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Is it possible to apply the law of the iterated logarithm (e.g. http://en.m.wikipedia.org/wiki/Law_of_the_iterated_logarithm) to derive non-trivial (i.e. bounded) 100% confidence intervals for population averages?

An abstract (http://www.tandfonline.com/doi/pdf/10.1080/10485250410001713963) gives at least such hint. However, its first reference by Robbins (http://projecteuclid.org/euclid.aoms/1177696786) does not seem to cover such result, as pointed out in @whuber's comment.

Edit: After justified comments by @whuber, I reformulated the question.

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    $\begingroup$ You need to say more about the context and the assumptions. Non-trivial 100% confidence intervals are usually impossible to achieve except when sampling from finite populations. $\endgroup$
    – whuber
    Commented Jul 11, 2014 at 19:58
  • $\begingroup$ Could you tell us precisely where the Robbins paper discusses "100% confidence intervals"? I could find no such reference in it, but maybe I overlooked something. $\endgroup$
    – whuber
    Commented Jul 11, 2014 at 21:02
  • $\begingroup$ You are right, I also can't find it. I was following this reference from tandfonline.com/doi/pdf/10.1080/10485250410001713963 (only abstract). $\endgroup$
    – Michael M
    Commented Jul 11, 2014 at 21:13

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100% confidence is only feasible if you have a bounded distribution or if you have sampled all members of the population (and then it becomes trivial for that distribution).

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  • $\begingroup$ That's also what I think. But I don't understand why the LIL is mentioned in one of above mentiones articles as basis to form such 100% c.i. I see the LIL's usefulness in sequential c.i.s $\endgroup$
    – Michael M
    Commented Jul 13, 2014 at 10:35
  • $\begingroup$ Michael, are you perhaps misreading 100% power as 100% confidence? $\endgroup$
    – whuber
    Commented Jul 14, 2014 at 15:50
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    $\begingroup$ As a technical note, 100% confidence can be achieved in many other circumstances. For instance, suppose the underlying set of distributions consists of all Normal$(\mu, 1)$ distributions with $-1\le \mu\le 1$. Then $[-1,1]$ is a 100% confidence interval for $\mu$ (regardless of the data). The point is that the kind of boundedness that is relevant here is that of the parameter rather than of the support of the distribution. $\endgroup$
    – whuber
    Commented Nov 3, 2014 at 19:01
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    $\begingroup$ Another example where a nontrivial 100% CI can be found is in sampling from a finite population. For instance, suppose a population of three elements is known to have a mean between 0 and 100. You sample one element and find it equals 150. You now know with 100% confidence that the mean is between 50 and 100, even though it is not the case that "you have sampled all members of the population." $\endgroup$
    – whuber
    Commented Nov 8, 2015 at 23:52
  • $\begingroup$ @whuber: thats look somewhat bayesian, it is not frequentist in the usual sense, since it is conditional on obtaing that special observation ... $\endgroup$ Commented Oct 26, 2017 at 18:25

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