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Let $X_1, X_2, \ldots, X_n$ be a sample from: $$f(x) = \begin{cases} e^{-(x-\theta)} & x > \theta \\ 0 & \text{otherwise} \end{cases} $$ Find the shortest-length confidence interval for $\theta$ at level $1−\alpha$ based on a sufficient statistic for $\theta$.

The answer to the following problem is given as $$\bigg(X_{(1)} - \frac{χ_{2, \alpha}^2}{2n},X_{(1)}\bigg)$$

How chi-squared distribution with two degrees of freedom gets involved here? I know that have to use the pivot method here but I am not getting the final, answer anyway. Any help is appreciated.

Note: I have tried to solve this question using the CDF and Chebyshev inequality method. But those method provides an answer which is totally different from this result. Also, I know proved that a chi-squared distribution with 2 degrees of freedom is an exponential distribution with mean 2 and vice versa.

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  • $\begingroup$ Because $2n(X_{(1)}-\theta)$ is exponential with mean $2$, or equivalently $\chi^2_2$. $\endgroup$ – StubbornAtom Aug 21 '20 at 19:07
  • $\begingroup$ Do we have to straight away remember such results or it appears as during derivation of a CI in case of inequality methods? In this case, I knew that $X_{(1)}$ is a pivot element, so I could have thrown some constants here and there to estimate it's distribution. What about general cases? $\endgroup$ – AxyuS Aug 21 '20 at 19:32
  • $\begingroup$ You don't have to remember anything as long as you see that $X_i-\theta \sim \text{Exp}(1)$ and the minimum of that is again exponential. So automatically $X_{(1)}-\theta$ is a pivot. As for general cases, we usually consider pivots based on a sufficient statistic when the sample size is fixed. $\endgroup$ – StubbornAtom Aug 21 '20 at 19:45
  • $\begingroup$ I can see now how it can be generalized. Thank you so much for your help. $\endgroup$ – AxyuS Aug 21 '20 at 19:51
  • $\begingroup$ Did you solve the problem with the hint below? No way to know if an answer is helpful unless the asker interacts. $\endgroup$ – StubbornAtom Aug 24 '20 at 10:14
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First note that $X_i-\theta \stackrel{\text{i.i.d}}\sim \text{Exp}(1)$, from which you can show that $\min\limits_{1\le i\le n}(X_i-\theta)=X_{(1)}-\theta$ has an exponential distribution with mean $1/n$. In other words, $$P(X_{(1)}-\theta\le t)=1-e^{-nt}\quad,\,t\ge 0$$

So an appropriate choice of pivot here is $T=X_{(1)}-\theta$.

Now for a confidence interval of $\theta$ based on $T$ with confidence coefficient $1-\alpha$, you have some $(\ell_1,\ell_2)$ with $0\le \ell_1< \ell_2$ such that

$$P_{\theta}\left\{\ell_1<T<\ell_2\right\}=P_{\theta}\left\{X_{(1)}-\ell_2\le \theta\le X_{(1)}-\ell_1 \right\}=1-\alpha\quad,\forall\,\theta $$

Or, $$e^{-n\ell_1}-e^{-n\ell_2}=1-\alpha \tag{$\star$}$$

So a two-sided confidence interval for $\theta$ is of the form $\left[X_{(1)}-\ell_2, X_{(1)}-\ell_1\right]$.

Length of this interval is $\ell_2-\ell_1=f(\ell_1,\ell_2)$ (say) and you have to minimize $f$ subject to $(\star)$.

This is a constrained optimization problem can be solved by usual calculus methods. Show that $f$ is increasing in $\ell_1$, so that $f$ is minimum when $\ell_1$ is minimum. Obtain the corresponding value of $\ell_2$ and you would have a $100(1-\alpha)\%$ shortest length confidence interval for $\theta$ based on $T$. We could have also worked with the pivot $2nT\sim \chi^2_2$ but that is not necessary.

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