I need to show that the standard error of the $n^n$ bootstrap means is $SE^*(\bar{Y^*}) = \frac{S\sqrt{n-1}}{n}$, where $\bar{Y^*}$ is the sample mean of a randomly drawn bootstrap sample, and $S^2 = \frac{1}{n-1}\sum_{i=1}^{n}(Y_i - \bar{Y})^2$. I know that $SE^*(\bar{Y^*}) = \sqrt{\frac{\sum_{b=1}^{n^n}(\bar{Y}^*_b - \bar{Y})^2}{n^n}}$, and have a hint that says I "should exploit the fact that the mean is a linear function of the observations."

Note that the "$n^n$ bootstrap means" is simply the bootstrap procedure in which all $n^n$ possible bootstrap samples are enumerated.

Thank you for any help!


In the the original sample of $n$ items $Y_i$ you have $\sum Y_i = n \bar Y$ and $\sum (Y_i-\bar Y)^2 = (n-1) S^2$.

In effect the original sample is the population for the bootstrap samples and you can treat each bootstrap sample as $n$ i.i.d. samples from a population with mean $\bar Y$ and variance $\frac{n-1}{n}S^2$ (as this is now being treated as a population you need to uncorrect for sample bias). So the bootstrap sums have mean $n\bar Y$ and second moment about $n\bar Y$ of $(n-1)S^2$

This implies that the bootstrap means have $E[\bar Y_b^*] = \bar Y$ and second moment about $\bar Y$ of $E[(\bar Y_b^*- \bar Y)^2] = \frac{n-1}{n^2}S^2$ i.e. with $\sqrt{E[(\bar Y_b^*- \bar Y)^2]} = \sqrt{ \frac{n-1}{n^2}S^2} = \frac{\sqrt{n-1}}{n}S$.

Since you have all $n^n$ possible bootstrap samples from the original sample, these expectations are realised, so $\frac{\sum_{b=1}^{n^n}(\bar{Y}^*_b - \bar{Y})^2}{n^n} = \frac{n-1}{n^2}S^2$ and $SE^*(\bar{Y^*}) = \sqrt{\frac{\sum_{b=1}^{n^n}(\bar{Y}^*_b - \bar{Y})^2}{n^n}} = \frac{\sqrt{n-1}}{n}S$

  • $\begingroup$ What exactly do you mean when you say "the bootstrap sums have mean $n\bar{Y}$ and second moment about $n\bar{Y}$ of $(n-1)S^2$? Are you saying that $E(\sum_i\bar{Y^*_i}) = n\bar{Y}$ and $E((n\bar{Y})^2) = (n-1)S^2$? $\endgroup$ – Jake Dec 5 '19 at 19:05
  • $\begingroup$ @Jake Not quite and I am not sure about your notation. A bootstrap sample is of $n$ items with replacement from the original sample, and the bootstrap sum is the sum of all $n$. So perhaps $E(\sum_i Y^*_i) = n\bar{Y}$ and $E(((\sum_i Y^*_i)-n\bar{Y})^2) = (n-1)S^2$. Then a bootstrap mean is $\bar{Y}^* = \frac1n \sum_i Y^*_i$ $\endgroup$ – Henry Dec 5 '19 at 19:20
  • $\begingroup$ Ah I understand what you mean now, thanks for the clarifications! $\endgroup$ – Jake Dec 5 '19 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.