LM的覆盖概率计算
Coverage probability calculation for LM
我正在尝试计算我在回归的截距和斜率上生成的一组残差 bootstrap 复制的覆盖概率。谁能告诉我如何计算置信区间的覆盖概率?非常感谢。
请注意,我使用 Qr 分解手动 运行 回归,但如果这样更容易,您可以使用 lm()
。我只是觉得手动做会更快。
set.seed(42) ## for sake of reproducibility
n <- 100
x <- rnorm(n)
e <- rnorm(n)
y <- as.numeric(50 + 25*x + e)
dd <- data.frame(id=1:n, x=x, y=y)
mo <- lm(y ~ x, data=dd)
# Manual Residual Bootstrap
resi <- residuals(mo)
fit <- fitted(mo)
ressampy <- function() fit + sample(resi, length(resi), replace=TRUE)
# Sample y values:
head(ressampy())
# Qr decomposition of X values
qrX <- qr(cbind(Intercept=1, dd[, "x", drop=FALSE]), LAPACK=TRUE)
# faster than LM
qr.coef(qrX, dd[, "y"])
# One Bootstrap replication
boot1 <- qr.coef(qrX, ressampy())
# 1000 bootstrap replications
boot <- t(replicate(1000, qr.coef(qrX, ressampy())))
编辑
结合jay.sf的回答,我重写了运行 lm()
方法的代码,并比较了jay.sf共享的link中第一种和第二种计算覆盖概率的方法=]:
library(lmtest);library(sandwich)
ci <- confint(coeftest(mo, vcov.=vcovHC(mo, type="HC3")))
ci
FUNInter <- function() {
X <- model.matrix(mo)
ressampy.2 <- fit + sample(resi, length(resi), replace = TRUE)
bootmod <- lm(ressampy.2 ~ X-1)
confint(bootmod, "X(Intercept)", level = 0.95)
}
FUNBeta <- function() {
X <- model.matrix(mo)
ressampy.2 <- fit + sample(resi, length(resi), replace = TRUE)
bootmod <- lm(ressampy.2 ~ X-1)
confint(bootmod, "Xx", level = 0.95)
}
set.seed(42)
R <- 1000
Interres <- replicate(R, FUNInter(), simplify=FALSE)
Betares <- replicate(R, FUNBeta(), simplify=FALSE)
ciinter <- t(sapply(Interres, function(x, y) x[grep(y, rownames(x)), ], "X\(Intercept\)"))
cibeta <- t(sapply(Betares, function(x, y) x[grep(y, rownames(x)), ], "Xx"))
#second approach of calculating CP
sum(ciinter[,"2.5 %"] <=50 & 50 <= ciinter[,"97.5 %"])/R
[1] 0.842
sum(cibeta[,"2.5 %"] <=25 & 25 <= cibeta[,"97.5 %"])/R
[1] 0.945
#first approach of calculating CP
sum(apply(ciinter, 1, function(x) {
all(data.table::between(x, ci[1,1], ci[1,2]))
}))/R
[1] 0.076
sum(apply(cibeta, 1, function(x) {
all(data.table::between(x, ci[2,1], ci[2,2]))
}))/R
[1] 0.405
根据 Morris et. al 2019, Table 6,覆盖概率 定义为真实 theta 位于 bootstrapped 置信区间 (CI)(即基于实际数据应用于许多样本的模型的置信区间,或者换句话说,新实验):
因此,我们要根据 OP 提出的 i.i.d 计算 CIs。 bootstrap R
次并计算 theta 在这些 CI 中出现或不出现的频率的比率。
首先,我们使用实际数据估计模型 mo
。
mo <- lm(y ~ x)
为了避免重复中不必要的解包拟合值 yhat
、残差 u
、模型矩阵 X
和系数 coef0
,我们预先提取它们。
yhat <- mo$fitted.values
u <- as.matrix(mo$residuals)
X <- model.matrix(mo)
theta <- c(50, 25) ## known from data generating process of simulation
在 bootstrap 函数 FUN
中,我们将要执行的所有步骤包装在一次复制中。为了快速应用 .lm.fit
,我们必须手动计算白色标准误差(与 lmtest::coeftest(fit, vcov.=sandwich::vcovHC(fit, type="HC1"))
相同)。
FUN <- function() {
## resampling residuals
y.star <- yhat + sample(u, length(u), replace=TRUE)
## refit model
fit <- .lm.fit(X, y.star)
coef <- fit$coefficients[sort.list(fit$pivot)]
## alternatively using QR, but `.lm.fit` is slightly faster
# qrX <- qr(X, LAPACK=TRUE)
# coef <- qr.coef(qrX, y.star)
## white standard errors
v.cov <- chol2inv(chol(t(X) %*% X))
meat <- t(X) %*% diag(diag(u %*% t(u))) %*% X
## degrees of freedom adjust (HC1)
d <- dim(X)
dfa <- d[1] / (d[1] - d[2])
white.se <- sqrt(diag(v.cov %*% meat %*% v.cov)*dfa)
## 95% CIs
ci <- coef + qt(1 - .025, d[1] - d[2])*white.se %*% t(c(-1, 1))
## coverage
c(intercept=theta[1] >= ci[1, 1] & theta[1] <= ci[1, 2],
x=theta[2] >= ci[2, 1] & theta[2] <= ci[2, 2])
}
现在我们使用 replicate
.
执行 bootstrap
R <- 5e3
set.seed(42)
system.time(res <- t(replicate(R, FUN())))
# user system elapsed
# 71.19 28.25 100.28
head(res, 3)
# intercept x
# [1,] TRUE TRUE
# [2,] FALSE TRUE
# [3,] TRUE TRUE
两列中的 TRUE
的 mean
同时跨行,或分别在每一列中,给出我们正在寻找的 覆盖概率 .
(cp.t <- mean(rowSums(res) == ncol(res))) ## coverage probability total
(cp.i <- colMeans(res)) ## coverage probability individual coefs
(cp <- c(total=cp.t, cp.i))
# total intercept x
# 0.8954 0.9478 0.9444
## values with other R:
# total intercept x
# 0.90700 0.95200 0.95200 ## R == 1k
# 0.89950 0.95000 0.94700 ## R == 2k
# 0.89540 0.94780 0.94440 ## R == 5k
# 0.89530 0.94570 0.94680 ## R == 10k
# 0.89722 0.94694 0.94777 ## R == 100k
这是重复 10 万次后的样子
剧情代码:
r1 <- sapply(seq(nrow(res)), \(i) mean(rowSums(res[1:i,,drop=FALSE]) == ncol(res)))
r2 <- t(sapply(seq(nrow(res)), \(i) colMeans(res[1:i,,drop=FALSE])))
r <- cbind(r1, r2)
matplot(r, type='l', col=2:4, lty=1, main='coverage probability', xlab='R',
ylab='cum. mean',ylim=c(.89, .955))
grid()
sapply(seq(cp), \(i) abline(h=cp[i], lty=2, col=i + 1))
legend('right', col=2:4, lty=1, legend=names(cp), bty='n')
数据:
set.seed(42)
n <- 1e3
x <- rnorm(n)
y <- 50 + 25*x + rnorm(n)
我正在尝试计算我在回归的截距和斜率上生成的一组残差 bootstrap 复制的覆盖概率。谁能告诉我如何计算置信区间的覆盖概率?非常感谢。
请注意,我使用 Qr 分解手动 运行 回归,但如果这样更容易,您可以使用 lm()
。我只是觉得手动做会更快。
set.seed(42) ## for sake of reproducibility
n <- 100
x <- rnorm(n)
e <- rnorm(n)
y <- as.numeric(50 + 25*x + e)
dd <- data.frame(id=1:n, x=x, y=y)
mo <- lm(y ~ x, data=dd)
# Manual Residual Bootstrap
resi <- residuals(mo)
fit <- fitted(mo)
ressampy <- function() fit + sample(resi, length(resi), replace=TRUE)
# Sample y values:
head(ressampy())
# Qr decomposition of X values
qrX <- qr(cbind(Intercept=1, dd[, "x", drop=FALSE]), LAPACK=TRUE)
# faster than LM
qr.coef(qrX, dd[, "y"])
# One Bootstrap replication
boot1 <- qr.coef(qrX, ressampy())
# 1000 bootstrap replications
boot <- t(replicate(1000, qr.coef(qrX, ressampy())))
编辑
结合jay.sf的回答,我重写了运行 lm()
方法的代码,并比较了jay.sf共享的link中第一种和第二种计算覆盖概率的方法=]:
library(lmtest);library(sandwich)
ci <- confint(coeftest(mo, vcov.=vcovHC(mo, type="HC3")))
ci
FUNInter <- function() {
X <- model.matrix(mo)
ressampy.2 <- fit + sample(resi, length(resi), replace = TRUE)
bootmod <- lm(ressampy.2 ~ X-1)
confint(bootmod, "X(Intercept)", level = 0.95)
}
FUNBeta <- function() {
X <- model.matrix(mo)
ressampy.2 <- fit + sample(resi, length(resi), replace = TRUE)
bootmod <- lm(ressampy.2 ~ X-1)
confint(bootmod, "Xx", level = 0.95)
}
set.seed(42)
R <- 1000
Interres <- replicate(R, FUNInter(), simplify=FALSE)
Betares <- replicate(R, FUNBeta(), simplify=FALSE)
ciinter <- t(sapply(Interres, function(x, y) x[grep(y, rownames(x)), ], "X\(Intercept\)"))
cibeta <- t(sapply(Betares, function(x, y) x[grep(y, rownames(x)), ], "Xx"))
#second approach of calculating CP
sum(ciinter[,"2.5 %"] <=50 & 50 <= ciinter[,"97.5 %"])/R
[1] 0.842
sum(cibeta[,"2.5 %"] <=25 & 25 <= cibeta[,"97.5 %"])/R
[1] 0.945
#first approach of calculating CP
sum(apply(ciinter, 1, function(x) {
all(data.table::between(x, ci[1,1], ci[1,2]))
}))/R
[1] 0.076
sum(apply(cibeta, 1, function(x) {
all(data.table::between(x, ci[2,1], ci[2,2]))
}))/R
[1] 0.405
根据 Morris et. al 2019, Table 6,覆盖概率 定义为真实 theta 位于 bootstrapped 置信区间 (CI)(即基于实际数据应用于许多样本的模型的置信区间,或者换句话说,新实验):
因此,我们要根据 OP 提出的 i.i.d 计算 CIs。 bootstrap R
次并计算 theta 在这些 CI 中出现或不出现的频率的比率。
首先,我们使用实际数据估计模型 mo
。
mo <- lm(y ~ x)
为了避免重复中不必要的解包拟合值 yhat
、残差 u
、模型矩阵 X
和系数 coef0
,我们预先提取它们。
yhat <- mo$fitted.values
u <- as.matrix(mo$residuals)
X <- model.matrix(mo)
theta <- c(50, 25) ## known from data generating process of simulation
在 bootstrap 函数 FUN
中,我们将要执行的所有步骤包装在一次复制中。为了快速应用 .lm.fit
,我们必须手动计算白色标准误差(与 lmtest::coeftest(fit, vcov.=sandwich::vcovHC(fit, type="HC1"))
相同)。
FUN <- function() {
## resampling residuals
y.star <- yhat + sample(u, length(u), replace=TRUE)
## refit model
fit <- .lm.fit(X, y.star)
coef <- fit$coefficients[sort.list(fit$pivot)]
## alternatively using QR, but `.lm.fit` is slightly faster
# qrX <- qr(X, LAPACK=TRUE)
# coef <- qr.coef(qrX, y.star)
## white standard errors
v.cov <- chol2inv(chol(t(X) %*% X))
meat <- t(X) %*% diag(diag(u %*% t(u))) %*% X
## degrees of freedom adjust (HC1)
d <- dim(X)
dfa <- d[1] / (d[1] - d[2])
white.se <- sqrt(diag(v.cov %*% meat %*% v.cov)*dfa)
## 95% CIs
ci <- coef + qt(1 - .025, d[1] - d[2])*white.se %*% t(c(-1, 1))
## coverage
c(intercept=theta[1] >= ci[1, 1] & theta[1] <= ci[1, 2],
x=theta[2] >= ci[2, 1] & theta[2] <= ci[2, 2])
}
现在我们使用 replicate
.
R <- 5e3
set.seed(42)
system.time(res <- t(replicate(R, FUN())))
# user system elapsed
# 71.19 28.25 100.28
head(res, 3)
# intercept x
# [1,] TRUE TRUE
# [2,] FALSE TRUE
# [3,] TRUE TRUE
两列中的 TRUE
的 mean
同时跨行,或分别在每一列中,给出我们正在寻找的 覆盖概率 .
(cp.t <- mean(rowSums(res) == ncol(res))) ## coverage probability total
(cp.i <- colMeans(res)) ## coverage probability individual coefs
(cp <- c(total=cp.t, cp.i))
# total intercept x
# 0.8954 0.9478 0.9444
## values with other R:
# total intercept x
# 0.90700 0.95200 0.95200 ## R == 1k
# 0.89950 0.95000 0.94700 ## R == 2k
# 0.89540 0.94780 0.94440 ## R == 5k
# 0.89530 0.94570 0.94680 ## R == 10k
# 0.89722 0.94694 0.94777 ## R == 100k
这是重复 10 万次后的样子
剧情代码:
r1 <- sapply(seq(nrow(res)), \(i) mean(rowSums(res[1:i,,drop=FALSE]) == ncol(res)))
r2 <- t(sapply(seq(nrow(res)), \(i) colMeans(res[1:i,,drop=FALSE])))
r <- cbind(r1, r2)
matplot(r, type='l', col=2:4, lty=1, main='coverage probability', xlab='R',
ylab='cum. mean',ylim=c(.89, .955))
grid()
sapply(seq(cp), \(i) abline(h=cp[i], lty=2, col=i + 1))
legend('right', col=2:4, lty=1, legend=names(cp), bty='n')
数据:
set.seed(42)
n <- 1e3
x <- rnorm(n)
y <- 50 + 25*x + rnorm(n)