A Poisson distribution can measure events per unit time, and the parameter is $\lambda$. The exponential distribution measures the time until next event, with the parameter $\frac{1}{\lambda}$. One can convert one distribution into the other, depending on whether it is easier to model events or times.

Now, a gamma-poisson is a "stretched" poisson with a larger variance. A Weibull distribution is a "stretched" exponential with a larger variance. But can these two be easily converted into each other, in the same way Poisson can be converted into exponential?

Or is there some other distribution that is more appropriate to use in combination with the gamma-poisson distribution?

The gamma-poisson is also known as the negative binomial distribution, or NBD.


3 Answers 3


This is a fairly straight forward problem. Although there is a connection between the Poisson and Negative Binomial distributions, I actually think this is unhelpful for your specific question as it encourages people to think of negative binomial processes. Basically, you have a series of Poisson processes:

$$Y_i(t_i)|\lambda_i\sim Poisson(\lambda_i t_i)$$

Where $Y_i$ is the process and $t_i$ is the time you observe it, and $i$ denotes the individuals. And you are saying that these processes are "similar" by tying the rates together by a distribution:

$$\lambda_i\sim Gamma(\alpha,\beta)$$

On doing the integration/mxixing over $\lambda_i$, you have:

$$Y_i(t_i)|\alpha\beta\sim NegBin(\alpha,p_i)\;\;\; where \;\;p_i=\frac{t_i}{t_i+\beta}$$

This has a pmf of:

$$Pr(Y_i(t_i)=y_i|\alpha\beta) = \frac{\Gamma(\alpha+y_i)}{\Gamma(\alpha)y_i!}p_i^{y_i}(1-p_i)^\alpha$$

To get the waiting time distribution we note that:

$$Pr(T_i\leq t_i|\alpha\beta)=1-Pr(T_i> t_i|\alpha\beta)=1-Pr(Y_i(t_i)=0|\alpha\beta)$$ $$=1-(1-p_i)^\alpha=1-\left(1+\frac{t_i}{\beta}\right)^{-\alpha}$$

Differentiate this and you have the PDF:


This is a member of the generalized Pareto distributions, type II. I would use this as your waiting time distribution.

To see the connection with the Poisson distribution, note that $\frac{\alpha}{\beta}=E(\lambda_i|\alpha\beta)$, so that if we set $\beta=\frac{\alpha}{\lambda}$ and then take the limit $\alpha\to\infty$ we get:

$$\lim_{\alpha\to\infty}\frac{\alpha}{\beta}\left(1+\frac{t_i}{\beta}\right)^{-(\alpha+1)}=\lim_{\alpha\to\infty}\lambda\left(1+\frac{\lambda t_i}{\alpha}\right)^{-(\alpha+1)}=\lambda\exp(-\lambda t_i)$$

This means that you can interpret $\frac{1}{\alpha}$ as an over-dispersion parameter.

  • 1
    $\begingroup$ You can also note that the waiting time distribution is, roughly speaking, an exponential distribution with a Gamma random rate parameter and strictly speaking this is a Beta distribution of the second kind, as for any Gamma distribution with a Gamma random rate parameter. $\endgroup$ Sep 24, 2012 at 7:50
  • $\begingroup$ Using @probabilityislogic as a basis, I found the following article providing more detail on the relationship between NBD and Pareto: Gupta, Sunil and Donald G. Morrison. Estimating Heterogeneith in Consumers' Purchase Rates. Marketing Science, 1991, 10(3), 264-269. Thanks to all who helped me answer this question. $\endgroup$
    – zbicyclist
    Sep 24, 2012 at 16:31
  • $\begingroup$ +1, I guess this nice analytical form may no longer exist for $Poisson(\lambda_i t_i + c)$, where $c$ is a constant. $\endgroup$
    – Randel
    Oct 20, 2017 at 18:03
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    $\begingroup$ @randel - you could get a "nice-ish" form by noting this rv is the sum of two independent rvs...$Z_i=Y_i+X_i $ where $Y_i $ is the same as above and $X_i\sim poisson (c) $. As $X_i $ doesn't depend on $\lambda_i $ or $Y_i $ the pdf of $Z_i $ is the convolution of the above negative binomial pdf and a poisson pdf. To get the waiting time distribution just multiply $Pr(Y_i=0) $ in the above answer by $Pr(X_i=0)=e^{-c}$. You then get waiting time cdf of $1-e^{-c}\left(1+\frac{t_i}{\beta}\right)^{-\alpha} $ and pdf of $e^{-c}\frac {\alpha}{\beta}\left(1+\frac{t_i}{\beta}\right)^{-(\alpha+1)}$. $\endgroup$ Oct 22, 2017 at 7:03
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    $\begingroup$ This won't work in terms of the mixing distribution, because you need $\lambda_i <ct_i^{-1} $ (else the poisson mean is negative). The gamma mixing distribution would need to be truncated (I also assumed that $c>0$ in my previous answer). This would mean no nb distribution. $\endgroup$ Oct 25, 2017 at 6:56

One possibility: Poisson is to Exponential as Negative-Binomial is to ... Exponential!

There is a pure-jump increasing Lévy process called the Negative Binomial Process such that at time $t$ the value has a negative binomial distribution. Unlike the Poisson process, the jumps are not almost surely $1$. Instead, they follow a logarithmic distribution. By the law of total variance, some of the variance comes from the number of jumps (scaled by the average size of the jumps), and some of the variance comes from the sizes of the jumps, and you can use this to check that it is overdispersed.

There may be other useful descriptions. See "Framing the negative binomial distribution for DNA sequencing."

Let me be more explicit about how the Negative Binomial Process described above can be constructed.

  • Choose $p \lt 1$.

  • Let $X_1, X_2, X_3, ...$ be IID with logarithmic distributions, so $P(x_i = k) = \frac{-1}{\log(1-p)} \frac{p^k}{k}.$

  • Let $N$ be a Poisson process with constant rate $-\log(1-p)$, so $N(t) = \text{Pois}(-t \log(1-p)).$

  • Let $NBP$ be the process so that

$$NBP(t) = \sum_{i=1}^{N(t)} X_i.$$

$NBP$ is a pure jump process with logarithmically distributed jumps. The gaps between jumps follow an exponential distribution with rate $-\log(1-p).$

I don't think it is obvious from this description that $NBP(t)$ has a negative binomial $NB(t,p)$ distribution, but there is a short proof using probability generating functions on Wikipedia, and Fisher also proved this when he introduced the logarithmic distribution to analyze the relative frequencies of species.

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    $\begingroup$ No, any compound Poisson process has an exponential waiting time. This means you add $\text{Pois}(\lambda t)$ IID random variables with some distribution. $\endgroup$ Sep 22, 2012 at 23:44
  • $\begingroup$ No, that is not what is meant by a compound Poisson process. en.wikipedia.org/wiki/Compound_Poisson_process " The jumps arrive randomly according to a Poisson process and the size of the jumps is also random, with a specified probability distribution." I did not say IID Poisson variables. You take the $N$th partial sum of IID logarithmic random variables where $N$ is the value of a Poisson process. $\endgroup$ Sep 23, 2012 at 0:35
  • $\begingroup$ If you multiply a Poisson process by $2$, this is not a Poisson process and the waiting times remain exponential. $\endgroup$ Sep 23, 2012 at 0:47
  • $\begingroup$ let us continue this discussion in chat $\endgroup$ Sep 23, 2012 at 3:02

I am not able to comment yet so I apologize is this isn't a definitive solution.

You are asking for the appropriate distribution to use with an NB but appropriate isn't entirely defined. If an appropriate distribution means appropriate for explaining data and you are starting with an overdispersed Poisson then you may have to look further into the cause of the overdispersion. The NB doesn't distinguish between a Poisson with heterogeneous means or a positive occurrence dependence (that one event occurring increases the probability of another occurring). In continuous time there is also duration dependence, eg positive duration dependence means the passage of time increases the probability of an occurrence. It was also shown that negative duration dependence asymptotically causes an overdispersed Poisson[1]. This adds to the list of what might be the appropriate waiting time model.

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    $\begingroup$ cause of the overdispersion: This is consumer purchase data. Individual consumers are poisson, each with a rate of purchase lambda. But not every consumer has the same lambda -- that's the cause of the overdispersion. The lambda purchasing rates are considered to be distributed as gamma. This is a common model (traces back to A.S.C. Ehrenberg), but I haven't found anything in his writing that answers this question. $\endgroup$
    – zbicyclist
    Sep 22, 2012 at 23:04

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