Is there a analytical formula for Super- and Sub-Poissonian distributions? I'm currently wrtiting my Bachelors thesis on photon statistics. The way different sources of light can be classified is by Poissonian (coherent light), Super-Poissonian (thermal light) and Sub-Poissonian (n-Photon-states). However, both books that I'm using as primary sources (Rodney Loudon, The Quantum theory of light and Mark Fox, Quantum Optics - An introduction) don't mention explicitly a formula for Sub and Super-Poisson distribution, just that for Sub-Poissonian light the variance $\Delta n$ falls below the expectation $\langle n \rangle$ (that is, $\Delta n < \langle n \rangle$), and for Super-Poissonian $\Delta n > \langle n \rangle$.
For my work, I want to plot three plots with varying $\Delta n$, but same $\langle n  \rangle$. However, I can't do this with the standard Poisson distribution
$$
P(n)=\frac{\mu^n e^{-n}}{n!}
$$
However, I've read that for large expectation values $\mu$, I can approximate the Poisson distribution with a normal distribution. Given that my expectation values usually don't exceed 10, this doesn't seem to be an option.
So I'm here, looking for help from mathematicians and hope that someone can help me.
 A: Unlike Poisson distribution, sub-Poissonian and super-Poissonian are not distributions, but rather properties of distributions (properties to have larger or smaller variance than the Poisson distribution with the same mean). An example of super-Poissonian distribution is negative binomial distribution, which has two parameters, $r \in \mathbb{Z}_{+}$ and $p \in (0, 1)$, for which $P(n) = \begin{pmatrix}n + r - 1\\ n\end{pmatrix} (1 - p)^r p^n$, the mean $\langle n \rangle = \frac{p r}{1-p}$ and variance $\langle (\Delta n)^2 \rangle = \frac{p r}{(1 - p)^2}$.
Another option is to multiply the parameter of Poisson distribution and the resulting random variable by a factor $\alpha$. For this distribution, $P(x) = \frac{(\alpha \mu)^{\alpha x}}{(\alpha x)!} e^{-\mu \alpha}$ if $\alpha x$ is integer and $0$ otherwise, the mean is $\langle x \rangle = \mu$ and variance is $\langle (\Delta x)^2 \rangle = \frac{\mu}{\alpha}$, so for $\alpha > 1$ the distribution will be sub-Poissonian and for $\alpha < 1$ super-Poissonian.
Edit: correction of the distribution, thanks to @Alex
