I have a Poisson process that I would like to be able to compute the Posterior Predictive Distribution for.

I've seen many descriptions that say that the analytic form of the distribution that I want is the Negative Binomial distribution, but I'm immediately faced with the problem that I've selected the Jeffrey's Prior for the conjugate Gamma Distribution and so my hyperparameter $\alpha$ isn't an integer. I assume that I can't use the Negative Binomial if I want to persist with a prior of $Gamma(0.5, 0)$, but I haven't seen this limitation described anywhere so I am not confident in my conclusion. What am I missing?

For convenience sake, let's assume that I replace the Jeffrey's prior with $Gamma(1, 0)$. At least now I have integers so I can talk in terms of counts of successes, failures and trials, but I don't understand how the accumulation of observations and the number of trials over which that data was gathered maps to the parameters of the Negative Binomial distribution. In particular, I don't understand what is meant by the $p$, the probability of success of a single trial. I am confused because I thought that $p$ for a Poisson process approaches zero; what $p$ is revealed by my tally of counts over intervals? My final point is that Wikipedia presents a number of formulations for the Negative Binomial and I'm not sure which one is appropriate for my situation.

Thanks for your assistance. I'm sorry to put such a basic question on this site, but I really can't figure it out on my own.


Commenter Juho K has answered that the Negative Binomial distribution can be generalised to real parameters. Thanks for that. More significantly, the second part of my question doesn't seem to be making sense, so I'm going to try and illustrate by way of an example.

In this example, I have constant duration successive intervals in which I am counting events per interval. Say my first 5 intervals have counts of ${3, 1, 2, 2, 4}$. I am assuming that the process that generates these numbers is a Poisson process. I start with a prior of $Gamma(0.5, 0)$ so that the distribution of my rate parameter $\lambda$ given my observed data is now $Gamma(12.5, 5)$. Ok.

Now I want to know the distribution of my variable for the next interval given the data that I have already seen. Wikipedia (and other sources) say that it is the Negative Binomial Distribution, which is characterised as follows

$f(k; r, p)$

where $k$ is the random variable "number of successes", $r$ is the number of failures, and $p$ is the probability of success of a single trial.

Online derivations of the posterior predictive distribution that I have seen have been quoted in terms of $y_{n+1} \sim NB(\alpha + \Sigma^n{y_i}, 1/(\beta + n + 1))$ so in my example $y_{n+1} \sim NB(12.5, 1/6)$. One sixth is a reasonable probability, but I don't know how 12.5 can be considered the "number of failures". My understanding must be wrong because as the number of elements of data are considered, the "number of failures" gets longer and the "probability of success" gets lower. I'd consider that one of the "alternative formulations" listed by Wikipedia might be more appropriate, but I can't figure out which one, or whether I'm just completely off.

  • $\begingroup$ Negative binomial can be generalized to real parameters (which may be the answer to the question), but I don't understand the part about $p$ approaching zero. Can you be more precise about what kind of data you have adn about the posterior predictive you are after. (The posterior predictive of a process would presumably be a process, but this sounds like you are most likely looking for a posterior predictive of a count). $\endgroup$ Apr 26, 2018 at 5:10
  • $\begingroup$ Have you looked at the wikipedia page for the negative binomial? $\endgroup$
    – Glen_b
    Apr 26, 2018 at 5:19
  • $\begingroup$ Hi Glen_b, yes, I did check the Wikipedia page but tbh the Wikipedia page was part of the source of my confusion because it provided several alternative formulations, none of which (in my mind) clearly coincide with my problem description. I know that I am missing something, but I need help to identify where my misunderstanding is. $\endgroup$
    – Peter Du
    Apr 26, 2018 at 6:10
  • $\begingroup$ Do you mean Poisson distribution or Poisson process? Rather than a lengthy and convoluted text, could you provide maths formulae? $\endgroup$
    – Xi'an
    Apr 26, 2018 at 7:51
  • $\begingroup$ Xi'an. I thought that I meant Poisson process, but you might have discovered the basic flaw in my thinking. $\endgroup$
    – Peter Du
    Apr 26, 2018 at 8:07


Your Answer

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

Browse other questions tagged or ask your own question.