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I am looking for a formula $y_i = f(i, \kappa)$, for $i=1...K$ data points, that generates a convex function like this one:

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The function should contain a parameter, e.g. $\kappa$ that can handle the amount of convexity. Ideally, setting this parameter to a certain value should give equal weights (e.g. $y_i = y_j$ for all $i=1...K$)

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  • $\begingroup$ Are $y_i$ data points? Is so, they are given and the property of adding up to 1 is either given or not. $\endgroup$ – Hellen Sep 28 '17 at 14:35
  • $\begingroup$ No $y$ is not given. $x_i$ is just 1...K. Maybe I should rewrite it as $y_i = f(i, \kappa)$ $\endgroup$ – JohnAndrews Sep 28 '17 at 14:50
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You can combine basic convex functions, e.g. $$y_1(x) = \sum_{i=1}^K (x-i)^2$$ and then renormalise them to obtain $$y(x) = \frac{y_1(x)}{\sum_{i=1}^K y_1(i)}$$

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  • $\begingroup$ Makes sense. How do I include the convexity parameter? $\endgroup$ – JohnAndrews Sep 28 '17 at 14:56
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    $\begingroup$ @JohnAndrews depends on what you want exactly. You could try with weighting the terms in the sum, or scaling using $(cx-i)$, or increasing the powers to $(x-i)^{2+c}$ (for $c>0$), or in a million other ways. I'd go with the first option. $\endgroup$ – TZakrevskiy Sep 28 '17 at 15:09
  • $\begingroup$ For which c can I have that all y_i are equal? $\endgroup$ – JohnAndrews Sep 28 '17 at 15:21
  • $\begingroup$ Also, what exactly is the difference between x and i in your function. $\endgroup$ – JohnAndrews Sep 28 '17 at 15:23
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    $\begingroup$ If you are scaling $x$, then for $c=0$ the function $y(x)$ is constant. $\endgroup$ – TZakrevskiy Sep 28 '17 at 15:23

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