3
$\begingroup$

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 $k\rightarrow \infty$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.
$\endgroup$
3
  • $\begingroup$ @user136737 I noticed you accepted my answer. Have you finished off the problem yourself? $\endgroup$
    – Taylor
    May 14, 2018 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, 2018 at 22:31
  • $\begingroup$ @Taylor You can finish the derivation by using a Taylor expansion to the second order of the enclosed exponential term ($x^{-2}\left[\alpha x - \exp\left(\alpha x\right)\right] \approx x^{-2} + \alpha^2/2$) and the Stirling approximation formula. $\endgroup$
    – Aubergine
    Feb 14, 2022 at 4:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.