The Generalized Gamma Distribution has p.d.f. defined as follows (see e.g. here for reference):

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

It's reported almost everywhere (for example here) that the log normal distribution is a special case for $k\rightarrow\infty$.

Anyway no information is given on how this log normal is determined.

In fact, log normal distribution is determined by two parameters $\mu$ and $\sigma^2$.

Please, can you help me to understand how to determine these two parameters starting from a generalized gamma distribution with known parameters and $k\rightarrow \infty$?


This site here shows how a reparametrization can be helpful. They set

  1. $\lambda = k^{-1/2}$
  2. $\sigma = \beta^{-1}k^{-1/2}$
  3. $\mu = \ln(\theta) + \beta^{-1}\ln(\lambda^{-2})$

and because $k \to \infty$, $\lambda \to 0$.

If it saves some time, here are the inverse transformations:

  1. $k = \lambda^{-2}$,
  2. $\beta = \sigma^{-1}\lambda$,
  3. $\theta = \exp\left(\mu - \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)$. Plugging them in you get: \begin{align*} f(t) &= \frac{\beta }{\Gamma (k)\cdot \theta }{{\left( \frac{t}{\theta } \right)}^{k\beta -1}}{{e}^{-{{\left( \frac{t}{\theta } \right)}^{\beta }}}}\\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) } \exp\left[\left(-\mu + \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)k\beta\right] t^{k\beta -1}\exp\left[-t^{\beta } \exp\left(\mu - \lambda^{-1}\sigma\ln(\lambda^{-2}) [-\sigma^{-1}\lambda]\right)\right] \\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) } \exp\left[\left(-\mu + \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)k\beta\right] t^{k\beta -1}\exp\left[-t^{\beta} \exp\left( \ln(\lambda^{-2}) - \mu\sigma^{-1}\lambda \right)\right] \\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) } \exp\left[\left(-\mu + \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)k\beta\right] t^{k\beta -1}\exp\left[-t^{\beta} \lambda^{-2} \exp\left(- \mu\sigma^{-1}\lambda \right)\right] \\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) } \exp\left[\left(-\mu + \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)\sigma^{-1}\lambda^{-1}\right] t^{\sigma^{-1}\lambda^{-1} -1}\exp\left[-t^{\sigma^{-1}\lambda} \lambda^{-2} \exp\left(- \mu\sigma^{-1}\lambda \right)\right] \\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) }t^{-1} \exp\left[\ln(t)\sigma^{-1}\lambda^{-1} -\mu\sigma^{-1}\lambda^{-1} + \lambda^{-2}\ln(\lambda^{-2}) -t^{\sigma^{-1}\lambda} \lambda^{-2} \exp\left(- \mu\sigma^{-1}\lambda \right)\right] \\ &= \frac{\sigma^{-1}\lambda }{\Gamma (\lambda^{-2}) }t^{-1} \exp\left[\ln(t)\sigma^{-1}\lambda^{-1} -\mu\sigma^{-1}\lambda^{-1} + \lambda^{-2}\ln(\lambda^{-2}) -\lambda^{-2} \exp\left(\sigma^{-1}\lambda\ln(t)- \mu\sigma^{-1}\lambda \right)\right]. \end{align*} I'm having a hard time finishing it off, though.
  • $\begingroup$ @user136737 I noticed you accepted my answer. Have you finished off the problem yourself? $\endgroup$ – Taylor May 14 '18 at 1:35
  • $\begingroup$ sorry for the delayed response. It looks like the reparametrization you reported was the answer to my problem. For $k\rightarrow \infty$ I used successfully $\sigma$ and $\mu$ to approximate the GGD with $Lognormal (\mu, \sigma^2)$, that is what I was trying to figure out. $\endgroup$ – user136737 May 27 '18 at 22:31

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