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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 $\mu$?

mathcsguy
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    Are you allowing for a negative gap? – T.J. Gaffney May 05 '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 05 '16 at 21:22
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    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 06 '16 at 00:13
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    See, e.g. https://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 06 '16 at 03:40
  • 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 06 '16 at 17:14
  • @mathcsguy : If you assume the standard deviation is so small relative to the positive mean that ALL "gaps" will be positive, then why not just use a gamma distribution with the same expected value and variance? If the standard deviation of a gamma distribution is small compared to the positive expected value, then for many practical purposes that's a normal distribution. It is very very close to a normal distribution and it's always positive. And note that the waiting time until the $n$th arrival in a Poisson process has just that sort of gamma distribution. – Michael Hardy Aug 14 '17 at 19:43

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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.

Michael Hardy
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