问题
I am trying to get the five types of bootstrap intervals for linear and quantile regression. I was able to bootstrap and find the 5 boostrap intervals (Quantile,Normal,Basic,Studentized and BCa) for the linear regression using Boot from car and boot.ci from boot. When i tried to do the same for quantile regression using rq from quantreg, it throws up an error. Here is the sample code
Creating the model
library(car)
library(quantreg)
library(boot)
newdata = Prestige[,c(1:4)]
education.c = scale(newdata$education, center=TRUE, scale=FALSE)
prestige.c = scale(newdata$prestige, center=TRUE, scale=FALSE)
women.c = scale(newdata$women, center=TRUE, scale=FALSE)
new.c.vars = cbind(education.c, prestige.c, women.c)
newdata = cbind(newdata, new.c.vars)
names(newdata)[5:7] = c("education.c", "prestige.c", "women.c" )
mod1 = lm(income ~ education.c + prestige.c + women.c, data=newdata)
mod2 = rq(income ~ education.c + prestige.c + women.c, data=newdata)
Booting linear and quantile regression
mod1.boot <- Boot(mod1, R=999)
boot.ci(mod1.boot, level = .95, type = "all")
dat2 <- newdata[5:7]
mod2.boot <- boot.rq(cbind(1,dat2),newdata$income,tau=0.5, R=10000)
boot.ci(mod2.boot, level = .95, type = "all")
Error in if (ncol(boot.out$t) < max(index)) { :
argument is of length zero
1) Why does boot.ci not work for quantile regression
2)Using this solution I got from stackexchange, I was able to find the quantile CI.
Solution for quantile(percentile CI) for rq
t(apply(mod2.boot$B, 2, quantile, c(0.025,0.975)))
how do i obtain other CI for bootstrap (normal, basic, studentized, BCa).
3) Also, my boot.ci command for linear regression produces this warning
Warning message:
In sqrt(tv[, 2L]) : NaNs produced
What does this signify?
回答1:
Using summary.rq
you can calculate boostrap standard errors of model coefficients.
Five boostrap methods (bsmethods
) are available (see ?boot.rq
).
summary(mod2, se = "boot", bsmethod= "xy")
# Call: rq(formula = income ~ education.c + prestige.c + women.c, data = newdata)
#
# tau: [1] 0.5
#
# Coefficients:
# Value Std. Error t value Pr(>|t|)
# (Intercept) 6542.83599 139.54002 46.88860 0.00000
# education.c 291.57468 117.03314 2.49139 0.01440
# prestige.c 89.68050 22.03406 4.07009 0.00010
# women.c -48.94856 5.79470 -8.44712 0.00000
To calculate bottstrap confidence intervals, you can use the following trick:
mod1.boot <- Boot(mod1, R=999)
set.seed(1234)
boot.ci(mod1.boot, level = .95, type = "all")
dat2 <- newdata[5:7]
set.seed(1234)
mod2.boot <- boot.rq(cbind(1,dat2),newdata$income,tau=0.5, R=10000)
# Create an object with the same structure of mod1.boot
# but with boostrap replicates given by boot.rq
mod3.boot <- mod1.boot
mod3.boot$R <- 10000
mod3.boot$t0 <- coef(mod2)
mod3.boot$t <- mod2.boot$B
boot.ci(mod3.boot, level = .95, type = "all")
# BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
# Based on 10000 bootstrap replicates
#
# CALL :
# boot.ci(boot.out = mod3.boot, type = "all", level = 0.95)
#
# Intervals :
# Level Normal Basic Studentized
# 95% (6293, 6838 ) (6313, 6827 ) (6289, 6941 )
#
# Level Percentile BCa
# 95% (6258, 6772 ) (6275, 6801 )
回答2:
Thanks for everyone who helped. I was able to figure out the solution myself. I ran a loop calculating the coefficients of the quantile regression and then used boot and boot.ci respectively. Here is the code
Booting commands only, model creation from question
mod3 <- formula(income ~ education.c + prestige.c + women.c)
coefsf <- function(data,ind){
rq(mod3, data=newdata[ind,])$coef
}
boot.mod <- boot(newdata,coefsf,R=10000)
myboot.ci <- list()
for (i in 1:ncol(boot.mod$t)){
myboot.ci[[i]] <- boot.ci(boot.mod, level = .95, type =
c("norm","basic","perc", "bca"),index = i)
}
I did this as I wanted CI on all variables not just the intercept.
来源:https://stackoverflow.com/questions/46395907/confidence-interval-for-quantile-regression-using-bootstrap