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.
Best Answer
So far you've managed to state all the things you need.
You know what the distribution of $Y_i=X_i-\theta$ is exponential(1) (which is independent of $\theta$)
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.