Exponential is to Poisson as Normal is to ??? In a time series, if the gap between successive events follows an exponential distribution with PDF $\lambda e^{-\lambda}$, then a Poisson distribution with parameter $\lambda$ tells you the probability of finding 0, 1, 2, etc events in time frames of width 1.
Now suppose the gap between successive events follows a normal distribution with parameters $\mu$ and $\sigma$. Is there a corresponding discrete distribution telling us the probability of finding 0, 1, 2, etc events in time frames of width $\mu$? 
 A: We want $0 < \sigma \ll \mu$ so that for practical purposes the probability that this normally distributed random variable is negative is $0.$
Let $\alpha = \mu^2/\sigma^2$ and $\lambda = \mu/\sigma^2.$ Consider the Gamma distribution
$$
\frac 1 {\Gamma(\alpha)} (\lambda x)^{\alpha-1} e^{-\lambda x} (\lambda\, dx) \quad \text{for } x\ge 0.
$$
This has expected value $\alpha/\lambda = \mu$ and variance $\alpha/\lambda^2 = \sigma^2,$ and the fact that $0<\sigma \ll \mu$ means that it is so close to a normal distribution as to be normally distributed for any practical purposes for which one could say that the probability of that normal random variable being negative is $0.$
In cases where $\alpha = \mu^2/\sigma^2$ is an integer $n,$ the Gamma distribution above is precisely the distribution of the waiting time until the $n$ arrival in the Poisson process, when the average waiting time until the next arrival is $1/\lambda.$
That's where I'd start thinking about this. Maybe I'll be back with more later.
