and \(c\ =\ \frac{1}{\Gamma\left(\alpha+1\right)}\left[2^{\alpha}\left(1-e^{-\frac{1}{2}}\right)^{\alpha}+\alpha e^{-1}\right]\).
Algo\(\left(0<\alpha<1\right)\)
  1. Set \(a\ =\ \frac{\left(1-e^{-\frac{1}{2}}\right)^{^{^{\alpha}}}}{\left(1-e^{-\frac{1}{2}}\right)^{^{^{\alpha}}}+\frac{\alpha e^{-1}}{2^{\alpha}}}\) and \(b\ =\ \left(1-e^{-\frac{1}{2}}\right)^{^{^{\alpha}}}+\frac{\alpha e^{-1}}{2^{\alpha}}\).
  2. Generate \(U\) from uniform(0,1).
  3. If \(U\le a\), then \(x=-2\ln\left[1-\left(Ub\right)^{\frac{1}{\alpha}}\right]\), otherwise \(x=-\ln\left[\frac{2^{\alpha}}{\alpha}b\left(1-U\right)\right]\).
  4. Generate \(V\) from uniform(0,1). If \(x\le1,\ and\ V\le\frac{x^{\alpha-1}e^{-\frac{x}{2}}}{2^{\alpha-1}\left(1-e^{-\frac{x}{2}}\right)^{^{^{\alpha-1}}}}\) return x or go back to 2. If \(x>1\), and if \(V\le x^{\alpha-1}\) return x or go back to 2.

Algorithm 3\citep*{Kundu2007}

The above algorithm has two pieces of envelop with the change point being one, but as suggested by \ref{568991} the change might depend on \(\alpha\). Hence after using the idea of Best the majorization function becomes