In the literature, is there any asymmetric $S$-shaped function that maps the interval $[0, 1]$ to interval $[0, 1]$?

Unfortunately I can't post figure so I just describe what I mean in text. The function I want should be monotonic increasing. In the meanwhile, for a real number $c < 0.5$, the function

$$0 \mapsto 0, \quad c \mapsto 0.5, \quad 1 \mapsto 1.$$

Moreover, the function is smooth on its domain, concave on interval $[0, c]$ and convex on interval $[c, 1]$.

By the way, can it be seen as a sort of link-function?



And here is how you go about finding one.

For our purposes, convex means $F''(x)\ge 0$ and concave means $F''(x)\le 0$.

Ok, so let $F$ be such a function. If we also assume monotonicity, we have $F'(x)\ge 0$, and $F$ is a cumulative distribution function. Therefore, the convex and concave conditions are $f'(x)\ge 0$ for $x\le c$ and $f'(x) \le 0$ for $x\ge c$ (where $f=F'$ is the pdf of $F$).

In other words, now we are looking for density function on $[0,1]$ that is increasing on $[0,c]$ and decreasing on $[c,1]$. We go to a table of probability distributions on $[0,1]$ (e.g., Wikipedia's list) and see that the cumulative distribution function of the Beta distribution fits the bill.

Another approach would be to explicitly construct such a function (I believe a quartic would do the job).

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    $\begingroup$ Came to suggest Beta cdfs,... so all I can do is +1 your more extensive answer. $\endgroup$ – Glen_b Mar 18 '13 at 9:27
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    $\begingroup$ Bravo! Not only the answer but also the "derivation". Thanks also Glen_b! $\endgroup$ – pengsun.thu Mar 18 '13 at 9:59
  • $\begingroup$ Dear @pengsun.thu: I have rejected your proposed edit. I'm not sure whether or not you can see the comment I left when I did so, so let me repeat it here: In lieu of editing the answer, please clarify whether you really intended the function to be concave on $[0,c]$ and convex on $[c,1]$. Such a function will not look much like an $S$ (asymmetric or otherwise). Cheers. :-) $\endgroup$ – cardinal Mar 18 '13 at 13:03
  • $\begingroup$ @pengsun.thu The answer belongs to Har, not me. I just upvoted it. $\endgroup$ – Glen_b Mar 20 '13 at 11:10
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    $\begingroup$ @cardinal: Yes, I do mean concave on [0,c] and convex on [c,1]. To be precise, it looks like an "S" mirrored by the diagonal line y=x. I'm not sure how to call it as simple as "S-shaped function" so I simply say "S-shaped function" (Please do not hesitate to let me know if any). Excuse my English if that turns out to be a misleading description:) After all, it's a minor issue since the answer by Har has been good enough. I just try to make it consistent with my question. $\endgroup$ – pengsun.thu Mar 20 '13 at 11:20

I would comment instead, but I don't have the score yet, so here's my two cents:

The only thing I could think of is an asymmetric tangent function, yet I couldn't find anything about them except for this part of the Proceedings of the Estonian Academy of Sciences, Engineering. See if it helps...?

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  • $\begingroup$ Yes, it dose! Very inspiring!! BTW, what is the motivation of such an asymmetric tangent in the domain of the paper you point to? $\endgroup$ – pengsun.thu Mar 18 '13 at 10:02
  • $\begingroup$ To measure finger arterial volume, V, as a function of its pressure, P, the inverse of the usual relationship P = f(V), and to be free of the limitations of equal sensitivity in positive and negative pressure, which isn't preferred with a symmetric model. $\endgroup$ – RS18 Mar 18 '13 at 10:37
  • $\begingroup$ Interesting! It reminds me of the symmetric sigmoid function in neuro network, where the horizontal axis denotes "stimulus" while the vertical axis denotes "response". There should be a place for asymmetric tangent if the "response" to positive/negative "stimulus" is imbalanced. $\endgroup$ – pengsun.thu Mar 18 '13 at 11:34

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