# Shortest length confidence interval for $\theta$ in a shifted exponential distribution

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.

• Because $2n(X_{(1)}-\theta)$ is exponential with mean $2$, or equivalently $\chi^2_2$. – StubbornAtom Aug 21 '20 at 19:07
• 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? – AxyuS Aug 21 '20 at 19:32
• 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. – StubbornAtom Aug 21 '20 at 19:45
• I can see now how it can be generalized. Thank you so much for your help. – AxyuS Aug 21 '20 at 19:51
• Did you solve the problem with the hint below? No way to know if an answer is helpful unless the asker interacts. – StubbornAtom Aug 24 '20 at 10:14

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

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.