使用 bootstrapping 估计斜率系数的 95% 置信区间

Question

我有以下代码：

set.seed(1)
n <- 100
x <- rnorm(n)
y <- 5 + 3*x + rnorm(n, mean = 0, sd = 0.3)
df <- data.frame(x=x,y=y)
slope <- lm(y~x, data = df)$coefficients[2]

我想使用 100 个引导样本来估计斜率系数的 95% 置信区间。我以前从未做过引导，所以我有点卡住了。我知道我想使用 sample 从 1:n 采样 100 个索引并替换。我也知道我需要斜率的标准误差来计算 CI。但是，当我执行stderr<- sd(slope)/sqrt(length(slope)) 时，我得到<NA>。只是不确定如何将其转换为 R。我应该将 CI 设为 (2.95, 3.05)。

Answer 1

c(1, 2, 3, 4, 5) 的单个引导样本可以通过 sample(c(1, 2, 3, 4, 5), 5, TRUE) 进行采样。许多可以使用replicate 绘制：

# this was your original code
set.seed(1)
n <- 100
x <- rnorm(n)
y <- 5 + 3*x + rnorm(n, mean = 0, sd = 0.3)
df <- data.frame(x=x,y=y)

# before we do many bootstrap samples, first define what happens in each one of them    
one_regr <- function(){
  mysample <- sample(1:n, n, TRUE) # which observations pairs are in this 
                                   # bootstrap sample?
  my_df <- df[mysample, ]
  slope <- lm(y  ~x, data = my_df)$coefficients[2]
  return(slope)
}

# test run -- works!
one_regr()

# now we can do many replicates with replicate()
many_slopes <- replicate(5000, one_regr())
hist(many_slopes, breaks = 30)
quantile(many_slopes, c(.025, .975))

加法1：

如果您不仅对斜率进行采样，而且对截距以及 in 进行采样，则可以从中获得更多乐趣

one_regr <- function(){
  mysample <- sample(1:n, n, TRUE) # which observation pairs are in this 
                                   # bootstrap sample?
  my_df <- df[mysample, ]
  return( lm(y  ~x, data = my_df)$coefficients )
}

many_lines <- replicate(5000, one_regr())
intercepts <- many_lines[1,]
slopes <- many_lines[2,]
plot(df$x, df$y, main = "data and 500 regression lines")
for(i in 1:5000){   # just draw the regression lines
    abline(a = intercepts[i], b = slopes[i], col = "#00009001")
}
plot(slopes, intercepts, main = "bootstrapped coefficients", 
     pch = 16, col = "#00009010")
quantile(slopes, c(.025, .975))

加法2：

https://www.investopedia.com/terms/s/standard-error.asp 状态：

统计的标准误差 (SE) 是统计样本总体的近似标准差。

所以斜率的标准误差是我们自举斜率的标准差：

> sd(slopes)
[1] 0.02781162

粗略地说，这应该是 95%-CI 宽度的 4 倍：

> sd(slopes)
[1] 0.02781162
> abs(2.95 - 3.05)/4
[1] 0.025