Standard error of the $n^n$ bootstrap means

Question

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!

Answer 1

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 $SE^*(\bar{Y^*}) = \sqrt{\frac{\sum_{b=1}^{n^n}(\bar{Y}^*_b - \bar{Y})^2}{n^n}} = \frac{\sqrt{n-1}}{n}S$