我试图通过casella的统计推断来编写示例10.3.2。 (我在这里附上了例子)。
我遇到制作相同情节的问题。有什么帮助吗?
模拟数据和比较表: n=25
lam<-5
nsim<-10000
set.seed(442256)
poisson<-function(nsim,n,lam){
ratio<-c()
distributionMean = NULL
for (i in 1 : nsim) distributionMean = c(distributionMean, mean(rpois( n, lam)))
d<- 2*n*((lam-distributionMean)-distributionMean*log(lam/distributionMean))
ratio<-c(ratio,d)
return(ratio )
}
logLi<-poisson(10000,25,5)
m<-matrix(0,2, 4)
m[1,1]=quantile(p1,0.80)
m[2,1]=qchisq(.80, df=1)
m[1,2]=quantile(p1,0.90)
m[2,2]=qchisq(.90, df=1)
m[1,3]=quantile(p1,0.95)
m[2,3]=qchisq(.95, df=1)
m[1,4]=quantile(p1,0.99)
m[2,4]=qchisq(.99, df=1)
row.names(m)<-c("simulated", "Chi-square")
colnames(m)<-c("80_perc", "90_perc","95_perc","99_perc")
答案 0 :(得分:0)
simulated <- quantile(logLi, c(0.8, 0.9, 0.95, 0.99))
chisquare <- qchisq(c(0.8, 0.9, 0.95, 0.99), df = 1)
rbind(simulated, chisquare)
图表的代码如下。
a <- hist(logLi, freq=FALSE, xlim = c(0,4),
breaks = seq(0, ceiling(max(logLi)), by = 0.1))
lines(a$mids, dchisq(a$mids, df = 1))