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我正在研究 R 中的两阶段阈值分位数回归模型，我的目标是分两个阶段估计简化形式方程的阈值（称为 rhohat）和结构方程的阈值（称为 qhat） . 在第一阶段，我通过分位数回归估计 rhohat 并获得拟合值。我使用这些拟合值来估计第二阶段的 qhat。代码如下（感谢 Bruce Hansen 教授，我修改了他的代码）：

#************************************************************#
#Quantile Regression.
qr.regress <- function(y,x){
beta <- c(rq(y~x,tau)$coefficients[1],rq(y~x,tau)$coefficients[2])
beta
}

#Threshold Estimation with one independent variable + constant.
joint_thresh <- function(y,x,q){
n=nrow(y)
k=ncol(x)
e=y-x%*%rq(y~x,tau)$coefficients[2]-rq(y~x,tau)$coefficients[1]
s0 <- det(t(e)%*%e)    
n1 <- round(.05*n)+k
n2 <- round(.95*n)-k
qs <- sort(q)
qs <- qs[n1:n2]
qs <- as.matrix(unique(qs))
qn <- nrow(qs)
sn <- matrix(0,qn,1)
for (r in 1:qn){
   d <- (q<=qs[r])
  xx <- (x)*(d%*%matrix(1,1,k))
  xx <- xx-x%*%rq(xx~x,tau)$coefficients[2]-rq(xx~x,tau)$coefficients[1]
  ex <- e-xx%*%rq(e~xx,tau)$coefficients[2]-rq(e~xx,tau)$coefficients[1]
  exw <- ex*(tau-(ex<0))
  sn[r] <- sum(exw)   
}
r <- which.min(sn)
smin <- sn[r] 
qhat <- qs[r]
d <- (q<=qhat)
x1 <- x*(d%*%matrix(1,1,k))
x2 <- x*((1-d)%*%matrix(1,1,k))
beta1 <- rq(y~x1,tau)$coefficients[2]
beta2 <- rq(y~x2,tau)$coefficients[2]
yhat <- x1%*%beta1+x2%*%beta2
list(yhat=yhat,qhat=qhat)
}

#Threshold Estimation with two independent variables + constant.
joint_thresh_2 <- function(y,x,q){
n <- nrow(y)
k <- ncol(x)
e=y-x[,1]%*%t(rq(y~x-1,tau)$coefficients[3])-x[,2]%*%t(rq(y~x-1,tau)$coefficients[2])-x[,3]%*%t(rq(y~x-1,tau)$coefficients[3])
s0 <- det(t(e)%*%e)    
n1 <- round(.05*n)+k
n2 <- round(.95*n)-k
qs <- sort(q)
qs <- qs[n1:n2]
qs <- as.matrix(unique(qs))
qn <- nrow(qs)
sn <- matrix(0,qn,1)
for (r in 1:qn){
  d <- (q<=qs[r])
  xx <- (x)*(d%*%matrix(1,1,k))
  xx <- xx[,1]-x%*%rq(x[,1]~xx-1,tau)$coefficients
  ex <- e-xx%*%qr.regress(e,xx)[2]-xx%*%qr.regress(e,xx)[1]
  exw <- ex*(tau-(ex<0))
  sn[r] <- sum(exw)  
}
r <- which.min(sn)
smin <- sn[r]
qhat <- qs[r]
d <- (q<=qhat)
x1 <- x*(d%*%matrix(1,1,k))
x2 <- x*((1-d)%*%matrix(1,1,k))
beta1 <- rq(y~x1-1,tau)$coefficients[1]
beta2 <- rq(y~x2-1,tau)$coefficients[3]
c1 <- rq(y~x1-1,tau)$coefficients[2]
c2 <- rq(y~x2-1,tau)$coefficients[2]
yhat <- x1[,1]%*%t(beta1)+x2[,3]%*%t(beta2)+c1+c2
list(yhat=yhat,qhat=qhat)
}

#Threshold Reduced-form eqn. 
tau=0.50

stqr_thresh_loop <- function(n,reps){
qhat=as.vector(reps)
rhohat=as.vector(reps)
kx <- 1 
sig <- matrix(c(1,0.5,0.5,1),2,2)  
x<- matrix(rnorm(n*kx),n,kx)
q <- matrix(rnorm(n),n,1) 
z2 <- cbind(matrix(1,n,1),q)
for(i in 1:reps){
e <- matrix(rnorm(n*2,quantile(rnorm(n),tau),1),n,2)%*%chol(sig)
z1=0.5+0.5*x*(q<=0)+1*x*(q>0)+e[,2] 
y=0.5+1*z1*(q<=1)+1.5*z1*(q>1)+1*z2[,2]+e[,1]
    out1 <-  joint_thresh(y=z1,x=x,q=q)
    z1hat<- out1$yhat
    rhohat[i] <- out1$qhat
zhat <- cbind(z1hat,z2)

out2 <- joint_thresh_2(y=y,x=zhat,q=q)
qhat[i] <- out2$qhat
        } #Close for loop.
list(rhohat=rhohat,qhat=qhat)
}

#************************************************************#

您可以轻松地自己运行它。问题是当我写作时，

stqr_thresh_loop(n=200,reps=500)

代码崩溃了，从来没有给我任何结果。我究竟做错了什么？非常感谢！！

4

1 回答 1

2

尝试一些具有较小值的时间：

> system.time({z = stqr_thresh_loop(n=10,reps=5)})
   user  system elapsed 
  0.612   0.000   0.615 
> system.time({z = stqr_thresh_loop(n=20,reps=5)})
   user  system elapsed 
  1.248   0.000   1.249 
> system.time({z = stqr_thresh_loop(n=40,reps=5)})
   user  system elapsed 
  2.740   0.000   2.743 
> system.time({z = stqr_thresh_loop(n=10,reps=10)})
   user  system elapsed 
  1.228   0.000   1.233 
> system.time({z = stqr_thresh_loop(n=10,reps=20)})
   user  system elapsed 
  2.465   0.000   2.472 
> system.time({z = stqr_thresh_loop(n=10,reps=40)})
   user  system elapsed 
  4.968   0.000   4.969

这在 n 和 reps 的经过时间中看起来非常线性。因此，n=200,reps=500 的时间至少是 n=20,reps=50 的 100 倍，即（在我的系统上）大约需要 20 分钟。你等了那么久吗？

尝试在循环中从 1:reps 打印“i”的值以掌握进度。

于 2012-07-17T15:43:34.183 回答