The negative binomial distribution can parameterized with $\mu$ (mean) and $\sigma$ (standard deviation) with $\text{NB}(\mu,\sigma)$. While this parameterization is a bit unusual, it sheds light on $ \lim_{\sigma \to \sqrt \mu} \text{NB}(\mu, \sigma) = \text{Poisson}(\mu)$.

Is there a way to generalize the notion of negative binomial to go beyond the $\sqrt \mu$ limit to the standard deviation? While the negative binomial can be interpreted as a over dispersed Poisson, the generalization would be akin to an under dispersed Poisson when $\sigma < \sqrt \mu$.

Assuming that $\mu$ is an integer, it would be reasonable to expect the generalization to converge to a Dirac on $\mu$ when the standard deviation converges to zero. It would also be reasonable to expect the generalization to exhibit continuous variations of probabilities on both sides around $\sqrt \mu$ .

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    $\begingroup$ It is much more convenient to use the index of dispersion $\text{ID} = \sigma^2 / \mu$ as the second parameter (along with the mean) because the the limit is for ID $\to 1$. $\endgroup$ – Yves Aug 13 '20 at 9:15

The answer is yes: the prolongating distribution is the Binomial distribution. The trilogy: Binomial|Poisson|Negative Binomial can be regarded as one single two-parameter distribution for a non-negative integer r.v. $N$, each probability $\text{Pr}\{N = n\}$ for $n \geqslant 0$ being a smooth function of the parameter vector. I will first recall some facts about a well-known trilogy forming a single distribution.

The Generalised Pareto Distribution (GPD)

Recall that the two-parameter GPD for a r.v. $X \geqslant 0$ involves a scale parameter $\sigma_X >0$ and a shape parameter $\xi_X$. The survival $S_X(x) := \text{Pr}\{X > x \}$ given by

$$ \tag{1} S_{X}(x) = \begin{cases} \left[ 1 + \xi_X \, x /\sigma_X \right]_{+}^{-1/\xi} & \text{ if }\xi_X \neq 0, \\ \exp\{ - x / \sigma_X \} & \text{ if }\xi_X = 0, \end{cases} \qquad x \geqslant 0, $$ where $z_+ := \max\{0,\, z\}$ for a real number $z$.

The GPD represents a trilogy of distributions corresponding to the possible signs of the shape parameter $\xi_X$. For $\xi_X <0$ we get a distribution with a finite upper end-point $-\sigma_X / \xi_X$, and with no attractive name. The cases $\xi_X = 0$ and $\xi_X >0$ correspond to the famous exponential distribution and -up to a re-parameterisation- to the Lomax distribution. A striking point is that the likelihood depends smoothly on the parameter vector which is consistently used for the three distributions. However there is no harm at considering the three distributions for themselves. The three distributions correspond to three ranges for the coefficient of variation $\text{CV}$: when $\xi_X < 0$ we get $\text{CV} < 1$ or underdispersion, when $\xi_X >0$ we get $\text{CV} > 1$ overdispersion, while $\text{CV} = 1$ in the exponential case $\xi_X = 0$.

Another trilogy: Binomial|Poisson|Negative Binomial

Back to our three famous distributions: Binomial, Poisson and Negative Binomial. I will not write what $\Pr\{N = n\}$ is in each of the three cases, but instead recall that the parameters are as follows

  • Binomial: size $\nu$ and probability $p$, with expectation $\lambda := \nu p$.

  • Poisson: rate $\lambda$, which is also the expectation.

  • Negative binomial: size $\nu$ and probability $p$, with expectation $\lambda := \nu p /q$ where $q := 1 - p$.

I retain here the parameterisation of the negative binomial distribution of the stats R package (?NegBinomial).

Remind that a random variable $N$ with integer value has no physical dimension. For such a r.v. we can usefully consider the index of dispersion $\text{ID}$: the ratio variance / mean, which is dimensionless. This leads to the terminology of under/over-dispersion for integer-valued r.vs, which must not be confused with that for 'ordinary' non-negative variables having a dimension which was used in the former section. Binomial, Poisson and negative binomial correspond to $\text{ID} < 1$ (underdispersion), $\text{ID} = 1$ and $\text{ID}>1$ (overdispersion). This can be viewed as an analogy with the GPD trilogy.

Now let us show that these three discrete distributions can be regarded as one. To see this, consider the probability generating function (p.g.f.) $G_N(z) := \mathbb{E}[z^N]$ which is given by

$$ \tag{2} G_N(z) = \begin{cases} [1 - (1- z) \, p]^\nu & \text{binomial}, \\ \exp\{-(1 - z) \,\lambda \} & \text{Poisson}, \\ \left[1 + (1 - z) \,p /q\right]^{-\nu} & \text{negative binomial}, \end{cases} $$

which holds at least for for $z$ complex with $|z| < 1$.

Keeping in mind the expression for the expectation $\lambda$ corresponding to the three cases, it transpires that $G_N(z)$ relates to the GPD survival $S_X(x)$ defined above through

$$ \tag{3} G_N(z) = S_{X}(1-z), \qquad \text{for }z \text{ real } 0 < z < 1, $$

provided that the GPD scale is taken as $\sigma_N := 1/\lambda$ and that the shape $\xi_N$ is given by

$$ \xi_N := \begin{cases} -1/\nu & \text{binomial}, \\ 0 & \text{Poisson}, \\ 1 / \nu& \text{negative binomial}. \end{cases} $$

Now we can try to define a probability distribution for $N$ with two parameters $\sigma_N >0$ and $\xi_N$ by using the formula

$$ G_N(z) = \left[ 1 + \xi_N \,\dfrac{1 - z}{\sigma_N}\right]^{-1/\xi_N} \qquad \text{if } \xi_N \neq 0. $$

For that aim, we will impose the condition: $\sigma_N + \xi_N >0$. In the binomial case when $\xi_N < 0$, this imposes that $p< 1$. This condition tells as well that $x = 1$ is an interior point of the support of the GPD with parameters $\sigma_N$ and $\xi_N$, and it allows using the principal determination of the logarithm to correctly define $G_N(z)$. While a non-integer value of $\nu > 0$ makes sense in the negative binomial case, a non-integer $\nu$ is not possible in the binomial case because the coefficients of the series expansion of $G_N(z)$ would then fail to be non-negative. So the parameter "domain" $\Theta_N$ is formed by the couples $[\sigma_N, \, \xi_N]$ with $\sigma_N >0$ and $\xi_N \geq 0$ or $\xi_N$ being the inverse of a negative integer with then $\sigma_N + \xi_N > 0$ (see Figure, left panel). This is not an open set, but note that every point with $\xi_N = 0$ is a cluster point.

Provided that $[\sigma_N,\,\xi_N]$ is in $\Theta_N$, we claim that $G_N(z)$ is a p.g.f. This is quite obvious because we saw that for each of the three cases $\xi_N >0$, $\xi_N=0$ and $\xi_N >0$ we get the p.g.f. of a distribution of our trilogy as in (2). Yet the positivity of the coefficients of the power series at $z=0$ could have been obtained for $\xi_N >0$ as a consequence of the fact that the GPD survival is a completely monotone function. For each possible value $n \geq 0$ of $N$, the value of the density $p_N(n; \sigma_N,\,\xi_N) := \text{Pr}\{N = n\}$ if infinitely diffferentiable w.r.t. $[\sigma_N,\, \xi_N]$ (see Figure, right panel) so it makes sense to consider $G_N(z)$ as the p.g.f. of one single distribution which can be used for ML estimation. Why not call this distribution Generalised Binomial?

Alternative parameterisation

Instead of the two parameters $\sigma_N$ and $\xi_N$, we can use the mean and the index of dispersion $$ \mathbb{E}[N] = 1/\sigma_N, \quad \text{ID}(N) = 1 + \xi_N / \sigma_N, $$ which leads to the inverse formula $$ \sigma_N = 1 / \mathbb{E}[N], \quad \xi_N = \left\{\text{ID}(N) - 1 \right\} / \mathbb{E}[N]. $$

The constraint $\sigma_N + \xi_N >0$ tells that $\text{ID}$ is positive. For any given value $\nu$ of $\mathbb{E}(N)$ can have $\text{ID} \approx 0$: this corresponds to the binomial distribution with probability $p \approx 1$ i.e. to a Dirac distribution with it mass at $\nu$, which hence must be an integer.

Remark: Maximum-Likelihood

Interestingly, if a sample $[X_i]$ of the GPD is available, the sign of the ML estimate $\widehat{\xi}_X$ of the shape parameter depends in a very simple way on the sample Coefficient of Variation $\widehat{\text{CV}} := \{M_2/M_1^2 -1\}^{1/2}$, where $M_r$ is the non-central sample moment of order $r$. Indeed, it can be shown that $\widehat{\xi}_X>0$ corresponds to the overdispersed case $\widehat{\text{CV}} > 1$, while $\widehat{\xi}_X<0$ corresponds to the underdispersed case $\widehat{\text{CV}} < 1$. In the case where $\widehat{\text{CV}}$ would exactly be equal to $1$, we would get the exponential distribution $\widehat{\xi}_X = 0$. If we consider each of the three distributions for itself, we may regard the ML estimation as impossible: for instance the estimation for the Lomax distribution when $\widehat{\text{CV}} < 1$.

Now consider the Generalised Binomial with an unknown size parameter - although this is quite uncommon in the binomial setting. The possibility of the ML estimation using a sample $[N_i]$ depends on the sample index of dispersion $\widehat{\text{ID}} := M_2/ M_1$. It is given by the conditions: $\widehat{\text{ID}} < 1$ for the binomial case - see Blumenthal S. and Bahiya R.C., and by $\widehat{\text{ID}} > 1$ in the negative binomial case. The later statement has been known for some years as Anscombe's conjecture for the Negative Binomial.


  • $\begingroup$ Awesome! I really like your perspective of 'Generalised Binomial', this is exactly what I was looking for. $\endgroup$ – Joannes Vermorel Aug 14 '20 at 13:19
  • $\begingroup$ Thank you. This question puzzled me a couple of years ago. $\endgroup$ – Yves Aug 14 '20 at 15:53

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