My homework question requires me to construct a 90% confidence for $\theta \ge 0$ where $X_1, ..., X_8$ iid $U(0, \theta)$ and the $8$ data values are given. However, I am not sure where to start with this. The question suggests using $\frac{X_{(n)}}{\theta}$ as a pivot variable (where $X_{(n)}=$ max{$X_1, ..., X_n$}), and I already have used $P(a \le \frac{X_{(n)}}{\theta} \le b)=0.9$ to get $a=0.05^{\frac{1}{n}}$ and $b=0.95^{\frac{1}{n}}$. I recognise that the solutions for $a$ and $b$ are not unique, however I just set each equal to the 5%-tile and 95%-tile respectively. I also know that I would let $n=8$ since there are $8$ observations, but am unsure of how to use this.

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    $\begingroup$ This is fine as long as you are asked for any $90\%$ confidence interval. But this choice of $a,b$ is not optimal in the sense that this does not yield the shortest length interval for $\theta$ based on $X_{(n)}$. $\endgroup$ – StubbornAtom Oct 1 '20 at 14:31

Using the hint you got, the pivotal quantity is $Y=\frac{X_{(n)}}{\theta}$

You surely know that


thus the optimum confidence interval is $[y;1]$


$$X_{(8)}\leq \theta \leq \frac{X_{(8)}}{\sqrt[8]{0.1}}$$


the density of your pivotal quantity is the following

enter image description here

It is evident that there are $\infty$ way to find the extremes of a CI at $(1-\alpha)\%$ but in this case the most reasonable solution is to minimize the range of your CI so the interval is of the form $[a;1]$

  • $\begingroup$ There are $8$ data values, ranging from $0.476$ to $2.346$, however I did not include this latter value in the question which is what $X_{(8)}$ is. So I don't understand how you got $(y;1)$. $\endgroup$ – Viv4660 Oct 1 '20 at 12:01
  • $\begingroup$ @Viv4660 : did some edits to clarify my answer $\endgroup$ – tommik Oct 1 '20 at 12:10
  • $\begingroup$ Ok - just to check my understanding, would $[\frac{X_{(8)}}{b}, \frac{X_{(8)}}{a}]$ also work where $a=0.05^{1/8}$ and $b=0.95^{1/8}$? $\endgroup$ – Viv4660 Oct 1 '20 at 12:16
  • $\begingroup$ @Viv4660 : this is an equiprobable tail CI that is not a reasonalbe CI in this case. You have to take into consideration to find the best CI you can.... IMHO $\endgroup$ – tommik Oct 1 '20 at 12:21
  • $\begingroup$ But essentially your solution is saying that 90% of the time, $\theta$ is at minimum the largest value drawn from the sample and doesn't that seem not very reasonable? $\endgroup$ – Viv4660 Oct 1 '20 at 12:34

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