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I am encountering a difficulty with the following task. Have I made a mistake, or is this an inherent flaw in the notion of confidence intervals? (Other such flaws exist.)

Consider a random sample $X_1,\ldots , X_n$ from a Uniform($\theta$, $\theta + a$) distribution, where $\theta$ is unknown and $a$ is known. We wish to determine a confidence interval for $\theta$.

The reader may verify the following details: The statistics $Y=\text{min}_i X_i$ and $Z=\text{max}_i X_i$ are jointly sufficient for $\theta$. For $\theta \le c_1 \le c_2 \le \theta + a$, $P\{c_1 \le Y \le Z \le c_2\} = [(c_2 - c_1)/a]^n$. For $0 < \gamma < 1$, set $d_1 =(1-\sqrt[n]\gamma)/2$ and $d_2 =(1+\sqrt[n]\gamma)/2$. Then $\gamma = P\{\theta + ad_1 \le Y \le Z \le \theta + ad_2\} = P\{Z -ad_2 \le \theta \le Y-ad_1\}$. Thus, $[Z -ad_2, Y -ad_1]$ is a $\gamma$ confidence interval for $\theta$.

Now here's the difficulty: If we observe $Z - Y > a\sqrt[n]\gamma$, then $Z -ad_2 > Y -ad_1$, so our formula yields a nonsensical answer. Have I made an error in my calculations? Or is this one of those problems with confidence intervals?

(Homework? I guess so - but a homework problem that I wrote for my students. Inspired by another problem in DeGroot & Schervish.)

Stephen Herschkorn
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  • I'm confused by your final parenthetical. Was the homework problem "spot the fallacy", or was it "give a $\gamma$-confidence interval for this distribution" and you later noticed this trouble? (Either way, see answer. This was a fun brain-teaser.) – Daniel Franke Jul 30 '13 at 02:38
  • The latter. [This won't post with just that. I think I need to enter more text.] – Stephen Herschkorn Jul 30 '13 at 02:44

3 Answers3

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The key point here is that confidence intervals are not uniquely determined by the distribution, but can be chosen in a number of different ways, although there are standard ways in which it may be done. These choices gear the test to look for, or emphasise, particular kinds of deviation from the hypotheses being tested.

The test you have formulated is valid, but chosen very badly in that it rejects all cases where $Z-Y$ gets sufficiently close to $a$.

While it is true that the pair $Y,Z$ are sufficient statistics, you may note that the distribution of $U=Z-Y$ is independent of $\theta$. A common approach for constructing good confidence intervals is to condition on $U$: i.e., to find $c_U$ so that $$\Pr[Z-a+c_U\le\theta\le Y-c_U\mid U=u]=1-\alpha$$ for all $u$. Unless I'm very mistaken, the distribution of $Y$ conditional on $U$ is uniform in $[\theta,\theta+a-U]$. If we let $c_U=(a-U)\cdot\alpha/2$, we get $$\Pr[Z-a+c_U\le\theta\le Y-c_U]=1-\alpha$$ which is true conditional on $U=u$ for any $u$, and hence is true in general.

Einar Rødland
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There's nothing wrong with your calculations; this is indeed a flaw in the notion of confidence intervals.

Your logic is all correct up until the "Thus" at the start of the last sentence of your third paragraph. Given $\theta$ and $a$, I think the following conclusions are correct:

  1. Given $c_1$ and $c_2$ such that $\theta \leq c_1 \leq c_2 \leq \theta + a$, the probability that $Y$ and $Z$ will satisfy $c_1 \leq Y \leq Z \leq c_2$ is $\left(\frac{c_2-c_1}{a}\right)^n$.

  2. Given $\gamma$ such that $0 < \gamma < 1$, and letting $d_1 \triangleq \frac{1 - \sqrt[n]\gamma}{2}$ and $d_2 \triangleq \frac{1 + \sqrt[n]\gamma}{2}$, the probability that $Y$ and $Z$ will satisfy $Z - ad_2 \leq \theta \leq Y - ad_1$ is $\gamma$.

Note the direction of the reasoning, however: you've made an assumption about $\theta$, and drawn inferences about the probability that $Y$ and $Z$ will satisfy certain predicates. But by trying to set a confidence interval on $\theta$, you're trying to go in the other direction: measuring $Y$ and $Z$ and trying to use them to draw inferences about $\theta$. If you want to reason in that direction, you need a Bayesian prior!

It's entirely possible that you'll end up with $Y$ and $Z$ such that $Z - ad_2 > Y - ad_1$, thus making nonsense of the inequality $Z - ad_2 \leq \theta \leq Y - ad_1$. But the probability of that outcome won't exceed $(1 - \gamma)$!

Daniel Franke
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  • While I prefer the Bayesian approach, frequentists will take issue with your comment that a prior is necessary to draw inferences about $\theta$. Confidence intervals, a fundamental notion, are defined without respect to priors. I think you are agreeing that this is one of those cases where the incoherence of frequentist methodology leads to an unreasonable consequence, no? – Stephen Herschkorn Jul 30 '13 at 02:54
  • Correct. More broadly, I'm arguing that any time a frequentist draws a reasonable conclusion about a distribution parameter, it's by accident :-) – Daniel Franke Jul 30 '13 at 03:13
  • By the way, the exact limit (as $n$ grows large) that we run into this difficulty is $1 - \gamma + \gamma \text{log} \gamma$. – Stephen Herschkorn Jul 30 '13 at 03:16
  • To my mind, a better resoluiton was reached over at http://stats.stackexchange.com/questions/66407/confidence-interval-for-uniform-theta-theta-a . In brief, it is possible to get nonsensical results with a wrong choice of random variables. In the case at hand, one gets a more reliable formula if one uses (the min and) the range rather than the max. – Stephen Herschkorn Aug 05 '13 at 21:12
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    I'm surprised that you're satisfied with that answer; it seems deeply unconvincing to me absent a clear explanation of why using min and range is objectively preferable to using min and max, or to using a spinner that returns either $\mathbb R$ or the empty set. After all, they all produce valid confidence intervals for the same parameter. I agree with whuber's assertion that confidence intervals should not be interpreted as Bayesian credibilities; the crux of my argument is that when interpreted correctly, confidence intervals tell us nothing of interest. – Daniel Franke Aug 05 '13 at 22:55
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    To preempt some obvious objections to the spinner example: replace the spinner with a cryptographic hash of $X_1,\ldots,X_n$. Now the procedure is deterministic and doesn't throw away any data. Common sense tells us that this is nonetheless a terrible test procedure, but how can we justify that claim? – Daniel Franke Aug 05 '13 at 23:38
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Just as further verification that you didn't just make a silly algebra mistake somewhere, here's what Mathematica has to say:

Mathematica notebook

Daniel Franke
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  • How do you enter all that fancy notation (e.g., roots,$\gamma$'s) in Mathematica®? – Stephen Herschkorn Jul 31 '13 at 05:09
  • Everything has a linear input form that you can convert to pretty print using Shift+Ctrl+N, but it's easier to open "Palettes > Basic Math Assistant" and just click the symbols you want to insert. – Daniel Franke Jul 31 '13 at 13:19