Template:Generalized gamma probability density function

Generalized Gamma Probability Density Function
The generalized gamma function is a three-parameter distribution. One version of the generalized gamma distribution uses the parameters k, $$\beta$$, and $$\theta $$. The pdf for this form of the generalized gamma distribution is given by:


 * $$f(t)=\frac{\beta }{\Gamma (k)\cdot \theta }{{\left( \frac{t}{\theta } \right)}^{k\beta -1}}{{e}^{-{{\left( \frac{t}{\theta } \right)}^{\beta }}}}$$

where $$\theta >0$$ is a scale parameter, $$\beta >0$$ and $$k>0$$ are shape parameters and $$\Gamma (x)$$ is the gamma function of x, which is defined by:


 * $$\Gamma (x)=\int_{0}^{\infty }\cdot {{e}^{-s}}ds$$

With this version of the distribution, however, convergence problems arise that severely limit its usefulness. Even with data sets containing 200 or more data points, the MLE methods may fail to converge. Further adding to the confusion is the fact that distributions with widely different values of k, $$\beta$$, and $$\theta $$ may appear almost identical [21]. In order to overcome these difficulties, Weibull++ uses a reparameterization with parameters k, $$\beta$$, and $$\theta $$ [21] where:


 * $$\begin{align}

& \mu =\ln (\theta )+\frac{1}{\beta }\cdot \ln \left( \frac{1} \right) \\ & \\  & \sigma =\frac{1}{\beta \sqrt{k}} \\ & \\  & \lambda =\frac{1}{\sqrt{k}} \\ \end{align}$$

where $$-\infty <\mu <\infty ,\text{ }\sigma >0\text{, }0<\lambda .$$

While this makes the distribution converge much more easily in computations, it does not facilitate manual manipulation of the equation. By allowing $$\lambda $$ to become negative, the pdf of the reparameterized distribution is given by:


 * $$f(t)=\left\{ \begin{matrix}

\tfrac{|\lambda |}{\sigma \cdot t}\cdot \tfrac{1}{\Gamma \left( \tfrac{1} \right)}\cdot {{e}^{\left[ \tfrac{\lambda \cdot \tfrac{\text{ln}(t)-\mu }{\sigma }+\text{ln}\left( \tfrac{1} \right)-{{e}^{\lambda \cdot \tfrac{\text{ln}(t)-\mu }{\sigma }}}} \right]}}\text{ if }\lambda \ne 0 \\ \tfrac{1}{t\cdot \sigma \sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}(t)-\mu }{\sigma } \right)}^{2}}}}\text{                           if }\lambda =0  \\ \end{matrix} \right.$$