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I am looking for a function $f(x)$, defined on $[0,1]$ and depending on a parameter $\lambda \in (0,1)$, that approximately has the attached shape(s). One requirement is that $\int_0^1 f(x)dx = \lambda$. In my sketch I drew the function for different example values of $\lambda$.

Which functions take such a form? What would be their expression? Or which method could I use to find such a function?

I would be grateful for any help!

goldfinch
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  • If $p(t)$ is any continuous probability density defined in $[0, 1]$ then $1 - \int_{0}^{x}p(t)dt$ will have this shape. You can rescale it to get an integral of $\lambda$ – Paul Mar 23 '22 at 14:39
  • Thank you very much, Paul, for your answer! This seems like a great idea. I just think that this doesn't work for every continuous probability density in [0,1]. Maybe I should have specified what I mean by "this shape" - namely that f(x) has first a concave part and then a convex part. Which probability density would you take, and how could I "rescale it"? Thanks again – goldfinch Mar 23 '22 at 16:03
  • $p(x)=6(\frac{1}{4}-(x-\frac{1}{2})^2)$ is a probability density in 0 to 1, I have not worked out the function $h(x)$ you will get when you integrate and subtract from 1.. This in turn will have an integral from 0 to 1. Whatever value this integral has you can rescale your $h(x)$ so that the integral from 0 to 1 is $\lambda$. – Paul Mar 23 '22 at 21:08
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    Many thanks, Paul. That was very helpful!! – goldfinch Mar 24 '22 at 11:42

1 Answers1

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The function you are seeking can be easily described by my previous development of a generalized conics function called superconics, as described here. Alternatively, you may get away with using a subset of that which is the superparabola, which can be found in Wikipedia here. There are figures in both citations that you will see are what you are looking for.

Cye Waldman
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