I needed a probability density function which worked on the interval $[0,1]$, had kind of a bell shape, and had an adjustable mode / peak $p$.

I thought of a pdf $f(x|p)$, given by \begin{equation} f(x|p) = \left\{ \begin{array}{l l l } \frac{ (x^{- \ln 2/\ln p})^2 \cdot (1-x^{- \ln 2/\ln p})^2 }{ \log(p) \left( \frac{x^{1-\frac{4 \log 2}{\log p}}}{\log(p) - 4\log(2)} -\frac{2 x^{1-\frac{3 \log 2}{\log p}}}{\log(p) - 3\log(2)} +\frac{x^{1-\frac{2 \log 2}{\log p}}}{\log(p) - 2\log(2)} \right) } &\quad \text{ for } 0<x<1 \\ 0 &\quad \text{ otherwise} \end{array} \right. \end{equation} which

  • has a peak at $x=p$ for $0<p<1$.
  • $P(X\le 0) = P(X\ge 1) = 0$
  • has a shape similar to the bell shape
  • looks skewed to the left for $p>1$ and to the right for $0<p<1$

Or equivalently: \begin{equation} f(x|p) = \left\{ \begin{array}{l l l } \frac{ (x^a)^2 \cdot (1-x^a)^2 } { (4a+1)^{-1} - 2(3a+1)^{-1} + (2a+1)^{-1} } &\quad \text{ for } 0<x<1 \\ 0 &\quad \text{ otherwise} \end{array} \right. \end{equation} which

which has its peak at $p=-\frac{\log 2}{\log x}$

Is there a similar pdf (or exactly this one) used in literature? What is it called?

PS: note that the given pdf is not symmetric: $f(x|1-p) \neq f(1-x |p) $

  • 2
    $\begingroup$ Are you familiar with the beta distribution? It's only defined on the [0,1] interval and its mode can be shifted with alternative parameter specifications. $\endgroup$
    – Sycorax
    Mar 7, 2014 at 17:04
  • 2
    $\begingroup$ Do you have a particular definition of "bell shaped" in mind? $\endgroup$
    – P Schnell
    Mar 7, 2014 at 17:18
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    $\begingroup$ All beta distributions with both parameters exceeding $2$ have $f'(0)=f'(1)=1$, are unimodal, and can be adjusted to place that mode anywhere within $(0,1)$. $\endgroup$
    – whuber
    Mar 7, 2014 at 17:23
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    $\begingroup$ Angelorf - Given @whuber had already clearly stated the conditions under which your claim about the beta distribution was false, your last comment needed to address those facts. You don't simply reiterate a contradiction. Explain why you think he's wrong (hint: he isn't, but attempting to explain it will help reveal what seems may be an important gap in your understanding - one that will be worth your time to explore and set right). $\endgroup$
    – Glen_b
    Mar 7, 2014 at 23:44
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    $\begingroup$ All has been forgiven. :) Thanks for commenting. Thanks to your remarks I found out that using the normal Beta distribution already suffices for what I was trying. $\endgroup$
    – Angelorf
    Mar 9, 2014 at 2:04

1 Answer 1


That pdf is similar to the generalized Beta distribution of the first kind, which is obtained by raising a Beta variable to a power.

  • $\begingroup$ My distribution seems to be a special case of the generalized Beta distribution. With parameters $p=2+1/a$ , $q=3$ and $b=1$, the remaining part should be equal, since both distributions have a volume of 1. So $|a| \cdot(b^{ap} B(p,q))^{-1} = ((4a+1)^{-1} - 2(3a+1)^{-1} + (2a+1)^{-1})^{-1}$ $\endgroup$
    – Angelorf
    Mar 7, 2014 at 18:01

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