Obtaining a log-normal waiting time via sequential exponential or gamma distributions - is it possible? Assume you've use a accelerated failure time model to find that the transition of subjects from State A to State B is log-normally distributed with parameters $\mu$ = X and $\sigma$ = Y.
This now needs to be used in a differential equation model as the rate at which subjects move from State A to State B. However, just using $\mu$ as the average probability of moving from A to B results in an exponentially distributed waiting time, not a log-normally distributed time.
I know you can however use a series of sequential independent exponential distributions to obtain an overall gamma distributed waiting time. For example, if it takes 2 days to move from A to B, four equally spaced exponential distributions results in an overall gamma distributed waiting time with $\kappa$ = 4 and $\theta$ = 2.
The question is, is there a gamma distribution whose shape and scale parameters approximate a log-normal? I know if I used many exponential distributions for a high $\kappa$ the central limit theorem allows the gamma distribution to approximate a log-normal, but I'm not sure if there's a way to obtain a log-normal waiting time from that. Essentially, is there some $\kappa$ and some $\theta$ that yields something approximating a log-normal with $\mu$ = X and $\sigma$ = Y?
 A: As mentioned in my comment, a way to approximate the log-normal by a gamma distribution is through the use of the KL-divergence.  That is, we choose the parameters of the gamma distribution to minimise
$$KL(\kappa,\theta)=\int_{0}^{\infty}p(z|\mu,\sigma)\log\left(\frac{p(z|\mu,\sigma)}{q(z|\kappa,\theta)}\right)dz$$
Where $p(z|\mu,\sigma)$ is the log-normal density, and $q(z|\kappa,\theta)$ is the gamma density.  Let $C(\mu,\sigma)$ denote the terms which don't depend on $\kappa,\theta$ we have:
$$KL(\kappa,\theta)=C(\mu,\sigma)+\log[\Gamma(\kappa)]+\kappa\log(\theta)-(\kappa-1)E_p[\log(z)]+\frac{1}{\theta}E_p[z]$$
Taking derivatives gives the following equations to be solved:
$$\psi^{(0)}(\kappa)+\log(\theta)=E_p[\log(z)]=\mu$$
$$\kappa\theta=E_p[z]=\exp\left(\mu+\frac{1}{2}\sigma^2\right)$$
Where $\psi^{(m)}(x)$ is the polygamma function.  These are not solvable analytically, although we can eliminate $\theta$ and we have:
$$\kappa= \exp\left(\psi^{(0)}(\kappa)+\frac{1}{2}\sigma^2\right) $$
A starting value can be obtained by using the approximation $\psi^{(0)}(\kappa)\approx\log(\kappa)-(2\kappa)^{-1}$ which gives $\kappa\approx\sigma^{-2}$.  This approximation gets better for smaller values of $\sigma^2$.
Choosing this KL-divergence ensures that the log-likelihood ratio between p and q is small near the regions of high density in the exact log-normal distribution.
Doing the divergence the other way $q\log\left(\frac{q}{p}\right)$ results in the log-likelihood ratio is small near the regions of high density in $q$.  This is known as Variational Bayes (although it is a non-standard application of it) and will typically understate the variance of the exact distribution.  This results in the moment matching I gave in the comments.  So the new equations to be solved are
$$\psi^{(0)}(\kappa)+\log(\theta)=\mu$$
$$\kappa\psi^{(1)}(\kappa)+\frac{\psi^{(2)}(\kappa)}{2\sigma^2}=1$$
This could be important as the right tail of a gamma are lighter than a log-normal - $\exp(-az)$ compared to $\exp(-b[\log(z)]^2)$.  To see this we get the same approximation (differentiating the approximation to the digamma function) for $\theta$ given the value of $\kappa$ but now $\kappa\approx\frac{1+\sqrt{1+4\sigma^2}}{2\sigma^2}>\sigma^{-2}$ - that is we have decreased the coefficient of variation compared to using the KL divergence used initially (which is also understated as the exact inverse of the squared CV is $\frac{1}{\exp(\sigma^2)-1}$).
