Count models beyond the Poisson

Section of Standard errors, and models for counts, part of Generalised linear models

Count models beyond the Poisson

Type Section
Taught in Standard errors, and models for counts
Part Generalised linear models
Questions using it 2

A quasi-Poisson variance is not, by itself, a distribution you can draw from. If you want a distribution, and you only need extra variation in the direction of overdispersion, the usual next model is the negative binomial. One construction is a gamma mixture of Poissons: draw a rate from a gamma distribution, then draw a Poisson count at that rate. The mean and variance are \mathrm{E}(Y)=\mu, \qquad \mathrm{Var}(Y)=\mu+\frac{\mu^2}{\nu}, \qquad \nu>0. This is quadratic in the mean, not a constant times the mean. It is always at least the Poisson variance. As \nu\to\infty the extra term vanishes and the Poisson returns. The parameter \nu is traditionally called a dispersion, but a larger \nu concentrates the counts, so it behaves as a concentration. The negative binomial is an exponential family in \mu only when \nu is held fixed. The fit in MASS::glm.nb is a profile: fix \nu, solve the GLM for the mean, move \nu, and repeat.

That mixture cannot produce underdispersion. Counts that are tighter than a Poisson need a different family. The Conway–Maxwell–Poisson distribution has been available since Conway and Maxwell (1962), from queueing theory with dependent waiting times, and it was brought back into statistical use by Shmueli and coauthors in 2005. In the original scaling, f(y;\lambda,\nu)\propto \frac{\lambda^y}{(y!)^\nu}, \qquad y=0,1,2,\ldots, with \nu<1 overdispersed, \nu=1 exactly Poisson, and \nu>1 underdispersed. Calling \nu a dispersion runs the name backwards: larger \nu means a tighter distribution. The serious defect is that \lambda is not the mean. There is no closed form for \mathrm{E}(Y) or \mathrm{Var}(Y), so a regression on \lambda cannot be compared with a Poisson or negative-binomial regression on \mu.

A repair from 2017 puts the mean in the parameter and solves backwards for the rate: f(y;\mu,\nu)\propto \frac{\lambda(\mu,\nu)^y}{(y!)^\nu}. Now \mu is exactly the mean, the same parameter as in the Poisson and the negative binomial, and the variance still has no closed form. That is acceptable if the mean is what you came to estimate. The variance is needed for standard errors, and those can be computed after the fit. The bankruptcy-bids example, which is strongly underdispersed once the covariates are in the model, is the next chapter. The point to keep from this chapter is why the mean parameterisation exists: without it, a better AIC on the \lambda scale is not a comparison with a model for the mean.

  1. For the beetle fit, recompute \exp(0.3456\pm 1.96\times 0.02926) and state the interval as a multiplicative change in odds per 0.01 of log-dose. Explain, in one sentence, why the unscaled coefficient 34.6 is a bad unit for that sentence.

  2. The Poisson score set to zero is \sum_i(y_i-\mu_i){\boldsymbol{X}}_i={\boldsymbol{0}} under the log link. Show the cancellation, starting from the general GLM score (y_i-\mu_i)\,\mu'({\boldsymbol{X}}_i^\top\boldsymbol{\beta})/\mathrm{Var}(Y_i).

  3. In the mine regression, the quasi-Poisson estimates match the Poisson estimates and the standard errors do not. Explain why, using the fact that \phi factors out of the score and is estimated afterwards.

  4. State one thing the negative binomial variance Equation (6) in Standard errors, and models for counts can represent that Equation (4) in Standard errors, and models for counts cannot, and one thing the Conway–Maxwell–Poisson can represent that the negative binomial cannot.

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See also

Standard errors in a GLM · Binomial counts in a GLM · Count responses · Quasi-Poisson · Reading a log-linear coefficient · Mine-injury data · Overdispersion · Underdispersion

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