# 100% confidence interval for mean

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.

• 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. – whuber Jul 11 '14 at 19:58
• 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. – whuber Jul 11 '14 at 21:02
• You are right, I also can't find it. I was following this reference from tandfonline.com/doi/pdf/10.1080/10485250410001713963 (only abstract). – Michael M Jul 11 '14 at 21:13

• 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. – whuber Nov 3 '14 at 19:01