Overview
For count response variables, the glm framework has two options. The Poisson family and the negative binomial family. In this vignette, we will consider both and learn when to use one or the other.
Getting familiar with the Poisson family
First lets get an idea of how the Poisson family looks. If lambda is near 1, the Poisson family is right skewed. As lambda increases, the Poisson family becomes more symmetrical.
library(stats)
library(MASS)
library(dplyr, warn.conflicts= FALSE)
library(ggplot2)
library(GlmSimulatoR)
set.seed(1)
#lambda = 1
poisson <- rpois(n = 10000, lambda = 1)
poissonDF <- as.data.frame(x=poisson)
ggplot(poissonDF, aes(x=poisson)) +
geom_histogram(bins = 100)
#lambda = 5
poisson <- rpois(n = 10000, lambda = 5)
poissonDF <- as.data.frame(x=poisson)
ggplot(poissonDF, aes(x=poisson)) +
geom_histogram(bins = 100)
#lambda = 10
poisson <- rpois(n = 10000, lambda = 10)
poissonDF <- as.data.frame(x=poisson)
ggplot(poissonDF, aes(x=poisson)) +
geom_histogram(bins = 100)
GLM Poisson
Lets create data and train a model.
set.seed(1)
simdata <- simulate_poisson(N = 10000, weights = c(.5, 1))
#Response looks similar to above histograms
ggplot(simdata, aes(x=Y)) +
geom_histogram(bins = 100)
glmPoisson <- glm(Y ~ X1 + X2, data = simdata, family = poisson(link = "log"))
summary(glmPoisson)
##
## Call:
## glm(formula = Y ~ X1 + X2, family = poisson(link = "log"), data = simdata)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.7852 -0.7074 -0.0209 0.6427 3.4563
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.996422 0.014963 66.59 <2e-16 ***
## X1 0.506335 0.006625 76.43 <2e-16 ***
## X2 0.996936 0.006816 146.26 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for poisson family taken to be 1)
##
## Null deviance: 38086 on 9999 degrees of freedom
## Residual deviance: 10164 on 9997 degrees of freedom
## AIC: 60925
##
## Number of Fisher Scoring iterations: 4
The estimated weights are close the the weights argument in simulate_poisson. The model estimates are accurate.
Poisson mean variance relationship
One important characteristic of the Poisson family is \(\mu\) equals \(\sigma^2\). We see the sample statistics are nearly equal for different values of lambda.
set.seed(1)
#lambda = 1
poisson <- rpois(n = 10000, lambda = 1)
mean(poisson)
## [1] 1.0055
## [1] 1.00617
#lambda = 5
poisson <- rpois(n = 10000, lambda = 5)
mean(poisson)
## [1] 4.9992
## [1] 4.999899
#lambda = 10
poisson <- rpois(n = 10000, lambda = 10)
mean(poisson)
## [1] 10.04
## [1] 10.13221
Poisson vs Negative Binomial
For modeling, the main difference between Poisson and the negative binomial is the extra parameter. The mean variance relationship is \(\sigma^2\) equals \(\mu + \mu^2/\theta\) for the negative binomial. Through \(\theta\) many relationships are possible.
When the response variable is a count, but \(\mu\) does not equal \(\sigma^2\), the Poisson distribution is not applicable. Over dispersion can be detected by dividing the residual deviance by the degrees of freedom. If this quotient is much greater than one, the negative binomial distribution should be used. There is no hard cut off of “much larger than one”, but a rule of thumb is 1.10 or greater is considered large.
Getting familiar with the negative binomial family
When \(\theta\) is large, \(\mu + \mu^2/\theta\) is roughly equal to \(\mu\) . Therefore the negative binomial will be similar to Poisson distribution.
set.seed(1)
poisson <- rpois(n = 10000, lambda = 1)
poissonDF <- as.data.frame(x=poisson)
ggplot(poissonDF, aes(x=poisson)) +
geom_histogram(bins = 100)
negBin <- rnegbin(n = 10000, mu = 1, theta = 1000)
negBinDF <- as.data.frame(x=negBin)
ggplot(negBinDF, aes(x=negBin)) +
geom_histogram(bins = 100)
If \(\theta\) is small, \(\mu + \mu^2/\theta\) does not approximately equal \(\mu\). The negative binomial will look different than the Poisson distribution. Note the difference in scale on the x axis and y axis.
set.seed(1)
poisson <- rpois(n = 10000, lambda = 1)
poissonDF <- as.data.frame(x=poisson)
ggplot(poissonDF, aes(x=poisson)) +
geom_histogram(bins = 100)
negBin <- rnegbin(n = 10000, mu = 1, theta = 1)
negBinDF <- as.data.frame(x=negBin)
ggplot(negBinDF, aes(x=negBin)) +
geom_histogram(bins = 100)
GLM Negative Binomial
Lets create data where the flexibility of the negative binomial is needed and compare the Poisson estimates to the negative binomial estimates.
set.seed(1)
simdata <- simulate_negative_binomial(N = 10000, weights = c(.5, 1), ancillary = 5) #ancillary is theta.
#Response looks like a negative binomial distribution.
ggplot(simdata, aes(x=Y)) +
geom_histogram(bins = 200)
glmPoisson <- glm(Y ~ X1 + X2, data = simdata, family = poisson(link = "log"))
glmNB <- glm.nb(Y ~ X1 + X2, data = simdata, link = "log")
summary(glmPoisson)
##
## Call:
## glm(formula = Y ~ X1 + X2, family = poisson(link = "log"), data = simdata)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -7.8711 -1.9642 -0.3617 1.3337 11.0001
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.995434 0.014925 66.70 <2e-16 ***
## X1 0.511899 0.006608 77.47 <2e-16 ***
## X2 0.995539 0.006797 146.46 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for poisson family taken to be 1)
##
## Null deviance: 91470 on 9999 degrees of freedom
## Residual deviance: 63327 on 9997 degrees of freedom
## AIC: 113055
##
## Number of Fisher Scoring iterations: 4
##
## Call:
## glm.nb(formula = Y ~ X1 + X2, data = simdata, link = "log", init.theta = 4.967762132)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -4.0892 -0.8275 -0.1503 0.5161 3.7115
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.98915 0.03658 27.04 <2e-16 ***
## X1 0.51132 0.01689 30.28 <2e-16 ***
## X2 1.00018 0.01707 58.60 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for Negative Binomial(4.9678) family taken to be 1)
##
## Null deviance: 14739 on 9999 degrees of freedom
## Residual deviance: 10364 on 9997 degrees of freedom
## AIC: 77883
##
## Number of Fisher Scoring iterations: 1
##
##
## Theta: 4.9678
## Std. Err.: 0.0835
##
## 2 x log-likelihood: -77874.7800
The estimated slopes are very similar. However, the standard errors are not. The fact \(\mu\) does not equal \(\sigma^2\) has caused a major under estimation of standard errors for the Poisson GLM. For Poisson models, it is important to look at residual deviance divided by the degrees of freedom. When this quotient is larger than 1, the negative binomial should be considered. In the above the quotient was 6.33 (63327 / 9997) for the Poisson glm. For these data, the negative binomial model is more appropriate.