infty$, for example we have with enumeration $y_1, y_2, y_3$ of your alternatives and $x$ the first: $$\begin{align*} d(x,y_1) & = 1 \\ d(x,y_2) & = 1 \\ d(x,y_3) & = 1 \end{align*}$$ so they are all "equally close".

However, any of these $p$ will yield a distance of $1$, since the changes are always to add one to one "component".

Another choice for converting $a:b:c$ to a vector containing "proportions", would be $$\vec{x} = \left(\begin{matrix}a/b\\b/c\\c/a\end{matrix}\right)$$

With the second option and $p=2$ we get, according to MATLAB with the code
x = [1 3 5];
Y = [x;x;x] + eye(3);
xx = x./x([2 3 1]);
YY = Y./Y(:,[2 3 1]);
Z = [norm(xx - YY(1,:)); norm(xx - YY(2,:)); norm(xx - YY(3,:))]

we have $$Z \approx \left(\begin{matrix}2.5221\\0.2167\\1.0050\end{matrix}\right)$$ So $y_2$ is "closest" in this interpretation.

AlexR
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  • So the idea is to turn my proportion into vector for calculation. And use Uniform Norm? – Shane Hsu Sep 19 '13 at 15:27
  • Well, essentially you can use any $\mathbb R^3$ norm you are fond of. $p=\infty$ or $p=2$ should yield the most intuitive results. I recommend the euclidean norm ($p=2$) because it "punishes" all deviations and not only the highest. – AlexR Sep 19 '13 at 20:04
  • @ShaneHsu I've added MATLAB code for my recommended example and given the output. That seems an "intuitive" result to me... – AlexR Sep 19 '13 at 20:14
  • I see. Thanks a lot! – Shane Hsu Sep 22 '13 at 05:20