The GUD lists a parameter constraint that $a_0 = 0 \implies β = 1$. This constraint seems unnecessarily restrictive.
Deriving the GUD as a Weibull-transformed standard Pearson distribution as in Issue #15, for example, shows that not only $β = 1$ but also $b_0 = 0$ is sufficient to yield $a_0^{(\text{GUD})} = 0$. If one is willing to consider the GUD as a Weibull-transformed standard extended Pearson distribution, then the constraint seems wholly unnecessary: any combination of extended-Pearson and Weibull parameters such that $a_0^{(\text{ExtPearson})} - \frac{β-1}{β} b_0^{(\text{ExtPearson})} = 0$ yields $a_0^{(\text{GUD})} = 0$.
The GUD lists a parameter constraint that$a_0 = 0 \implies β = 1$ . This constraint seems unnecessarily restrictive.
Deriving the GUD as a Weibull-transformed standard Pearson distribution as in Issue #15, for example, shows that not only$β = 1$ but also $b_0 = 0$ is sufficient to yield $a_0^{(\text{GUD})} = 0$ . If one is willing to consider the GUD as a Weibull-transformed standard extended Pearson distribution, then the constraint seems wholly unnecessary: any combination of extended-Pearson and Weibull parameters such that $a_0^{(\text{ExtPearson})} - \frac{β-1}{β} b_0^{(\text{ExtPearson})} = 0$ yields $a_0^{(\text{GUD})} = 0$ .