10
$\begingroup$

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 $\mu$?

$\endgroup$
  • 3
    $\begingroup$ Are you allowing for a negative gap? $\endgroup$ – T.J. Gaffney May 5 '16 at 15:41
  • $\begingroup$ I'm assuming the standard deviation is small enough relative to the mean that all gaps can be assumed to be positive. $\endgroup$ – mathcsguy May 5 '16 at 21:22
  • 1
    $\begingroup$ 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 $\endgroup$ – Henry May 6 '16 at 0:13
  • 1
    $\begingroup$ 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. $\endgroup$ – BGM May 6 '16 at 3:40
  • $\begingroup$ 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. $\endgroup$ – mathcsguy May 6 '16 at 17:14
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.