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As the title implies, why can't $ 1 - \alpha = 1$ , so that we have a $100 \%$ confidence level?

I saw these two answers, but want a more mathematical proof:

How to estimate 100% confidence interval aka. what is the Z value of standard normal distribution at probability of 100%?

100% confidence interval for mean

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We have such an interval, it is just less than useful.

The confidence interval for the mean is $\bar{x} \pm z_{1-\alpha/2} \sigma/\sqrt{n}$.

As $\alpha \rightarrow 0$, $z_{1-\alpha/2} \rightarrow z_1$ (I'm being a bit fast and loose with my notation).

$z_k$ is the kth quantile of a standard normal. What would be the 100th quantile of a standard normal (equivalently, the point where there is 100% probability to the left of said point). That would be $\infty$ (again, being fast and loose for the purposes of exposition).

So your 100% CI exists and it is the real line. Well...that isn't useful.

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  • $\begingroup$ Agreed, with a pedantic generalisation to the entire outcome space, which may not be the real line. E.g. the 100% confidence limits for a proportion are 0 and 1. (The title of the question is general, for all the focus on normal distributions and/or sample means broadly.) $\endgroup$
    – Nick Cox
    Commented Oct 23, 2020 at 14:27

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