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Let $ U_1, \dots, U_n $ be a random sample of uniform distribution over $ [a,1] $. Construct a confidence interval for $ a $ with $ 1-\alpha = 0.95 $.

I managed to show that $ T = \min\{U_i\} $ is a sufficient statistic with $$ \mathbb{P}(T \le t) = 1 - \mathbb{P}(T > t) = 1 - \left( \frac{t-a}{1-a} \right)^n. $$

Now I would like to estimate confidence interval for $ T$ since it's a good estimator of $ a $: \begin{align*} & 1-\alpha = 1 - \mathbb{P}(T > t) \\ & \alpha = \mathbb{P}(T > t) = \mathbb{P}(1 > T > t) = \left( \frac{t-a}{1-a} \right)^n \\ & t = ((\alpha + a)(1-a))^{-n} \end{align*}

The problem is, $t$ depends on unknown parameter $a$. What am I doing wrong?

Edit: Following suggestions in the comments, I analyze statistic $ S = \frac{T-a}{1-a} $. Let $t$ be observed minimum of measurements.

\begin{align*} & 1-\alpha = \mathbb{P}(S \leq t) = 1 - \mathbb{P}(S > t) = 1 - t^n \\ & \alpha = t^n, \quad\alpha^{1/n}=t \\ \end{align*}

Going back from $S$ to $T$: \begin{align*} 1-\alpha &= \mathbb{P}(S \leq t) \\ &= \mathbb{P}\left(\frac{T-a}{1-a} \leq \alpha^{1/n}\right) \\ &= \mathbb{P}(T \leq (1-a)\alpha^{1/n} + a) \\ &= \mathbb{P}\left( \frac{T-\alpha^{1/n}}{1-\alpha^{1/n}} \leq a \right) \\ &= \mathbb{P}\left( \frac{T-\alpha^{1/n}}{1-\alpha^{1/n}} \leq a \leq 1 \right) \end{align*}

That would imply the confidence interval (with $1-\alpha = 0.95$) is $\left[\frac{t-\alpha^{1/n}}{1-\alpha^{1/n}}, 1\right] = \left[\frac{t-0.05^{1/n}}{1-0.05^{1/n}}, 1 \right]$. Is this reasoning correct? I'm worried that left side of the interval can quickly go below zero and it does not say anything useful about this distribution.

This problem belongs to my homework assignment.

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  • $\begingroup$ Wouldn't my interval still depend on $a$ then? If $ \mathbb{P}((T-a)(1-a)>t)=t^{-n}$ and $t=\alpha^n$, then still I have to move all $a$'s to the right side inside $\mathbb{P}$. And the right side is range end. $\endgroup$ Mar 15, 2020 at 13:22
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    $\begingroup$ $t$ is not unknown: it is the observed minimum of the data. You can therefore solve for $a$ in terms of $t$ and $\alpha.$ $\endgroup$
    – whuber
    Mar 15, 2020 at 13:30
  • $\begingroup$ I made a mistake; work with $S=\frac{T-a}{1-a}$ instead. $\endgroup$ Mar 15, 2020 at 16:46
  • $\begingroup$ Thank you @StubbornAtom, I updated the post. $\endgroup$ Mar 15, 2020 at 21:23
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    $\begingroup$ Have a good think about the upper bound you are presently giving for your interval. Is there any lower value that would be a better upper bound (e.g., without derogation of coverage probability)? $\endgroup$
    – Ben
    Mar 15, 2020 at 23:59

1 Answer 1

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The problem stems from calculation of the probability $P(T \le t)$; it should be

\begin{align} P(T\le t)&=1-P(T>t) \\&=1-(P(X_1>t))^n \\&=1-\left(\frac{1-t}{1-a}\right)^n\qquad,\,t\in(a,1) \end{align}

For every $0<a<1$ and $0<s<1$, this yields $$P_a\left(0 \le \frac{T-a}{1-a}\le s\right)=P_a\left(\frac{T-s}{1-s} \le a \le T \right)=1-(1-s)^n.$$

If you set that last probability to $1-\alpha$ and solve for $s$, you will have your confidence interval.

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    $\begingroup$ Thanks to @Ben for the correction. $\endgroup$ Mar 16, 2020 at 6:10

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