1
$\begingroup$

I am confused about the difference between modelling counts (i.e. catch) using a poisson distribution with an offset (i.e. effort) and modelling catch per unit effort using a gamma(link=log) distribution.

I have always used poisson models for count data, but I have lately analysed data that are highly dispersed and since the model did not converge (I have also tried ZIP, OLRE, negative binomial...) , I modelled catch per unit effort using a Gamma model, that gave me a very good fit.

I wonder if the Gamma model is a correct alternative approach and why it converges while the poisson does not.

$\endgroup$

1 Answer 1

2
$\begingroup$

(OP is probably long gone by now, but ...)

tl;dr you probably want to use a negative binomial model with an offset to account for exposure.

Offsets have a separate purpose from the choice of distribution; they model variation in the outcome based on sampling effort (exposure). They are necessary for count-data models (Poisson, negative binomial) where changing the scale or units of the response will mess up the mean-variance relationship, and merely convenient for continuous-data models (where they typically change the value of the estimated parameter but not the overall model fit/inference).

A Gamma distribution might give you a roughly equivalent fit to a negative binomial; for example, the variance of the Gamma for a sample with mean $\mu$ is $\mu^2/a$ (where $a$ is the shape parameter), while the variance of the negative binomial is $\mu + \mu^2/k$. (For low values of the mean, the variance of the negative binomial is a bit higher because of the effects of sampling a discrete variable.)

The negative binomial probably makes more sense because it is intended for modeling a discrete outcome.

For count data where the absolute size of the count is very large you'll probably get similar results from Gamma, negative binomial, and log-Normal models.

$\endgroup$
1
  • $\begingroup$ Thanks Ben, OP still around :) $\endgroup$ Apr 7 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.