R-最小二乘回归系数的非线性功能的标准误差和置信区间

Question

我在r中运行一个OLS回归，从中我得到了几个系数。这是代码的一部分：

Attacks <- Treat.Terr.Dataset$Attacks[2:30]
Attackslag <- Treat.Terr.Dataset$Attacks[1:29]
TreatmentEffect <- Treat.Terr.Dataset$TreatmentEffect[2:30]
TreatmentEffectlag <- Treat.Terr.Dataset$TreatmentEffect[1:29]
olsreg <- lm(TreatmentEffect ~ TreatmentEffectlag + Attacks + Attackslag)
coeffs<-olsreg$coefficients

然后我需要计算：(Attacks + Attackslag) / (1 - TreatmentEffectlag)。问题是我可以通过使用(coeffs[3] + coeffs[4]) / (1 - coeffs[2])在R上进行此操作，但是结果是一个固定的数字，没有任何p值或置信区间，就像计算器会向我展示。

有人知道我是否可以用置信区间使用任何功能来计算此功能？

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编辑器注意

如果目标数量是回归系数的线性函数，则该问题将减少到一般线性假设测试，以便在可能的地方进行精确推断。

Answer 1

## variance-covariance of relevant coefficients
V <- vcov(olsreg)[2:4, 2:4]
## point estimate (mean) of relevant coefficients
mu <- coef(olsreg)[2:4]
## From theory of OLS, coefficients are normally distributed: `N(mu, V)`
## We now draw 2000 samples from this multivariate distribution
beta <- MASS::mvrnorm(n = 2000, mu, V)
## With those 2000 samples, you can get 2000 samples for your target quantity
z <- (beta[, 2] + beta[, 3]) / (1 - beta[, 1])
## You can get Monte Carlo standard error, and Monte Carlo Confidence Interval
mean(z)
sd(z)
quantile(z, prob = c(0.025, 0.975))
## You can of course increase sample size from 2000 to 5000

Answer 2

这是一个使用'CAR'软件包的Delta方法的独立示例：

# Simulate data
dat <- data.frame(Attacks = rnorm(30), Trt=rnorm(30))
dat <- transform(dat, AttacksLag = lag(Attacks), TrtLag = lag(Trt))
dat <- dat[2:30,]
# Fit linear model
m1 <- lm(Trt ~  TrtLag + Attacks + AttacksLag, data=dat)
# Use delta method
require("car")
del1 <- deltaMethod(m1, "(Attacks + AttacksLag) / (1 - TrtLag)")
# Simple Wald-type conf int
del1$Est +  c(-1,1) * del1$SE * qt(1-.1/2, nrow(dat)-length(coef(m1)))
# [1] -0.2921529  0.6723991