[R-sig-ME] Zero-truncated Poisson distribution in glmmTMB

Erik Solbu er|k@@o|bu @end|ng |rom n|b|o@no
Sat Apr 25 09:24:37 CEST 2020


Hi everyone,

I'm trying to replace a package called 'poilog' with glmmTMB, but the results differ in such a way that I'm not sure how to interpret the results from glmmTMB.  I've written a short example below:

#### Start of code ####

# Required packages
library(poilog)
library(glmmTMB)

# Simulate using the 'poilog' package
# sim_01 <- rpoilog(S = 1000, mu = 0, sig = 2, keep0 = TRUE)
# This is the same as
sim_01 <- rpois(n = 1000, lambda = exp(rnorm(n = 1000, mean = 0, sd = 2)))

# estimate the parameters
# using the 'poilog' package, assume regular Poisson distribution
est_01 <- poilogMLE(n = sim_01, zTrunc = FALSE)
# using the 'poilog' package, assume zero truncated Poisson distribution
est_02 <- poilogMLE(n = sim_01[sim_01>0], zTrunc = TRUE)

# make data.frame for glmmTMB estimation
df_01 <- data.frame(abundance = sim_01,
                   species = 1:length(sim_01))

# using the 'glmmTMB' package, assume regular Poisson distribution
est_03 <- glmmTMB(abundance ~ (1 | species),
                      family = poisson(link = "log"),
                      data = df_01)
# using the 'glmmTMB' package, assume zero truncated Poisson distribution
est_04 <- glmmTMB(abundance ~ (1 | species),
                      family = truncated_poisson(link = "log"),
                      data = df_01[df_01$abundance>0, ])

# Compare estimates
est_df <- data.frame(method = rep(c("poilog", "glmmmTMB"), each = 2),
                     assumptions = rep(c("poisson", "truncated_poisson"), 2),
                     mu = c(est_01$par[1], est_02$par[1],
                            est_03$fit$par[1], est_04$fit$par[1]),
                     sigma = c(est_01$par[2], est_02$par[2],
                               exp(c(est_03$fit$par[2], est_04$fit$par[2]))))

est_df

#### End of code ####

When I fit the truncated_poisson model in glmmTMB, the mean increases and the standard deviation decreases, which is similar to what happens if you ignore the zeros and fit a regular poisson distribution, but it doesn't seem to be doing quite that either.

So, I'm hoping anyone knew how to interpret the mean and standard deviation from the truncated_poisson output and how they relate to the mean and standard deviation of a regular poisson model?

Thank you very much for reading this and any response is much appreciated.

Best regards,

Erik



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