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This post could be entitled as "Interpretation of confidence interval 2" according to the prior question. (Interpretation of confidence interval) I've read the QA but still not gotten to the point.

I am not the same person who asked the previous question, but I came upon quite a similar interpretation problem on confidence interval (CI). Mine was why CI needs to be an INTERVAL while we can find arbitrary candidate of 90% or 95% area that contains the true value. This question is partly solved.

Above in the very old post, the answer quoted

"Although the various methods are equal from a purely mathematical point of view, the standard method of computing confidence intervals has two desirable properties: each interval is symmetric about the point estimate and each interval is contiguous."

But these listed desirable properties of CI, 1) symmetry and 2) contiguity, do not seem theoretically apparent. Of course, CI is one of the expressions on confidence of an estimator, so these two properties should be full-filled so that CI is human-friendly. But don't we have any further reasons why we choose contiguous symmetric interval for representation of an estimation? (for frequentist?)

The reason why this thing is not so much discussed in the quoted article (http://onlinestatbook.com/2/estimation/confidence.html) might be partially because CI is often used to show the results of an research and decision making but not directly adopted for future experimental or statistical refinement. But, I want to make clear why the prefered summarized expression of confidence (which is Confidence Interval: CI) should be symmetric (in some sense, in some scale) about the point estimate and contiguous from theoretical, practical and decision making perspectives.

The topic may have something to do with Equal-Tail Interval (ETI, one of Credible Interval: CI) in Bayesian interpretation. I am quite a beginner in statistics, so the actual point might be "In what sense Equal-Tail Interval (for Bayesian) is desirable (for frequentist?)" or "Equal-Tail Interval preserves & well-summarize something of the posterior distribution (and if ETI were to be used as a kind of prior in frequentist perspectives, would preserve something?)". Maybe since I started statistics from Bayesian, I cannot catch up with frequentist approach, so I'm sorry that I only have very vague ideas about these things, but please mention anything.

Let me summarize. Why should Confidence Interval: CI (not Credible Interval) symmetric and contiguous?

Thank you.

P.S. In a declined answer, I found CI like $(-\infty, \alpha)$ is less informative than conventional CIs. But if we consider only the property of 95% confidence, there’s no difference between conventional CI and weird ones (or weird expression of confidence). Something is making conventional CI (contiguous & probability-symmetric) more informative & well-summarizing.

Then, I understood the point is “how CI could be good summary of statistical results”. CI by definition is equal-tail & probability-symmetric, so this property might be enough to decode the point estimate as well as variance of the estimator to some extent on a given distribution. (ex $\sigma^2$ estimation by $\chi^2$-dist) Is this part of reasons why CI as a summary of data need to be contiguous and probability-symmetric? Uh…

mriryt
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  • Confidence intervals for population mean based on z and t distribution are usually symmetrical. However, CIs for other parameters may not have the point estimate as center. The usual CIs for normal $\sigma^2$ and $\sigma$ are not symmetrical. Often they they are 'probability-symmetric' which means that the same probability (e.g, 0.025) is cut from each tail of some asymmetrical distribution (such as chi-squared) to make the CI. (Shorter intervals may be possible by cutting 2% from one tail and 3% from the other, but this is not often done.). // Discontiguous CIs are rare, but not "illegal" – BruceET Nov 23 '21 at 01:49
  • One-sided (very asymmetric!) confidence intervals are certainly used. (I've never heard of a "confident interval" -- it sounds like an interval that is relaxed at job interviews!) – David K Nov 23 '21 at 03:18
  • Thank you very much for replies. My first bolded comments "symmetric (in some sense, in some scale)" in the question was meant for cases of normal $\sigma^2$ confidence interval, of course because the interval measured by $\chi^2$-dist is not symmetric along the $\sigma$ axis. But it is still Equal-Tail Interval, so in some sense from probability it is symmetric. Let me excuse for confusing phrase, but the question does not focus only on $\mu$ estimation. The focus is upon the reason of contiguity and why the interval should contain the point estimate 'in a middle'. – mriryt Nov 23 '21 at 05:57
  • My first notion was that we certainly can make odd 95% confidence area (not interval) of a parameter $\theta$ that is splited around the point estimate but still contain 95% of the estimation. Thus in this question, I am considering fairly weired cases of such representation of confidence. One-sided confidence interval sounds interesting. I want more information. Also revised the "confident interval" to "confidence interval". Thanks. – mriryt Nov 23 '21 at 06:01

1 Answers1

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Consider a random sample of size $n=50$ from $\mathsf{NORM}(\mu=100, \sigma=15),$ Sampled in R:

set.seed(2021)
x = rnorm(50, 100, 15)
summary(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  70.54   86.43  100.89  100.24  113.67  131.80 
[1] 16.99206   # sample SD

stripchart(x, pch="|")

enter image description here

95% CI for population mean. In R the usual t-interval for $\mu$, based on Student's t distribution with 49 degrees of freedom, is given as part of R's t.test procedure. Only the 95% CI $(85.41, 105.07)$ is shown below:

t.test(x)$conf.int
[1]  95.40906 105.06723
attr(,"conf.level")
[1] 0.95

The midpoint of this interval is the sample mean $\bar X=100.2381,$ which is given in the summary above, rounded to 2 places.

A 95% CI for population variance $\sigma^2$ is based on 'pivoting' the relationship $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu=49).$ One can compute this 95% CI $(201.5,488.4)$ in R as follows:

(49 * var(x))/qchisq(c(.975,.025), 49)
[1] 201.4709 448.3540

The corresponding 95% CI for $\sigma$ is $(14.19,21.17),$ found by taking square roots of endpoints of the interval above for $\sigma^2:$

sqrt((49 * var(x))/qchisq(c(.975,.025), 49))
[1] 14.19404 21.17437

The midpoint of this interval is $17.68,$ which is not the same as the point estimate $S = 16.99,$ given in the summary. However, this is an example of a probability symmetric CI because it was formed by taking probability $0.025$ from each tail of the relevant chi-squared distribution.

Note: The length of the interval above for $\sigma$ is $6.980.$ A shorter 95% interval $(14.30, 20.98)$ of length $6.685$ can be found by cutting probability 2% from one tail of $\mathsf{Chis}(49)$ and 3% from the other.

sqrt((49 * var(x))/qchisq(c(.97,.03), 49))
[1] 14.29509 20.98057
diff(sqrt((49 * var(x))/qchisq(c(.97,.03), 49)))
[1] 6.685476

Still shorter intervals may be found, but the usual practice is to use probability-symmetric intervals. The existence of many different 95% CIs is one reason statistics texts usually say "a 95% confidence interval..." instead of "the" 95% CI ...."

BruceET
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  • Maybe shorter intervals compared to usual (equal-tail) CI is some kind of idea derived from Highest posterior density interval (of Credible Interval: CI) in Bayesian interpretation. Why not use this type interval instead in frequentist view (say, for $\sigma^2$ interval). For me, the reason why "the usual practice is to use probability-symmetric" is unclear. Is this merely conventional reason? – mriryt Nov 23 '21 at 06:18
  • This answer greatly supplemented my question. When I got stuck in this problem, CI for $\sigma^2$ was also under consideration. Usual CI for $\sigma^2$ is not symmetric along $\sigma$, but probability-symmetric (although I've never heard this term before) as you mentioned. And the heart of the question is "why CI (or 'a' representation of confidence) need to be probability-symmetric" about the point estimate and contiguous. – mriryt Nov 23 '21 at 06:23
  • Probability symmetric CIs are almost always used. Probably that's out of convenience because they're usually easier than searching for probability apportionment to make shortest possible interval, and for standardization. But I know of no rule that absolutely prohibits other styles of intervals. Maybe in the background of your question there is some unstated criterion that narrows the field of what's allowable. – BruceET Nov 23 '21 at 16:06
  • Not sure why @r.e.s deleted their Answer. My quick scan of it revealed nothing incorrect. – BruceET Nov 23 '21 at 16:09