# Is centering needed when bootstrapping the sample mean?

When reading about how to approximate the distribution of the sample mean I came across the nonparametric bootstrap method. Apparently one can approximate the distribution of $\bar{X}_n-\mu$ by the distribution of $\bar{X}_n^*-\bar{X}_n$, where $\bar{X}_n^*$ denotes the sample mean of the bootstrap sample.

My question then is: Do I need the centering? What for?

Couldn't I just approximate $\mathbb{P}\left(\bar{X}_n \leq x\right)$ by $\mathbb{P}\left(\bar{X}_n^* \leq x\right)$?

• I don't see why you we need to center anything. All the samples discussed here are of the same size right? Oct 12, 2012 at 17:25
• Same size, yes. I don't see the reason for the centering either. Would anybody be able to come up with a mathematical explanation why or why not we have to do that? I mean, can we prove that the bootstrap works or does not work if we do not center? Oct 12, 2012 at 18:20
• (Btw, a proof that the bootstrap works for the case where we centered can be found in Bickel, P.J. and D.A. Freedman (1981), Some asymptotic theory for the bootstrap.) Oct 12, 2012 at 18:24
• I'm curious: Why is this question downvoted? Oct 12, 2012 at 23:14
• Maybe we do the entering to be able to use the Central Limit Theorem which gives us that $n^{\frac{1}{2}}(\bar{X}_n-\mu)$ converges to the same distribution as $n^{\frac{1}{2}}(\bar{X}_n^*-\bar{X}_n)$, namely to $\mathcal{N}(0,\sigma^2)$. Maybe there are no asymptotics available for the case without centering that tell us whether it works.
– kelu
Oct 14, 2012 at 11:11

Yes, you can approximate $\mathbb{P}\left(\bar{X}_n \leq x\right)$ by $\mathbb{P}\left(\bar{X}_n^* \leq x\right)$ but it is not optimal. This is a form of the percentile bootstrap. However, the percentile bootstrap does not perform well if you are seeking to make inferences about the population mean unless you have a large sample size. (It does perform well with many other inference problems including when the sample size size is small.) I take this conclusion from Wilcox's Modern Statistics for the Social and Behavioral Sciences, CRC Press, 2012. A theoretical proof is beyond me I'm afraid.

A variant on the centering approach goes the next step and scales your centered bootstrap statistic with the re-sample standard deviation and sample size, calculating the same way as a t statistic. The quantiles from the distribution of these t statistics can be used to construct a confidence interval or perform a hypothesis test. This is the bootstrap-t method and it gives superior results when making inferences about the mean.

Let $s^*$ be the re-sample standard deviation based on a bootstrap re-sample, using n-1 as denominator; and s be the standard deviation of the original sample. Let

$T^*=\frac{\bar{X}_n^*-\bar{X}}{s^*/\sqrt{n}}$

The 97.5th and 2.5th percentiles of of the simulated distribution of $T^*$ can make a confidence interval for $\mu$ by:

$\bar{X}-T^*_{0.975} \frac{s}{\sqrt{n}}, \bar{X}-T^*_{0.025} \frac{s}{\sqrt{n}}$

Consider the simulation results below, showing that with a badly skewed mixed distribution the confidence intervals from this method contain the true value more frequently than either the percentile bootstrap method or a traditional inverstion of a t statistic with no bootstrapping.

compare.boots <- function(samp, reps = 599){
# "samp" is the actual original observed sample
# "s" is a re-sample for bootstrap purposes

n <- length(samp)

boot.t <- numeric(reps)
boot.p <- numeric(reps)

for(i in 1:reps){
s <- sample(samp, replace=TRUE)
boot.t[i] <- (mean(s)-mean(samp)) / (sd(s)/sqrt(n))
boot.p[i] <- mean(s)
}

conf.t <- mean(samp)-quantile(boot.t, probs=c(0.975,0.025))*sd(samp)/sqrt(n)
conf.p <- quantile(boot.p, probs=c(0.025, 0.975))

}

# Tests below will be for case where sample size is 15
n <- 15

# Create a population that is normally distributed
set.seed(123)
pop <- rnorm(1000,10,1)
my.sample <- sample(pop,n)
# All three methods have similar results when normally distributed
compare.boots(my.sample)


This gives the following (conf.t is the bootstrap t method; conf.p is the percentile bootstrap method).

          97.5%     2.5%
conf.t      9.648824 10.98006
conf.p      9.808311 10.95964


With a single example from a skewed distribution:

# create a population that is a mixture of two normal and one gamma distribution
set.seed(123)
pop <- c(rnorm(1000,10,2),rgamma(3000,3,1)*4, rnorm(200,45,7))
my.sample <- sample(pop,n)
mean(pop)
compare.boots(my.sample)


This gives the following. Note that "conf.t" - the bootstrap t version - gives a wider confidence interval than the other two. Basically, it is better at responding to the unusual distribution of the population.

> mean(pop)
 13.02341
> compare.boots(my.sample)
97.5%     2.5%
conf.t      10.432285 29.54331
conf.p       9.813542 19.67761


Finally here is a thousand simulations to see which version gives confidence intervals that are most often correct:

# simulation study
set.seed(123)
sims <- 1000
results <- matrix(FALSE, sims,3)
colnames(results) <- c("Bootstrap T", "Bootstrap percentile", "Trad T test")

for(i in 1:sims){
pop <- c(rnorm(1000,10,2),rgamma(3000,3,1)*4, rnorm(200,45,7))
my.sample <- sample(pop,n)
mu <- mean(pop)
x <- compare.boots(my.sample)
for(j in 1:3){
results[i,j] <- x[j,1] < mu & x[j,2] > mu
}
}

apply(results,2,sum)


This gives the results below - the numbers are the times out of 1,000 that the confidence interval contains the true value of a simulated population. Notice that the true success rate of every version is considerably less than 95%.

     Bootstrap T Bootstrap percentile          Trad T test
901                  854                  890

• Thank you, that was very informative. This .pdf (from a lesson) describes a caveat to your conclusion: psychology.mcmaster.ca/bennett/boot09/percentileT.pdf This is a summary of what Bennet says: Many datasets consists of numbers that are >=0 (i.e. data that can be counted), in which case the CI should not contain negative values. Using the bootstrap-t method this can occur, making the confidence interval implausible. The requirement that the data be >=0 is in violation of the normal distribution assumption. This is not a problem when constructing a percentile bootstrapped CI Aug 10, 2016 at 15:18