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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.