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Triangular Distribution

One shape parameter $c\in[0,1]$ giving the distance to the peak as a percentage of the total extent of the non-zero portion. The location parameter is the start of the non- zero portion, and the scale-parameter is the width of the non-zero portion. In standard form we have $x\in\left[0,1\right].$

$$\begin{eqnarray*} f\left(x;c\right) & = & \left\{ \begin{array}{ccc} 2\frac{x}{c} & & x < c \\ 2\frac{1-x}{1-c} & & x \geq c \end{array} \right.\\ F\left(x;c\right) & = & \left\{ \begin{array}{ccc} \frac{x^{2}}{c} & & x < c \\ \frac{x^{2}-2x+c}{c-1} & & x \geq c \end{array} \right.\\ G\left(q;c\right) & = & \left\{ \begin{array}{ccc} \sqrt{cq} & & q < c \\ 1-\sqrt{\left(1-c\right)\left(1-q\right)} & & q \geq c \end{array} \right. \end{eqnarray*}$$

$$\begin{eqnarray*} \mu & = & \frac{c}{3}+\frac{1}{3}\\ \mu_{2} & = & \frac{1-c+c^{2}}{18}\\ \gamma_{1} & = & \frac{\sqrt{2}\left(2c-1\right)\left(c+1\right)\left(c-2\right)}{5\left(1-c+c^{2}\right)^{3/2}}\\ \gamma_{2} & = & -\frac{3}{5}\end{eqnarray*}$$

$$\begin{eqnarray*} h\left(X\right) & = & \log\left(\frac{1}{2}\sqrt{e}\right)\\ & \approx & -0.19314718055994530942.\end{eqnarray*}$$

Implementation: scipy.stats.triang