问从glm和lmer中提取剩余方差

EN

library(lme4)
library(ggplot2)

setwd("~/Documents/r_code/multilevel_modelling/")

rm(list=ls())

oxford.df <- read.fwf("oxboys/OXBOYS.DAT",widths=c(2,7,6,1))
names(oxford.df) <- c("stu_code","age_central","height","occasion_id")
oxford.df <- oxford.df[!is.na(oxford.df[,"age_central"]),]
oxford.df[,"stu_code"] <- factor(as.character(oxford.df[,"stu_code"]))
oxford.df[,"dummy"] <- 1

chart <- ggplot(data=oxford.df,aes(x=occasion_id,y=height))
chart <- chart + geom_point(aes(colour=stu_code))

# see if lm and glm give the same estimate
glm.01 <- lm(height~age_central+occasion_id,data=oxford.df)
glm.02 <- glm(height~age_central+occasion_id,data=oxford.df,family="gaussian")
summary(glm.02)
vcov(glm.02)
var(glm.02$residual)
(logLik(glm.01)*-2)-(logLik(glm.02)*-2)
1-pchisq(-2.273737e-13,1)
# lm and glm give the same estimation
# so glm.02 will be used from now on

# see if lmer without level2 variable give same result as glm.02
mlm.03 <- lmer(height~age_central+occasion_id+(1|dummy),data=oxford.df,REML=FALSE)
(logLik(glm.02)*-2)-(logLik(mlm.03)*-2)
# 1-pchisq(-3.408097e-07,1)
# glm.02 and mlm.03 give the same estimation, only if REML=FALSE

> mlm.03
Linear mixed model fit by maximum likelihood 
Formula: height ~ age_central + occasion_id + (1 | dummy) 
   Data: oxford.df 
  AIC  BIC logLik deviance REMLdev
 1650 1667 -819.9     1640    1633
Random effects:
 Groups   Name        Variance Std.Dev.
 dummy    (Intercept)  0.000   0.0000  
 Residual             64.712   8.0444  
Number of obs: 234, groups: dummy, 1

Fixed effects:
            Estimate Std. Error t value
(Intercept)  142.994     21.132   6.767
age_central    1.340     17.183   0.078
occasion_id    1.299      4.303   0.302

Correlation of Fixed Effects:
            (Intr) ag_cnt
age_central  0.999       
occasion_id -1.000 -0.999

> var(resid(glm.01))
[1] 64.98952
> sd(resid(glm.01))
[1] 8.061608