重要性抽样对重要性分布的选择非常敏感。既然你选择了$\lambda = 20$，您绘制的样本rexp的平均值为$1/20$有方差$1/400$. 这是你得到的分布

但是，您要评估的积分从 0 到$\pi =3.14$. 所以你想使用一个$\lambda$这给了你这样的范围。我用$\lambda = 1$.

使用$\lambda = 1$我将能够探索0到的完整积分空间$\pi$, 并且似乎只有几次平局$\pi$将被浪费。现在我重新运行你的代码，只改变$\lambda = 1$.

# clear the environment and set the seed for reproducibility
rm(list=ls())
gc()
graphics.off()
set.seed(1)

# function to be integrated
f <- function(x){
  1 / (cos(x)^2+x^2)
}

# importance sampling
importance.sampling <- function(lambda, f, B){
  x <- rexp(B, lambda) 
  f(x) / dexp(x, lambda)*dunif(x, 0, pi)
}

# mean value of f
mu.num <- integrate(f,0,pi)$value/pi

# initialize code
means  <- 0
sigmas <- 0
error  <- 0
CI.min <- 0
CI.max <- 0
CI.covers.parameter <- FALSE

# set a value for lambda: we will repeat importance sampling N times to verify
# coverage
N <- 100
lambda <- rep(1,N)

# set the sample size for importance sampling
B <- 10^4

# - estimate the mean value of f using importance sampling, N times
# - compute a confidence interval for the mean each time
# - CI.covers.parameter is set to TRUE if the estimated confidence 
#   interval contains the mean value computed by integrate, otherwise
# is set to FALSE
j <- 0
for(i in lambda){
  I <- importance.sampling(i, f, B)
  j <- j + 1
  mu <- mean(I)
  std <- sd(I)
  lower.CB <- mu - 1.96*std/sqrt(B)  
  upper.CB <- mu + 1.96*std/sqrt(B)  
  means[j] <- mu
  sigmas[j] <- std
  error[j] <- abs(mu-mu.num)
  CI.min[j] <- lower.CB
  CI.max[j] <- upper.CB
  CI.covers.parameter[j] <- lower.CB < mu.num & mu.num < upper.CB
}

# build a dataframe in case you want to have a look at the results for each run
df <- data.frame(lambda, means, sigmas, error, CI.min, CI.max, CI.covers.parameter)

# so, what's the coverage?
mean(CI.covers.parameter)
#[1] .95

如果你玩$\lambda$，你会看到，如果你把它做得很小（.00001）或很大，覆盖概率会很差。

编辑 - - - -

关于一旦你离开，覆盖概率就会降低$B = 10^4$到$B = 10^6$，这只是一个随机事件，基于您使用的事实$N = 100$复制。覆盖概率的置信区间$B = 10^4$是，

所以你不能真的说增加$B = 10^6$ 显着降低覆盖概率。

事实上，在你的代码中为同一个种子，改变$N = 100$到$N = 1000$，然后与$B = 10^4$，覆盖概率为 0.123，并且$B = 10^6$覆盖概率是$.158$.

现在，0.123 附近的置信区间为

因此，现在有了$N = 1000$复制，您会发现覆盖概率显着增加。