I am interested in knowing whether there is any discrete probability distribution similar to Poisson but also extended in the negative value part (i.e., it can take negative values and there is no a lower limit for the values it can take).

Is there any family of discrete distributions meeting these requirements?


More requirements: The shape is not necessarily symmetrical. There is one more condition to be met: values close to $-1$ and $1$ are very likely to be observed, but value $0$ is very improbable. Besides this, the probabilities of observing values greater than $1$ or lower than $-1$ monotonically decrease (more or less like a Poisson distribution does, but also in the negative side).


Answering the comments below: yes, my data follow a sort of bimodal distribution with modes in $1$ and $-1$.

  • $\begingroup$ Are those the only requirements? For example: can the distribution have any shape (e.g. non-symmetric, skewed, or heavy tailed)? $\endgroup$ – Tim Jul 25 '16 at 9:53
  • 1
    $\begingroup$ By saying that $0$ is improbable while $-1$ and $1$ are very likely do you mean some kind of bimodal distribution with modes at $-1$ and $1$ ..? $\endgroup$ – Tim Jul 25 '16 at 10:06
  • $\begingroup$ Thanks for making it more clear with your edit. In this case you should follow @Glen_b suggestion to use some kind of mixture of distributions. $\endgroup$ – Tim Jul 25 '16 at 10:13

There's an infinite number of such distributions -- one need merely a way to get a probability for each value of $i$ and one has just such a distribution. It's a simple matter to spend a leisurely weekend afternoon or an evening inventing a hundred of the things.

However, few are explicitly named in the literature.

One example that does come up now and then is the Skellam distribution, which is the distribution of the difference between two independent Poisson variates. If the two Poisson parameters are equal, it is symmetric.

plot of three Skellam probability functions

  • One could as easily consider taking a difference of other distributions, such as geometric or negative binomial distributions, for example.

  • One might consider mixtures (of weights $w$ and $1-w$) of some distribution on the non-negative integers and the negative of some other distribution on the non-negative integers.

    e.g. one might take a 0.8 probability of a Poisson with mean 2 and 0.2 probability of the negative of a geometric with mean 2:

    p.m.f. of mixture of two discrete distributions

or indeed any other method that is of interest/relevance to some problem.

Edit: looking at the information in your edits, I'd suggest considering a three part mixture of a probability at 0, some distribution on the positive integers and some distribution on the negative integers. This could be geometric, negative binomial, Poisson etc, (zero-truncated as needed).

For example, if you took the positive and negative halves as geometric, you'd have 4 parameters in the distribution (the two geometric p's, and the two mixture probabilities on those geometric parts; the probability at zero being defined by the other two, since they must add to 1)

Here's an example:

mixture of geometric, flipped-geometric, 0

  • 1
    $\begingroup$ I mark this answer as favourite, although the one by @broncoAbierto also offers an interesting point of view. $\endgroup$ – Vicent Sep 19 '16 at 17:20
  • $\begingroup$ The multiplication by $Y$ in that answer simply takes a 50-50 mixture of the Poisson and minus a Poisson with the same parameter; it is particular example of the case in my second bullet point: "One might consider mixtures (of weights w and 1−w) of some distribution on the non-negative integers and the negative of some other distribution on the non-negative integers". However, the particular one mentioned in that answer is not suited to your present data set for several reasons -- ... ctd $\endgroup$ – Glen_b -Reinstate Monica May 13 '17 at 22:23
  • $\begingroup$ ctd... (i) the tiny size of your spike at 0 there's no Poisson that will match the rest and (ii) the asymmetry in your plot indicates you either need something other than 50-50 or two different Poisson parameters $\endgroup$ – Glen_b -Reinstate Monica May 13 '17 at 22:23

$$X \sim Y\times \mbox{Poisson}(\lambda)$$

where $Y$ follows a Rademacher distribution (i.e. $Y$ is $1$ with probability $\frac{1}{2}$ and $-1$ with probability $\frac{1}{2}$).

  • 4
    $\begingroup$ Actually $Y$ has a name, it's Rademacher distribution: en.wikipedia.org/wiki/Rademacher_distribution $\endgroup$ – Tim Jul 25 '16 at 10:16
  • $\begingroup$ Thanks for pointing that out. I'll edit to improve the answer. $\endgroup$ – broncoAbierto Jul 25 '16 at 10:17
  • $\begingroup$ $Y$ can also be described as $(-1)^Z$ where $Z$ is Bernoulli with parameter $1/2$. $\endgroup$ – Dilip Sarwate May 13 '17 at 17:30

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.