我正在分析数据计数的泊松回归。泊松要求方差和均值相等,因此我正在检查分散度以确保这一点。对于分散,我使用两种方法:
> pm <- glm(myCounts ~ campaign, d, family = poisson)
> summary(pm)
Call:
glm(formula = myCounts ~ campaign, family = poisson, data = d)
Deviance Residuals:
Min 1Q Median 3Q Max
-4.074 -1.599 -0.251 1.636 6.399
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 4.955870 0.032174 154.03 <2e-16 ***
campaign -0.025879 0.001716 -15.08 <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: 428.04 on 35 degrees of freedom
Residual deviance: 195.81 on 34 degrees of freedom
AIC: 426.37
Number of Fisher Scoring iterations: 4
> dispersiontest(pm)
Overdispersion test
data: pm
z = 3.1933, p-value = 0.0007032
alternative hypothesis: true dispersion is greater than 1
sample estimates:
dispersion
5.53987
> # Calculate dispersion with Negative Binomial
> nb_reg <- glm.nb(myCounts ~ campaign, data=d)
> summary.glm(nb_reg)
Call:
glm.nb(formula = myCounts ~ campaign, data = d, init.theta = 22.0750109,
link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.9235 -0.7083 -0.1776 0.6707 2.4495
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.914728 0.082327 59.697 < 2e-16 ***
campaign -0.023471 0.003965 -5.919 1.1e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for Negative Binomial(22.075) family taken to be 1.069362)
Null deviance: 76.887 on 35 degrees of freedom
Residual deviance: 35.534 on 34 degrees of freedom
AIC: 325.76
Number of Fisher Scoring iterations: 1
如您所见,NB提供了1.069362的色散。但是,dispersiontest()在5.5上具有明显的过度分散。如果我没有记错的话,AER并不是参数测试,那么我们只能知道是否存在过度/分散不足。然而,这两种方法是矛盾的。
有人知道为什么吗?
在glm.nb()中,方差被参数化为𝜇 + 𝜇 ^ 2 / 𝜃,其中𝜇是您的平均值(请参见this discussion处的更多信息),𝜃是theta,而在泊松中则是ϕ * 𝜇
,其中ϕ是您看到的色散5.53987。
在负二项式中,色散1.069362
没有意义,您需要查看负二项式()中的theta,在您的情况下为22.075。我没有您的数据,但是将您的截距用作均值的粗略估计:
mu = exp(4.914728)
theta = 22.0750109
variance = mu + (mu^2)/theta
variance
977.6339
variance / mu
[1] 7.173598
这会给您一些类似于分散的东西。您应该注意,您的离散度是根据完整模型估算的,而我只是从您的截距中猜测出一个。
最重要的是结果完全不同意。您可以使用负二项式对数据进行建模。
下面是一个示例,将说明上述关系。我们使用负二项式(这是最简单的)来模拟过度分散的数据:
y = c(rnbinom(100,mu=100,size=22),rnbinom(100,mu=200,size=22))
x = rep(0:1,each=100)
AER::dispersiontest(glm(y~x,family=poisson))
Overdispersion test
data: glm(y ~ x, family = poisson)
z = 8.0606, p-value = 3.795e-16
alternative hypothesis: true dispersion is greater than 1
sample estimates:
dispersion
8.200214
大致来说,这是通过将各组的方差除以各组的均值得出的:
mean(tapply(y,x,var)/tapply(y,x,mean))
[1] 8.283044
您可以看到分散显示为1,而实际上您的数据过于分散:
summary(MASS::glm.nb(y~x))
Call:
MASS::glm.nb(formula = y ~ x, init.theta = 21.5193965, link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-4.0413 -0.6999 -0.1275 0.5800 2.4774
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 4.66551 0.02364 197.36 <2e-16 ***
x 0.66682 0.03274 20.37 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for Negative Binomial(21.5194) family taken to be 1)