# Pivotal Quantity for the location parameter of a two parameter exponential distribution

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 $$\operatorname{Exp} \sim (\lambda)$$.

• 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. – StubbornAtom May 6 at 16:20

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 $$a<\overline X-\theta with the desired confidence coefficient $$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.