I need help with calculating the variance of a bootstrap sample mean. I want to answer the following question (question 15.4 in The Elements of Statistical Learning).

Suppose $x_i$, $i=1, \ldots , N$ are iid $(\mu, \sigma^2)$. Let $\bar{x}^\star_1$ and $\bar{x}^\star_2$ be two bootstrap realizations of the sample mean. Show that the sampling correlation $corr(\bar{x}^\star_1,\bar{x}^\star_2)=\frac{n}{2n-1} \approx 50\%$. Along the way, derive $\bar{x}^\star_1$ and the variance of the bagged mean $\bar{x}_{bag}$. Here, $\bar{x}$ is a linear statistic; bagging produces no reduction in variance for linear statistic.

This is the solution that I have so far:

The sample mean $\bar{x}$ is given by

$\bar{x}=\frac{1}{N} \sum_{i=1}^N x_i,$

and the bootstrap sample mean $\bar{x}^\star_1$ is computed from a resampled dataset, the same applies for $\bar{x}^\star_2$. The sampling correlation between $\bar{x}^\star_1$ and $\bar{x}^\star_2$ is given by

$corr(\bar{x}^\star_1,\bar{x}^\star_2)= \frac{Cov(\bar{x}^\star_1,\bar{x}^\star_2)}{\sqrt{Var(\bar{x}^\star_1)Var(\bar{x}^\star_2)}},$

so, we need to begin with computing the covariance term $Cov(\bar{x}^\star_1,\bar{x}^\star_2)$ and the respective variance terms in order to calculate their correlation. We start by denoting that

$cov(\bar{x}_1^\star, \bar{x}_2^\star)=\frac{\sigma^2}{n}$

since $E[\bar{x}_1^\star]=E[\bar{x}_2^\star]$ which is the population mean. Now we want to calculate the variances.

However, this is where I get stuck. How do one calculate the variance of a bootstrap realization sample mean?

My first thought was that since the bootstrap sample is obtained by resampling with replacement, the variance of the sample mean for a bootstrap sample is the same as the variance of the original sample mean, i.e. $\sigma^2/n$, but this cannot be correct since this would not obtain $corr(\bar{x}^\star_1,\bar{x}^\star_2)=\frac{n}{2n-1} \approx 50\%$.

Any help would be appreciated in order to help me see what it is that I am missing here!


I (hopefully) solved it using the following definition

$Var(\bar{x}_i^\star)=\frac{1}{n^2}\left[ \sum_{i=1}^n Var(x_i) + 2\sum_{i=1}^{n-1}\sum_{j=i+1}^n Cov(x_i^\star,x_j^\star)\right] = \frac{1}{n^2}\left[n\sigma^2+n(n-1)\frac{\sigma^2}{n}\right]$

and obtained the variance $\frac{\sigma^2(2n-1)}{n^2}$. Note that $x_i^\star$ and $x_j^\star$ are resamples of the original dataset.

From this, I could insert the answers into the formula for correlation and prove that $corr(\bar{x}^\star_1,\bar{x}^\star_2)=\frac{n}{2n-1} \approx 50\%$. What do you think about this solution? Do you think it is correct (even though I did not use the bagged mean to prove it)?


1 Answer 1


I would do this by conditioning on the sample.

Write $s^2=\frac{1}{n}\sum_{i=1}^n (x_i-\bar x)^2$, so $s^2$ is the variance of the empirical distribution of the sample. Also, write $\mathrm{var}_P$ and $\mathrm{var}_B$ for the variance under sampling of the original $x$ and the variance under bootstrap resampling. All the bootstrap resamples share the same initial sample but have independent resampling.

We have

  • $\mathrm{var}_P[\bar x]=\sigma^2/n$
  • $\mathrm{var}_B[\bar x_1^*]=s^2/n$ (because it's just iid sampling from the empirical distribution)
  • $\mathrm{var}_B[\bar x_2^*]=s^2/n$ (ditto)

Now, the total variance of $\bar x_1^*$ is given by the conditional variance formula $$\mathrm{var}_{PB}[\bar x_1^*]=E_P[\mathrm{var}_B[\bar x_1^*]]+\mathrm{var}_P[E_B[\bar x_1^*]]=E_P[s^2/n]+\mathrm{var}_P[\bar x]=\frac{n-1}{n}\frac{s^2}{n}+\sigma^2/n$$

Two bootstrap means share the sampling components of the variance but have independent resampling components. $$\mathrm{cov}[\bar x_1^*,\bar x_2^*]= \sigma^2/n$$ so $$\mathrm{corr}[\bar x_1^*,\bar x_2^*]=\frac{\mathrm{cov}[\bar x_1^\star,\bar x_2^\star]}{\sqrt{\mathrm{var}[\bar x_1^\star]\mathrm{var}[\bar x_2^\star]}}=\frac{\sigma^2/n}{\frac{n-1}{n}\frac{s^2}{n}+\sigma^2/n}\approx 1/2$$

And if you bag $m$ times you reduce the resampling component by a factor of $m$, because they are all conditionally independent: $$\mathrm{var}_{PB}[\bar x_\textrm{bag}]=\sigma^2/n+\frac{1}{m}\frac{n-1}{n}\frac{s^2}{n}$$ which gets closer to $\sigma^2/n$ as $m$ increases.


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