Generalized Gamma: log normal as limiting special case 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 $k\rightarrow \infty$?
 A: This site here shows how a reparametrization can be helpful. They set 


*

*$\lambda = k^{-1/2}$

*$\sigma = \beta^{-1}k^{-1/2}$

*$\mu = \ln(\theta) + \beta^{-1}\ln(\lambda^{-2})$


and because $k \to \infty$, $\lambda \to 0$. 
If it saves some time, here are the inverse transformations: 


*

*$k = \lambda^{-2}$,

*$\beta = \sigma^{-1}\lambda$,

*$\theta = \exp\left(\mu - \lambda^{-1}\sigma\ln(\lambda^{-2}) \right)$. 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.

