Confidence interval for the confidence interval?

I'm studying confidence intervals, and I'm curious about how one might generate a confidence interval for the confidence interval, if that even makes sense.

For example, let's say I draw simple random samples of n=100 from some population, calculate sample means and standard deviations, and construct 95% confidence intervals. I repeat this procedure 100 times. I know I expect about 95 of these intervals to capture the population mean, and about 5 of them not to. However, can I construct a confidence interval around this expectation? If I were to repeat this entire "100 samples of 100 samples" over and over again, what can I say about the distribution of how often the intervals captures?

Essentially, could I construct a confidence interval for the confidence interval? Would that even make any sense?

Thanks!

• In that situation, wouldn't you take the average of all confidence intervals to improve your estimate? As a theoretical answer to your question, you can always create a confidence interval as long as you can (1) make an assumption about the underlying distribution, (2) have a mean, (3) have a variance, (4) have a confidence level. – Jean-Paul Feb 7 '16 at 22:25
• and it's turtles all the way down – rep_ho Feb 7 '16 at 23:50
• I think what you essentially want to do is form a confidence interval for the coverage probability. The coverage probab9ility is the probability that any one confidence interval contains the population mean. – Mark L. Stone Feb 8 '16 at 1:44
• (continued) Presuming you have M i.i.d replications of forming confidence intervals based on n = 100 sample size, you can form the confidence interval for coverage probability based on binomial fraction of intervals containing the population mean. – Mark L. Stone Feb 8 '16 at 1:50
• Mark, that makes a ton of sense, and I think it's similar to Glen's answer below. Right? – Guy Davidson Feb 9 '16 at 20:23

Treating each interval containing the parameter as a Bernoulli process with each trial having some coverage probability $p$, the number of "coverages" should be $\text{Binomial}(n,p)$.
The potential problem is whether the $p$ one actually has is really the $p$ one was hoping for (whether due to the extent of the failure of assumptions or because of approximations involved in obtaining the intervals).