Generating a quasi Poisson distribution, version 2

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Here
and there, I mentioned two codes to generated quasiPoisson random
variables. And in both of them, the negative binomial approximation
seems to be wrong. Recall that the negative binomial distribution is

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg01.png

where

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg02.png

and in R, a negative binomial distribution can be parametrized using
two parameters, out of the following ones

  • the sizehttp://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg03.png
  • the probabilityhttp://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg04.png
  • the mean
http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg05.png

Recall also that the variance for a negative binomial distribution
should be

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg06.png

Here, we consider a distribution such that http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg07.png
and http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg08.png. In the previous posts, I used

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg09.png
http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg10.png

i.e.

rqpois = function(n, lambda, phi) {
mu = lambda
k = mu/(phi * mu - 1)
r1 = rnbinom(n, mu = mu, size = k)
r2 = rnbinom(n, size=phi*mu/(phi-1),prob=1/phi)
k = mu/phi/(1-1/phi)
r3 = rnbinom(n, mu = mu, size = k)
r4 = rnbinom(n, size=mu/phi/(1-1/phi),prob=1/phi)
r = cbind(r1,r2,r3,r4)
return(r)
}

but as we can see below, none of those two functions work,

> N=rqpois(1000000,2,4)
> mean(N[,1])
[1] 2.001992
> mean(N[,2])
[1] 8.000033
> var(N[,1])/ mean(N[,1])
[1] 7.97444
> var(N[,2])/ mean(N[,2])
[1] 4.002022

with the first one, the expected value is correct, while it is the
overdispersion parameter for the second. Now, if we do the maths
correctly, it comes

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg11.png

http://perso.univ-rennes1.fr/arthur.charpentier/latex/bineg12.png

which should now work,

> mean(N[,3])
[1] 2.001667
> mean(N[,4])
[1] 2.002776
> var(N[,3])/ mean(N[,3])
[1] 3.999318
> var(N[,4])/ mean(N[,4])
[1] 4.009647

So, finally it is better when we do the maths well.

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