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After some frantic googling I do believe the answer is yes, but more so I am frustrated that the relation between the two parameters seems to be nowhere described explicitely so I do it here. (I hope this isn't against the rules of stackexchange.)

This very nice article states: we will denote the random variable Y having a negative binomial distribution as Y ~ NB($\mu, \kappa$) with a parameterization such that E(Y) = $\mu$, var(Y) = $\mu + \kappa \mu^2$.

I take this latter equation as the definition of $\kappa$.

Apparently this kappa is implemented in SAS.

Now turning to R, the function glm.nb in the MASS package contains a parameter $\mu$ which is obviously the same $\mu$ as above and a parameter $\theta$. The question is how $\theta$ and $\kappa$ are related. The documentation for glm.nb only refers to it as an "additional parameter". The answers to this and this stackexchange questions directly imply that $\theta = 1/\kappa$, but this question [EDIT: since removed] seems to suggest that $\theta = \kappa$.

The help page for negative binomial in R is nice and introduces a parameter called size that equals $1/\kappa$. Fitting glm.nb on random data generated by rnbinom for various choices of $\mu$ and size seems to support the thesis that $\theta = 1/\kappa$ (i.e. that $\theta$ = size) but also that for large values of size the estimation is poor.

Summarizing: I do believe that $\theta = 1/\kappa$ but it would be nice if there were an easily googlable place on the internet stating this explicitly. Maybe one of the answers to this questions can serve as such a place?

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  • $\begingroup$ since you are right with your suspicion, why not answer your question yourself? I means you basically state the answer in the question anyways. $\endgroup$ – Momo Oct 9 '13 at 12:48
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Following the suggestion of @Momo I will answer the question myself. What I had forgotten yesterday when I posted this question, is that I can just see what glm.nb does by typing "glm.nb" into the console. From the code it returns it can be inferred that indeed the variance equals $\mu + \mu^2/\theta$ so that $\theta = 1/\kappa$.

Also I'd like to use the opportunity to advertise this article I found since then, also addressing these matters.

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