3
$\begingroup$

There are great posts on confidence intervals, a subject that was brought up recently on this question, leading to an endogamous and circular surfing between CV classics, such as this one and this one - all of them truly remarkable. Yet there is one point that I don't get, and haven't been able to find a good explanation.

On this post, the second of the CV entries quoted in the first paragraph, there is what could be considered a secondary, or supportive narrative to the question, making reference to the fact that the posterior probability of a CI (confidence interval, I presume, as opposed to credible interval) is either zero or one.

In one of the comments a very credible statement is made that proving this is a trivial mathematical exercise. Here's the exact quote:

I do not think there is any issue with the statement "It's probability is either zero or one" in reference to the (posterior) probability that a CI contains a (fixed) parameter. (This does not even really rely on any frequentist interpretation of probability!). It also does not rely on "unknown states". Such a statement refers precisely to the situation in which one is handed a CI based on a particular sample. It is a simple mathematical exercise to show that any such probability is trivial, i.e., takes values in {0,1}. – cardinal♦

I sort of see how the actual parameter (the mean of the imaginary population) in the confidence interval is fixed, and hence it is not subject to being a varying outcome of a conceivably infinite random experiment. This, however, is far from a mathematical prove; further it is limited as a self-convincing reasoning in two ways: 1. A probability of $0$ or $1$ is still a probability; and 2. As part of the same column of comments in the post referenced, note is made that this fact (that the probability of the true mean is contained in the CI being $\{0, 1\}$) is independent of having a frequentist or Bayesian approach to statistics.

So I understand that what I just typed is incomplete, erroneous and misleading (with a high degree of confidence), but I'm showing what I've found, and where I got stuck.

Now the actual question: What is the mathematical proof of the statement in the title?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ "Third of the linked posts" does not tell us what the post is, because the order in which answers are presented can change and is controlled by the user. Please link directly to the post you are referring to. $\endgroup$
    – whuber ♦
    Commented Dec 2, 2015 at 1:52
  • 1
    $\begingroup$ All your links go to the questions. Specifically what text are you referring to for the assertion that the posterior probability that a CI covers the mean is either $0$ or $1$? It looks like it's going to take some kind of unusual or contorted interpretation to make sense of such a statement, so we need to see precisely what was written. $\endgroup$
    – whuber ♦
    Commented Dec 2, 2015 at 1:56

1 Answer 1

Reset to default
6
$\begingroup$

Here it goes:

Suppose the mean $m$ exists and is constant. Suppose further that you already have calculated a confidence interval $CI = [a,b]$. Since $m$, $a$ and $b$ are all constant, it follows that either

i) $a \leq m \leq b$, in which case $P(m \in CI) = 1$, or

ii) $m < a$ or $b < m$, in which case $P(m \in CI) = 0$

qed

So you see, it is almost silly, but the trick lies in the fact that all the numbers are actually constant (or assumed constant): the mean because it is so supposed in frequentist statistics and the confidence interval because you already calculated it and so is not random at all.

Improve this answer
$\endgroup$
2
  • $\begingroup$ It is the definition of anticlimatic.... I'm sure you are correct and I'll come back and accept the answer shortly, but even seeing how there isn't much to the whole mystery, I'm still wondering if we shouldn't the talk about a probability of 1/2. $\endgroup$
    – Antoni Parellada
    Commented Dec 2, 2015 at 11:53
  • $\begingroup$ This is a common debate. In frequentist stats you model "true" randomness and so these things happen, because in "truth" there is nothing random to a few numbers having some characteristic or other. In bayesian stats, however, you model your own uncertainty as opposed to "real" randomness using probability distributions. You should check out any introductory text book to bayesian statistics if you're interested in these sorts of things! $\endgroup$
    – Felipe Gerard
    Commented Dec 2, 2015 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.