From this https://content.wolfram.com/uploads/sites/19/2013/04/Zwilling.pdf :

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But I don't know how to solve it .

I know let the derivative of the function L with respect to $\alpha$ and $\beta$ equal to zero , then calculate the $\alpha$ and $\beta$ . But the equation is too complicated .

I wrote the derivative to $\alpha$ and let it equal to 0: $$ \begin{array}{l} \frac{\partial \ln L}{\partial \alpha}=\sum_{i=1}^{n}(\frac{y_{i}}{\alpha}+\frac{1}{\alpha^{2}} \ln (1+\alpha e^{x_{i} \cdot \beta})+(y_{i} + \frac{1}{\alpha})\frac{e^{x_{i} \cdot \beta}}{1+ \alpha e^{x_{i} \cdot \beta}} \\ -\frac{\Gamma^{\prime}(y_{i}+\frac{1}{\alpha})}{\alpha^{2} \Gamma(y_{i}+\frac{1}{\alpha})}+\frac{\Gamma^{\prime}(\frac{1}{\alpha})}{\alpha^{2} \Gamma(\frac{1}{\alpha})} ) = 0 \end{array} $$

But have no idea how to calculate the $\alpha$ along with $\beta$ , and even without of $\beta$ the left equation is complicated , do you have some tricks or technical to solve such equation ?

  • $\begingroup$ Are you looking for a closed-form solution, like with OLS? $\endgroup$ – Dimitriy V. Masterov Aug 28 '20 at 7:48
  • $\begingroup$ @Dimitriy V. Masterov I am curious on the closed-form, and grandient descent doesn't looks like to apply well on this equation. Actually, any solution is welcome , I haven't seen a workable one . $\endgroup$ – Mithril Aug 28 '20 at 9:01
  • $\begingroup$ There is no closed-form solution, but here is a good paper by Greene. I believe your model is a special case, with P=2. $\endgroup$ – Dimitriy V. Masterov Aug 28 '20 at 18:00

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