Question

我在R中运行OLS回归，从中得到几个系数。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)。问题是我可以使用(coeffs[3] + coeffs[4]) / (1 - coeffs[2])在R上执行此操作，但结果是固定数字，没有任何p值或置信区间，就像计算器会显示我一样。

有没有人知道我是否可以用置信区间来计算这个函数？

编辑注释

如果目标数量是回归系数的线性函数，则问题会减少到可能进行精确推理的一般线性假设检验。

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

这是一个使用来自“汽车”的delta方法的独立示例。包：

# 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

最小二乘回归系数非线性函数的标准误差和置信区间

2 个答案: