Can anybody show me, why the dispersion parameter of the negative binomial distribution is taken to be one? In the Poisson case you can show that $E(y)/V(y)=\mu/\mu=1$ which is called equidispersion. But how can you show that the dispersion parameter of the negative binomial distribution is 1 as well?

  • $\begingroup$ Because actual parameter values are a matter of convention, and conventions vary, please define your dispersion parameter or at least provide a link to the definition you are using. $\endgroup$
    – whuber
    May 19, 2011 at 16:27

2 Answers 2


For a general exponential family, we have the variance in the following form:

$$Var(Y_{i})=\phi h(E[Y_{i}])$$

for some function $h(.)$. Using the wikipedia definition of negative binomial we have a pdf of:

$$p(Y_{i}=y|r,p)={r+y-1 \choose y}p^{y}(1-p)^{r}\;\;\;\;\;\;y=0,1,2,\dots$$

And this has expectation $E[Y_{i}]=\frac{pr}{1-p}$ and a variance equal to $Var[Y_{i}]=\frac{pr}{(1-p)^{2}}$. Note that this cannot be written in the usual form for a generalised linear model, but has the form:


And as such it can be seen as taking the function $h(x)=x+\frac{1}{r}x^{2}$. hence dispersion equal to "1". although neg binomial technically not a member of exponential family (as it is a mixture of exponential family, similar to student distribution).

  • $\begingroup$ I don't think you are allowed to have one of the parameters in $h$. Otherwise you could say that $Var[Y_i]=\frac{1}{1-p}E[Y_i]$, and the dispersion parameter is $\frac{1}{1-p}$. If $r$ is fixed, however, then the argument goes through. And that is what glm.nb uses for fitting: rerunning glm with a fixed value of $r$ and dispersion parameter 1. $\endgroup$
    – Aniko
    May 21, 2011 at 18:26
  • $\begingroup$ That is what I was hinting at with my last statement. You can also show that the negative binomial can't be written in exponential form unless $r$ is fixed, as you have $p(y)=\exp\left(y\log(p)+r\log(1-p)+\log([r+y-1]!)-\log(y!)-\log([r-1]!)\right)$ $\endgroup$ May 23, 2011 at 1:28

Based on your previous question using glm.nb, I'll take a wild guess that you are referring to the text in the output of that function:

> library(MASS)
> a <- glm.nb(Days ~ Eth + Sex, data=quine)
> summary(a)

glm.nb(formula = Days ~ Eth + Sex, data = quine, init.theta = 1.171409701, 
    link = log)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.5901  -1.0235  -0.3985   0.3414   2.3438  

            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   2.9630     0.1351  21.936  < 2e-16 ***
EthN         -0.5738     0.1588  -3.613 0.000303 ***
SexM          0.2135     0.1594   1.340 0.180380    
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

(Dispersion parameter for Negative Binomial(1.1714) family taken to be 1)

    Null deviance: 182.14  on 145  degrees of freedom
Residual deviance: 168.12  on 143  degrees of freedom
AIC: 1112.9

Number of Fisher Scoring iterations: 1

              Theta:  1.171 
          Std. Err.:  0.145 

 2 x log-likelihood:  -1104.855 

In this output, the sentence about the dispersion parameter is an artifact of the approach that glm.nb is using for fitting the negative binomial distribution. It repeatedly calls glm for fixed values of the shape parameter theta, then updates theta based on the results of the fit. The sentence means that the distribution used by the final model is not overdispersed compared to the NegBin(1.1714) distribution, which is what you want, since that is the final fitted distribution. It says nothing about the overdispersion of NegBin(1.1714) compared to anything else.

  • $\begingroup$ Sorry for being unprecisely. So first of all theta is the shape parameter of the negbin distribution (it's not a scale parameter) [books.google.de/…. I just need a proof that the dispersionparamter is taken to be one. In other words, can anybody show, that the negbin distribution is member of the exponential family with $\phi=1? $\endgroup$
    – MarkDollar
    May 20, 2011 at 7:48
  • 1
    $\begingroup$ Sorry about the shape/scale mistake, I fixed it. Please edit your original question to make it clear to everyone what you are after. $\endgroup$
    – Aniko
    May 20, 2011 at 13:45

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