回归分析作业1
作业内容：
1、重现《线性统计模型》（王松桂等）例3.1.3
2、模拟最小二乘法，并根据模拟结果得出估计系数与估计方差的一些结论，如无偏性等
作业实现（仅供参考）
#1
y<-c(10.98,11.13,12.51,8.40,9.27,8.73,6.63,8.50,7.82,9.14,8.24,12.19,11.8,
9.57,10.94,9.58,10.09,8.11,6.83,8.88,7.68,8.47,8.86,10.36,11.08)
x<-c(35.30,29.70,30.80,58.8,61.4,71.3,74.4,76.7,70.7,57.5,46.4,28.90,28.1,
39.1,46.80,48.5,59.30,70.0,70.0,74.5,72.1,58.1,44.6,33.40,28.60)
lm(y~x)
summary(lm(y~x))
Rss<-deviance(lm(y~x))
Rss/23
#最小二乘法系数估计：
#beta0=13.57290,beta1=-0.07873,sigma^2=0.760774
#中心化结果
x1<-x-mean(x)
lm(y~x1)
summary(lm(y~x1))
Rss<-deviance(lm(y~x1))
Rss/23
#最小二乘法系数估计：
#beta0= 9.43160,beta1=--0.07873,sigma^2=0.760774
plot(y~x)
abline(lm(y~x))#see picture1
plot(y~x1)
abline(lm(y~x1))#see picture2
#2真值模型：y=3.2*x+0.5+e
x<-rexp(100,3)
simulation<-function(x){
beta0<-rep(0,1000)
beta1<-rep(0,1000)
sigma<-rep(0,1000)
Rss<-rep(0,1000)
for(i in 1:1000){
set.seed(7*i)
e<-rnorm(100)
y<-3.2*x+0.5+e
beta0[i]<-lm(y~x)$coef[1]
beta1[i]<-lm(y~x)$coef[2]
Rss[i]<-deviance(lm(y~x))
sigma[i]<-deviance(lm(y~x))/98
}
print("beta0=")
print(mean(beta0))
print("beta1=")
print(mean(beta1))
print("sigma^2=")
print(mean(sigma))
ks.test(Rss,"pchisq",98)
}
simulation(x)
#输出结果：
#[1] "beta0="
#[1] 0.4965268
#[1] "beta1="
#[1] 3.199646
#[1] "sigma^2="
#[1] 1.003083
#
# One-sample Kolmogorov-Smirnov test
#data: Rss
#D = 0.0231, p-value = 0.6619
#alternative hypothesis: two-sided
#结果分析：验证了定理3.2.1,3.2.3