r - 最小二乘回归系数的非线性函数的标准误差和置信区间

Question

我在 R 中运行 OLS 回归，从中我得到了几个系数。这是部分代码:

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 上执行此操作，但结果是一个固定数字，没有任何 p-值或置信区间，就像计算器会告诉我的那样。

有谁知道我是否可以使用任何函数来计算置信区间？

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编者注

如果目标量是回归系数的线性函数，那么问题就简化为可以进行精确推断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