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I've just come across a statement in

Good, P. I.: Resampling methods. Springer, 2001

and wondered if someone could explain it to me.
If you want to construct a 95% confidence interval for, let's say the mean of a normal distribution then the bounds would be $\bar{X}\pm 1.96s/\sqrt{n}$ with $\hat X$ being the sample mean and $s$ being the standard deviation of the sample.

So obviously if you want to halve the size of your confidence interval you'd have to multiply the sample size by 4 (i.e. take 400 samples instead of 100). So the width of the CI is proportional to $n^{-\frac{1}{2}}$. CI's with this feature are called first order exact (I guess most of you know that, I'm just repeating it because I didn't)

Now what Good says is that $BC_\alpha$ intervals are second order exact, that is their width for large samples is proportional to $n^{-1}$. Why is that so?

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    $\begingroup$ First versus second order accuracy is not about the width of the confidence interval; it is about the coverage probability. The usual 95% confidence interval is such that $P(\mu \in \hat{C}) = 0.95 + O(1/\sqrt{n})$. A $BC_\alpha$ 95% interval is such that $P(\mu \in \hat{C}_{BC_\alpha}) = 0.95 + O(1/n)$. Both intervals will have widths proportional to $n^{-1/2}$. $\endgroup$ – paul Jun 18 '14 at 16:33
  • $\begingroup$ So the statement "Such confidence intervals, whose width for large samples is proportional to $n^{−1/2}$ are said to be first-order exact. The percentile bootstrap confidence interval is also first-order exact. BCα intervals are second-order exact in that their width for large samples is proportional to $n^{−1}$" taken from the book by Good is false? $\endgroup$ – Barkas Jun 18 '14 at 16:41
  • $\begingroup$ That statement is absolutely incorrect. For example, consider i.i.d. observations from a normal distribution. For any $n$, the 95% confidence interval with the smallest width is the usual $\bar{X} \pm 1.96 s /\sqrt{n}$. Any smaller interval will have coverage probability below 0.95. It is usually impossible to create a valid confidence interval with width proportional to $n^{-1}$. $\endgroup$ – paul Jun 18 '14 at 17:14
  • $\begingroup$ There are special situations where you can get confidence intervals with width proportional to $n^{-1}$, such as when estimating the maximum of a random variable with bounded support. However, in those situations the $BC_\alpha$ interval will not be valid. $\endgroup$ – paul Jun 18 '14 at 17:15

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