# Confidence interval for Uniform($\theta$, $\theta + a$)

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.)

• – whuber Aug 3 '13 at 20:11
• @Stephen - I think of CIs as inverted hypothesis tests: the $1-\alpha$ CI is the set of values that would not be rejected by an $\alpha$-level test. It's not obvious to me what your test is. If $\theta = \theta_1$ and we hypothesize $\theta = \theta_0$, what is the rejection probability when $\theta_0 = \theta_1$? Does it rise monotonically as $\theta_0$ moves away from $\theta_1$? – Ray Koopman Aug 4 '13 at 7:31
• Stephen, when a CI is correctly applied, it cannot fail. Ergo, we are discussing questions of application and interpretation. It seems to me that several of the answers in the related threads address these very points. – whuber Aug 4 '13 at 15:12
• I have not claimed that you are wrong, Stephen. However, I do not see what (technically) is the problem with a "nonsensical" answer, because it does not necessarily violate the defining properties of a confidence interval. That's why I maintain this question is one about the understanding and interpretation of confidence intervals. It's fine with me if you see no connection between this discussion and the ideas presented in the other threads, but I hope that other readers will find the possible connections to be interesting and constructive. – whuber Aug 5 '13 at 0:42
• You seem to be using "confidence" in your comment in the sense of a Bayesian credibility rather than as defined in the statistical literature. As an extreme example of the distinction, a procedure that uses a spinner to return $\mathbb{R}$ with 95% probability and the empty set with 5% probability is a valid 95% confidence interval (albeit a terrible one, obviously, because it entirely ignores the data!). That shows why there is more to evaluating the quality of a CI procedure than checking its coverage, but it does not demonstrate any inherent flaw with CIs. – whuber Aug 5 '13 at 19:43

It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "confidence interval methodology" (whatever that is)—you could have constructed a more appropriate one for your purposes.

A more orthodox approach to this particular problem:

First, note that the sufficient statistic $(Y,Z)$ has an obvious ancillary component: the sample range $Z-Y$. It's a very strong precision index: as it approaches $a$, $\theta$ is known almost exactly. So any inference should be conditional on its observed value $z-y$.

Second, the sample minimum $Y$ conditional on the observed value of the range is uniformly distributed between $\theta$ and $\theta+a-r$. Confidence intervals based on this distribution could be constructed & would be well behaved. Still, the likelihood for $\theta$ is flat between $z-a$ and $y$, so it wouldn't be possible to say for any confidence interval with less than 100% coverage that it contained values of $\theta$ less discrepant with the data than those outside it. Best to stick with the 100% interval.

(2) You're right that $\left(Y-\left(1-\frac{1-\gamma} 2\right)(a-r),Y-\frac{1-\gamma} 2(a-r)\right)$ is a valid C.I., conditional on $r$, but why not use $(Y-\gamma(a-r),Y)$? The likelihood is $\frac{1}{a-r}$ for all values of $\theta$ between $y+r-a$ and $y$, & I don't see any generally compelling reason to honour an arbitrary 95% of those values with inclusion in my confidence interval. The 100% interval $(y+r-a,y)$ is preferable as it separates zero-likelihood values of $\theta$ from those with positive likelihood.
*At least for inference on $\theta$ from a single sample: there may be some applications—say in quality control, where consideration of coverage over repeated samples is more than a thought experiment—for which it could have some use.]
• As for the flaw, see my comment towards the bottom above: The possibility of an empty confidence interval surely provides evidence of a flaw in the frequentist methodology. It seems you have to fudge here to say to the client, "Well, the confidence interval is useless in this case, but we know it's between $Z-a$ and $Y$..." – Stephen Herschkorn Aug 5 '13 at 19:50
• Oh, I see. Your approach yields $[Y-(1-\frac\gamma 2)(a-R),Y-\frac\gamma 2(a-R)]$, where $R$ is the range, as the CI. Much better. Thanks! – Stephen Herschkorn Aug 5 '13 at 21:01
• Response to addition: Why "conditional on r"? The unconditional probability $P\{Y-(1-\frac{\gamma}2(a-R) \le \theta \le Y-\frac{\gamma}2(a-R)\} = \gamma$. Also, I am not saying this is the only $\gamma$ CI, just as the usual formulae for a 95% confidence interval for a mean, say, do not provide the only 95% confidence interval for that parameter. – Stephen Herschkorn Aug 6 '13 at 20:39
• Why "conditional on $r$"? Just to emphasize that the CI's valid for each and every observed value of $r$. And I didn't think you were saying that only one CI was possible, but I wanted to explain why only the 100% one is good in this case. – Scortchi - Reinstate Monica Aug 6 '13 at 22:14