where d is chosen such that \(c = \int_{0}^{\infty} t(x; \alpha) dx\) should be minimum, which boils down to the approximation \(d = 0.07+0.75\sqrt{1-\alpha}\).
Algo\(\left(0<\alpha<1\right)\)
- Initialize \(z\ =\ 0.07+0.75\left(1-\alpha\right)^{\frac{1}{2}}\) and \(b=\ 1+\frac{e^{-z}\alpha}{z}\).
- Generate \(U\) and set \(p\ =\ bU\). If \(p>1\), go to 5.
- Set \(x\ =\ zp^{\frac{1}{\alpha}}\). Generate \(v\). if \(v\le\frac{\left(2-x\right)}{2+x}\), deliver \(x\).
- If \(v>e^{-x}\) go to 2, otherwise deliver \(x\).
- Set \(x=-\ln\left(\frac{z\left(b-p\right)}{a}\right)\), \(y=\ \frac{x}{z}\). Generate \(v\). If \(v\left(\alpha+y-\alpha y\right)<1\), deliver \(x\).
- If \(v>y^{\alpha-1}\) go to 2, otherwise deliver \(x\).