# 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 $\mu$?

• Are you allowing for a negative gap? – T.J. Gaffney May 5 '16 at 15:41
• I'm assuming the standard deviation is small enough relative to the mean that all gaps can be assumed to be positive. – mathcsguy May 5 '16 at 21:22
• If the standard deviation is small enough relative to the mean to make that assumption, then the number of events in a time frame of length $t$ is likely to be close to $t/\mu$. You will also lose the memorylessness property so the probability may be affected by the timing of the previous event before the interval – Henry May 6 '16 at 0:13
• See, e.g. en.wikipedia.org/wiki/Renewal_theory for the generalization. The holding time should be positive as said, and the most fundamental relationship between the $n$th jump time $J_n$ and the number of arrivals at time $t$ is $J_n \leq t$ iff $N(t) \geq n$ and that is how you relate the distribution of jump time with the distribution of the number of arrivals. – BGM May 6 '16 at 3:40
• Excellent feedback. @Henry, through simulation, with a frame size of $\mu$, the normally distributed wait times lead to discrete distribution which is a discrete approximation of a $N(1,\sigma)$ with peak in the 1-per bin and the 0-per bin value symmetric with the 2+-per bin value. BGM, I had not heard of Renewal Theory and will look into it. – mathcsguy May 6 '16 at 17:14

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 $1/\lambda.$