变系数参数估计

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n<-100
t0<-runif(n,0,1)
t<-seq(0.01,1,0.01)
x1<-seq(0.1,10,0.1)
x2<-seq(0.2,20,0.2)
e<-rnorm(n,0,1)
y<-1+t*t+sin(t)*x1+x2*exp(t)+e
h<-1/100
u1<-matrix(0,nrow=n,ncol=n)
for(i in 1:n)
{
for(j in 1:n)
{u1[i,j]<-t[j]-t0
}
}
u2<-x1
u3<-u1*x1
u4<-x2
u5<-u1*x2
T<-u1/h
K<-exp(-(T)^2/2)/(2*pi)^(1/2)
for(k in 1:n)
{
a<-data.frame(t0,t,u1[k,],u2,u3[k,],u4,u5[k,],T[k,],K[k,],y)
}
for(m in 1:n)
{
fit<-glm(y~u1[m,]+u2+u3[m,]+u4+u5[m,],w=K[m,],data=a[m])
}
用R估计变系数模型的参数，就是那个a老报错。求大神！！

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关键词：参数估计 变系数 matrix 模型

逻辑有问题啊 方法要变通

Hamasakia 发表于 2015-11-23 10:57
逻辑有问题啊 方法要变通

您能说清楚点麽

Your problem is merely a weighted least square optimization. There are many packages available in R on weighted least squared estimate. Nonparametric methods like local polynomial, smoothing spline, etc,  have been integrated into R. You may use like the "locpol" package. From your code, I guess that you could be just a beginner. When doing programming, you'd better to reduce the number of loops, otherwise, computation time would be a big trouble when the sample size is very large.

楚韵荆风 发表于 2015-11-24 19:07
Your problem is merely a weighted least square optimization. There are many packages available in R  ...

对的呀，我刚开始学R。这个程序是想模拟出变系数模型的一系列参数，然后与实际值对比，可是就是出不来

同学你好，请问你的这个CODING 最后改对了没有？求分享下哇，吼吼，另外你是半参系数，可以直接用glm的package?