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Suppose we have $n$ iid samples from $Exp(1,\eta)$

This distribution is $e^{-x+\eta}$ for $x \ge \eta$

I want to understand why the following is a correct symmetric $100 \gamma $ confidence interval for $\eta$

I know that

$Q=X_{1:n}-\eta$ is a pivotal quantity where $X_{1:n}$ is the minimum order statistic. it is pivotal because it has distribution $exp(\frac{1}{n})$

The answer to the confidence interval is

[$X_{1:n}+\frac{1}{n}ln(\frac{\alpha}{2}),X_{1:n}+\frac{1}{n}ln(1-\frac{\alpha}{2})]$ with $\alpha=1-\gamma$

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  • $\begingroup$ its on the second line $\endgroup$
    – Quality
    Commented Apr 1, 2019 at 22:39
  • $\begingroup$ @Taylor it's a shifted exponential; the first parameter is either scale or rate (we can't tell from the question but it won't change anything) and the second is the shift. $\endgroup$
    – Glen_b
    Commented Apr 1, 2019 at 22:46

1 Answer 1

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Hint:

$X_{(1)}=\min\{X_1,\cdots ,X_n\}$

$Z=X_{(1)}-\eta \sim Exp(\frac{1}{n})$

$F_Z(z)=1-e^{-nz}$ and $p(Z>z)=e^{-nz}$

$P(u_1<Z<u_2)=1-\alpha$

$\frac{\alpha}{2}=p(Z < u_1)=1-e^{-nu_1}$

$-nu_1=\ln(1-\frac{\alpha}{2})$ so $u_1=-\frac{1}{n}\ln(1-\frac{\alpha}{2})$

$\frac{\alpha}{2}=p(Z>u_2)=e^{-nu_2}$

$u_2=-\frac{1}{n}\ln(\frac{\alpha}{2})$

$-\frac{1}{n}\ln(1-\frac{\alpha}{2})<X_{(1)} -\eta<-\frac{1}{n}\ln(\frac{\alpha}{2})$

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