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Let $(X_{1}, \dots X_{n})$ be a random sample of a random variable $X$ with pdf:
$f(x|\theta) = \exp{(-(x-\theta))}\mathbb{1}_{{(\theta},{\infty)}}(x), \enspace \theta > 0$.

How do I find the pivotal quantity and an approximated confidence interval with level of confidence $\gamma \in (0,1)$ for $\theta$ based on a sufficient statistic?

By the way, I know a sufficient statistic $T(x)$ for $X$ is the $\min ({X_{1}, \dots, X_{n}}) = X_{(1)} \geqslant \theta$.

I got stuck trying to find the pivotal quantity, any help is appreciated.

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  • $\begingroup$ 1. What do you know about pivotal quantities? 2. If $\theta$ were 0, what's the distribution of $X_{(1)}$? 3. What's the distribution of $X_i-\theta$? ... you need to contribute a lot more to solving this $\endgroup$
    – Glen_b
    Oct 2, 2016 at 2:17
  • $\begingroup$ 1. As far as i know, pivotal quantities are functions of the variable of interest that don't depend on the unknown parameter. 2. $\theta$ can't be 0, 0 doesn't belong to it's domain. 3. The distribution of $Y = X_{i} - \theta$ is a exponential(1) $\endgroup$
    – z.schaap
    Oct 2, 2016 at 16:48
  • $\begingroup$ The restriction of $\theta$ to be positive is arbitrary, not necessary -- as a shift parameter $\theta$ could be defined anywhere on the real line -- positive or negative. Try considering the question in my point 2. again. It was not an idle question. $\endgroup$
    – Glen_b
    Oct 2, 2016 at 17:07
  • $\begingroup$ So, we have $X_{(1)} = \Pr{\min(X_{1}, \dots, X_{n}) \geqslant \theta}$ that applied independence result in $ F(\min(X_{1}, \dots, X_{n})) = 1 - [1 - F(x|\theta = 0)]^{n}$, the fdp though is $n f(x|\theta = 0) [1-F(x)]^{n-1}$. Making the right substituitions, it results that the distriution of the minimum is $n\exp{(-nx)}$ which is exponential with parameter $n$. $\endgroup$
    – z.schaap
    Oct 2, 2016 at 22:40

1 Answer 1

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So far you've managed to state all the things you need.

  1. You know what the distribution of $Y_i=X_i-\theta$ is exponential(1) (which is independent of $\theta$)

  2. You know the distribution of $X_{(1)}$ when $\theta=0$.

The obvious thing to consider (and I am surpriseded this hasn't occurred to you by now, since you're obviously reasonably adept at the required manipulations) is to apply he same idea in (1.) to (2.) (noting that $Y_i$ is the same thing as $X_i$-when-$\theta=0$ ... ); the answer should be clear by inspection, and then the relevant pivotal quantity should be immediately obvious.

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  • $\begingroup$ I believe my pivotal quantity is $Y_{i}$, then. And to find my confidence interval I should only do $\Pr(a \leqslant Y_{i} \leqslant b) = \gamma$? $\endgroup$
    – z.schaap
    Oct 3, 2016 at 2:06
  • $\begingroup$ You should show it, then there's no need to apply belief (facts are facts). And then yes that's an interval for the pivot, from which you should then be able to back out an interval for $\theta$. $\endgroup$
    – Glen_b
    Oct 3, 2016 at 4:24

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