H1Z1怎么备份到U盘_H1Z1备份数据恢复方法介绍

Question

This was something I thought about:

1. If you have data from a coin with a very low probability of success, it is possible that the Wald confidence interval can contain negative numbers and be illogical? But if you do bootstrap, the ci should never be illogical, i.e. worse case scenario the lower bound will always be exactly 0:

I verified this in R:

set.seed(123)
p_true <- 0.001
n <- 1000
n_sims <- 100
n_boot <- 100

wald_negative <- 0
bootstrap_negative <- 0

for(i in 1:n_sims) {
  x <- rbinom(n, 1, p_true)
  successes <- sum(x)
  
  p_hat <- successes / n
  
  se <- sqrt(p_hat * (1 - p_hat) / n)
  wald_lower <- p_hat - 1.96 * se
  wald_upper <- p_hat + 1.96 * se
  
  if(wald_lower < 0) {
    wald_negative <- wald_negative + 1
  }
  
  boot_estimates <- numeric(n_boot)
  for(j in 1:n_boot) {
    boot_x <- sample(x, n, replace = TRUE)
    boot_estimates[j] <- sum(boot_x) / n
  }
  
  boot_lower <- quantile(boot_estimates, 0.025, 
                         names = FALSE)
  boot_upper <- quantile(boot_estimates, 0.975, 
                         names = FALSE)
  
  if(boot_lower < 0) {
    bootstrap_negative <- bootstrap_negative + 1
  }
}

cat("Results from", n_sims, "simulations:\n")
cat("Wald CI negative:", wald_negative, 
    "times (", round(100 * wald_negative / n_sims, 1), 
    "%)\n")
cat("Bootstrap CI negative:", bootstrap_negative, 
    "times (", round(100 * bootstrap_negative / n_sims, 1), 
    "%)\n")

2. If you have data from an exponential random variable with a rate parameter very close to being 0, it is possible that the Wald confidence interval can contain negative numbers and be illogical? But if you do bootstrap, the ci should never be illogical. i.e. worse case scenario the lower bound will always be exactly 0

I also verified this in R:

set.seed(123)
lambda_true <- 0.001
n <- 100
n_sims <- 100
n_boot <- 100

wald_negative <- 0
bootstrap_negative <- 0

for(i in 1:n_sims) {
    x <- rexp(n, rate = lambda_true)
    
    lambda_hat <- 1 / mean(x)
    
    se <- lambda_hat / sqrt(n)
    wald_lower <- lambda_hat - 1.96 * se
    wald_upper <- lambda_hat + 1.96 * se
    
    if(wald_lower < 0) {
        wald_negative <- wald_negative + 1
    }
    
    boot_estimates <- numeric(n_boot)
    for(j in 1:n_boot) {
        boot_x <- sample(x, n, replace = TRUE)
        boot_estimates[j] <- 1 / mean(boot_x)
    }
    
    boot_lower <- quantile(boot_estimates, 0.025, 
                           names = FALSE)
    boot_upper <- quantile(boot_estimates, 0.975, 
                           names = FALSE)
    
    if(boot_lower < 0) {
        bootstrap_negative <- bootstrap_negative + 1
    }
}

cat("Results from", n_sims, "simulations:\n")
cat("True lambda:", lambda_true, "\n")
cat("Wald CI negative:", wald_negative, 
    "times (", round(100 * wald_negative / n_sims, 1), 
    "%)\n")
cat("Bootstrap CI negative:", bootstrap_negative, 
    "times (", round(100 * bootstrap_negative / n_sims, 1), 
    "%)\n")

If this is true, is the bootstrap CI always more advantageous than the Wald CI? Now with modern computers where simulation is not a problem, wont bootstrapping CI almost always be better (ie avoid illogical problem) if not same as Wald?

Answer 1

There are a few kinds of bootstrap confidence intervals, and it appears you're using the percentile method. Yes, the percentile bootstrap confidence intervals will never cover infeasible parameter space (assuming the statistic you're using doesn't ever become infeasible. For example, the sample mean can't be negative if all the data are non-negative). This is because the point estimate can never be infeasible, and the percentile method calculates the confidence bounds from the bootstrapped point estimates.

Just as there are a few bootstrap confidence intervals, so too are there several binomial confidence intervals. If you are studying a rare outcome, it may be advantageous to use something like a Wilson Interval or a Clopper Pearson interval.

Similarly, for a random variable supported on the positive reals -- such as an exponential random variable -- it may make sense to calculate the confidence interval in log space and then transform the interval via the exponential.

Bootstrapping is a great invention, but analytic results extend beyond the Wald interval, especially for cases which are bounded like the ones you provide. These analytic results can have good coverage in some circumstances, and the percentile confidence interval is by no means the best of the bootstrap confidence intervals.

Answer 2

I will just add a few comments, on the margins, to @DemetriPananos' very good answer.

There are many (really, many) methods for computing a binomial proportion CI; the Wald method is about as poor as it gets (the first section of Wikipedia's article on binomial CI's is titled "Problems with using a normal approximation or "Wald interval"". A normal is a very poor approximation for a binomial when p is close to 0, or 1. This is well known, and is a reason why many other methods came about (Wilson, Agresti-Coull, Jeffreys, Clopper-Pearson, Blaker, etc.). Wald had a place in days before computers; I am not sure why it is still being taught today? Clopper Pearson, or Blaker, should be the default (it used to be painful to compute by hand, in part because of factorials).

I also want to address your final sentence; "Now with modern computers where simulation is not a problem, wont bootstrapping CI almost always be better (ie avoid illogical problem) if not same as Wald?". Bootstrapping is certainly a major advance for statistics, but it is not a "silver bullet". The major assumption for any of the many bootstrapping methods to provide "reasonable" answers is that the sample be representative of the population. This is fundamentally untestable (since we do not know the population's distribution), so it really means that the sample size should be large, and typically larger than what would be required for parametric methods.