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Let $X$ be a random variable with probability density function $f(x,\theta, \beta)=\beta e^{-\beta(x-\theta)} \mathbb{1}_{(\theta,\infty)}$ with $\beta>0, \theta \in \mathbb{R}$ (a two parameter exponential distribution) from which a random sample is taken. If $\beta$ is known and $\theta$ unknown, find an optimal confidence interval for $\theta$.

So I need help finding the pivotal quantity for this example. I thought using $U_{i}=X_{i}-\theta \sim \operatorname{Exp}(\beta)$ then $\overline{U}=\overline{X}-\theta \sim \operatorname{Gamma}(n,\beta)$ to homologate the procedure to find a confidence interval for $\lambda$ from $\operatorname{Exp} \sim (\lambda)$.

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  • $\begingroup$ You did find a pivotal quantity $U$ for $\theta$. Alternatively, since $X_i-\theta$ is i.i.d $\mathsf{Exp}$ with mean $1/\beta$, we have $2\beta(X_i-\theta)$ i.i.d $\chi^2_2$, thus giving the pivot $2\beta\sum (X_i-\theta)\sim \chi^2_{2n}$. Now the confidence interval can be obtained using chi-square fractiles. $\endgroup$ – StubbornAtom May 6 at 16:20
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Indeed $\overline X-\theta$ is a valid pivot for $\theta$ when $\beta$ is known. All you have to do now is find $a,b$ such that $a<\overline X-\theta<b\iff \theta\in(\overline X-b,\overline X-a)$ with the desired confidence coefficient $P_{\theta}\left[\theta\in(\overline X-b,\overline X-a)\right]$ (for a two-sided interval). This is easily done using software.

Note that you have $X_i-\theta\stackrel{\text{i.i.d}}\sim\mathsf{Exp}$ with mean $1/\beta$, which implies $X_{(1)}-\theta\sim \mathsf{Exp}$ with mean $1/(n\beta)$ where $X_{(1)}=\min\limits_{1\le j\le n}X_j$. So yet another pivotal quantity is $$T(\mathbf X,\theta)=2n\beta(X_{(1)}-\theta)\sim \chi^2_2$$

We expect a confidence interval based on this pivot to be 'better' (in the sense of shorter length, at least for large $n$) than the one based on $\sum\limits_{i=1}^n X_i$ as $X_{(1)}$ is a sufficient statistic for $\theta$.

Now you can derive a two-sided confidence interval with confidence level $1-\alpha$ starting from

$$P_{\theta}(\chi^2_{1-\alpha/2,2}< T< \chi^2_{\alpha/2,2})=1-\alpha\quad\forall\,\theta$$

Here $\chi^2_{\alpha,2}$ is of course the $(1-\alpha)$th fractile of $\chi^2_2$, i.e. $P(\chi^2_2>\chi^2_{\alpha,2})=\alpha$. Notably if you have an observed sample at hand, then calculations involving chi-square fractiles for the confidence interval can be done by hand since printed chi-square tables are readily available.

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