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Let $X_1,...,X_n$ be a random sample on $\text{Uniform}(\theta -1/2, \theta +1/2).$ I need to find a confidence interval for $\theta$ with ($1-\alpha$) of confidence.

I have this:

$\max(X_i)-1/2<\theta<\min(X_i) +1/2,$ hence I say $\mathbb{P}(\max (X_i)-1/2<\theta<\min(X_i) +1/2)\ge 1-\alpha $

or using the statistics $\min(X_i) , \max(X_i),$

$\mathbb{P}(\min(X_i) \le\theta\le\max(X_i))\ge 1-\alpha$
$\mathbb{P}(\theta\le \max(X_i)) -\mathbb{P}(\min(X_i) \le\theta)$ $\ge 1-\alpha. $

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Definitely not the most efficient approach (doesn't condition on the sufficient statistic), but the mean is unbiased for $\theta$. And the variance of a single observation from the above distribution is 1/12. So, by the CLT:

$$ \sqrt{n} \left( \bar{X} - \theta \right) \rightarrow_d \mathcal{N}\left(0, 1/12 \right)$$

Which means that a $1-\alpha$ confidence interval for $\theta$ can be given by

$$\left(\bar{X}+(12n)^{-.5}\mathcal{Z}_{\alpha/2}, \bar{X}+(12n)^{-.5} \mathcal{Z}_{1-\alpha/2} \right)$$

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    $\begingroup$ Right about 'not the most efficient'. For $n = 5, 20$ your CIs are of widths $0.506, 0.253,$ respectively; while CIs based on max and min have average widths $0.451, 0.139,$ respectively. $\endgroup$
    – BruceET
    Jun 15 '21 at 21:17
  • $\begingroup$ @BruceET well using the min and max would be biased. Otherwise, every instance of the German Tank problem would claim you have the last tank that was made. $\endgroup$
    – AdamO
    Jun 15 '21 at 22:03
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    $\begingroup$ Not advocating use of max or min, but of midrange (their average). So if a sample of size 12 has midrange 9.95, the 95% CI would be approx (9.89, 10.02) of length 0.14. $\endgroup$
    – BruceET
    Jun 15 '21 at 23:08
  • $\begingroup$ thanks that was so helpfull $\endgroup$
    – Hug ma
    Jun 16 '21 at 0:25
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I too will avoid analytic derivations of distributions of $\max(X_i)$ and $\min(X_i)$ because I guess that is the main point of this assignment. [Also see this page.]

However, results from a simulation for the case $n = 20, \theta = 10,$ based on the sample midrange, are shown below.

set.seed(2021)
n=20
mr = replicate(10^6, mean(range(runif(n, 9.5, 10.5))))
CI = quantile(mr, c(.025,.975)); CI
     2.5%     97.5% 
 9.930509 10.069571 

hdr = "Simulated Distributions of Midrange"
hist(mr, prob=T, br=50, col="skyblue2", main=hdr)
 abline(v = CI, col="red", lwd=2, lty="dashed")

enter image description here

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Frequentist interval via pivotal quantity: An improvement over the interval suggested by @AdamO but still suboptimal solution can be obtained using almost exactly the same method as the one I give here so I omit the details of the following derivation. The pdf of $$ Z=\frac{X_{(1)}+X_{(n)}}2-\theta, $$ is $$ f(z)=n(1-2|z|)^{n-1} $$ for $-1/2 \le z \le 1/2$. Since the distribution of $Z$ doesn't depend on $\theta$, $Z$ is a pivotal quantity.

This pdf is symmetric and the upper $\alpha/2$-quantile of $Z$ is $\frac{1-\alpha^{1/n}}2$. Thus $$ P\left(-\frac{1-\alpha^{1/n}}2<\frac{X_{(1)}+X_{(n)}}2-\theta<\frac{1-\alpha^{1/n}}2\right)=1-\alpha. $$ Inverting the double inequality, we find that $$ \frac{X_{(1)}+X_{(n)}}2 \pm \frac{1-\alpha^{1/n}}2 \tag{1} $$
is a $1-\alpha$ confidence interval for $\theta$. The midrange $(X_{(1)}+X_{(n)})/2$ is not a sufficient statistics for $\theta$, however.

Whittinghill and Hogg: As pointed out by @COOLSerdash, by inverting a likelihood ratio statistic these authors derive the interval $$\left(x_{(n)}-\frac{(1-\alpha)^{1/n}}2,x_{(1)}+\frac{(1-\alpha)^{1/n}}2\right)\tag{2}$$ which is a function of the sufficient statistic $(X_{(1)},X_{(2)})$ for $\theta$. However, simulations (see below) suggest that this interval is also suboptimal.

A Bayesian credible interval: An alternative is to represent our prior ignorance about $\theta$ by a uniform improper prior $\pi(\theta)=1$. The posterior density of $\theta$ is then $$ \pi(\theta|\mathbf{x})\propto \prod_{i=1}^n I_{(\theta-\frac12,\theta+\frac12)}(x_i)=I_{(x_{(n)}-\frac12,x_{(1)}+\frac12)}(\theta), $$ that is, conditional on the observations, $\theta$ is uniformly distributed on the interval from $(x_{(n)}-\frac12,x_{(1)}+\frac12)$. A $1-\alpha$ credible interval for $\theta$ is therefore $$ \left(x_{(n)}-\frac12 + \frac{\alpha}2L, x_{(1)}+\frac12 - \frac{\alpha}2L\right) \tag{3} $$ where $L=1-(x_{(n)}-x_{(1)})$. Interestingly, judged by frequentist criteria, based on the following simulation, this interval appear to have the exact nominal coverage but is considerably shorter on average than both (1) and (2):

ci.normal <- function(x, alpha) {
  n <- length(x)
  mean(x) + c(-1,1)*(12*n)^(-.5)*qnorm(alpha/2, lower.tail = FALSE)
}
ci.pivot <- function(x, alpha=.05) {
  n <- length(x)
  (min(x)+max(x))/2 + c(-1,1)*(1 - alpha^(1/n))/2
}
ci.wh <- function(x, alpha) {
  n <- length(x)
  c <- (1-alpha)^(1/n)/2
  c(max(x)-c, min(x)+c)
}
ci.bayes <- function(x, alpha=.05) {
  L <- 1 - (max(x)-min(x))  
  c(max(x) - .5 + L*alpha/2, min(x) + .5 - L*alpha/2)
}
coverage <- function(fn, theta=0, nsim=100000, n, alpha=0.05) {
  hits <- 0
  ci.lengths <- numeric(nsim)
  for (i in 1:nsim) {
    x <- runif(n, theta-.5, theta+.5) 
    ci <- fn(x,alpha)
    ci.lengths[i] <- ci[2] - ci[1]
    if (ci[1] < theta & ci[2] > theta)
      hits <- hits + 1
  }
  list(coverage = hits/nsim, meanlength = mean(ci.lengths))
}
> coverage(ci.normal, n=5)
$coverage
[1] 0.95315

$meanlength
[1] 0.5060605

> coverage(ci.pivot, n=5)
$coverage
[1] 0.95004

$meanlength
[1] 0.4507197

> coverage(ci.wh, n=5)
$coverage
[1] 0.94968

$meanlength
[1] 0.3226174

> coverage(ci.bayes, n=5)
$coverage
[1] 0.94991

$meanlength
[1] 0.3169024
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    $\begingroup$ The confidence interval $(Y_n - c, Y_1 + c)$ with $c=(1/2)(1-\alpha)^{1/n}$ has similar length and coverage as the Bayesian credible interval. It still seems to be a bit wider though. $\endgroup$ Jun 18 '21 at 7:37
  • $\begingroup$ @COOLSerdash Indeed it does. How did you derive this interval? $\endgroup$ Jun 18 '21 at 9:32
  • $\begingroup$ I found it and its derivation in this paper. It's based on a likelihood ratio test (section 5). $\endgroup$ Jun 18 '21 at 9:51
  • $\begingroup$ Ah, thanks! That's interesting. $\endgroup$ Jun 18 '21 at 9:55

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