From textbook, I know that:
If each $X_i$ is distributed as negative binomial $(r_i,p)$ then $\sum X_i$ is distributed as negative binomial $(\sum r_i ,p)$.
In R: dnbinom(x, size=r, prob=p)
However, I use a parametrization of negative binomial using mean and aggregation parameter [in R: dnbinom(x, size, mu) ]
What can I say from the sum of $X_i$ with such parametrization ?
[edit]
I come back to this question because finally the previous discussion does not solve my problem. From textboook, I know that NB(size1, prob)+NB(size2, prob)=NB(size1+size2, prob). But I use the parametrization: NB(size, mu) and then I need to know: NB(size, mu1)+NB(size, mu2) with size being constant. Note that mu=size*(1-prob)/prob and then prob=size/(size+mu)
Then prob is not a constant because it varies with mu and I cannot apply the solution with constant prob.
Based on simulations, it seems that : NB(size, mu1)+NB(size, mu2) = NB(size?, mu1+mu2)
Have you an idea about the solution to estimate the value of size?
Exemple of simulation in R language
size <- 1
mu1 <- 10
V1 <- rnbinom(10000, size=size, mu=mu1)
cat("prob1=", size/(size+mu1))
mu2 <- 20
V2 <- rnbinom(10000, size=size, mu=mu2)
cat("prob2=", size/(size+mu2))
hist(V1)
hist(V2)
hist(V1+V2)
library(MASS)
fitdistr(V1, "negative binomial")
fitdistr(V2, "negative binomial")
fitdistr(V1+V2, "negative binomial")
Marc
