# How to get this confidence interval from a pivotal quantity

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 $$\alpha=1-\gamma$$

• its on the second line Commented Apr 1, 2019 at 22:39
• @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. Commented Apr 1, 2019 at 22:46

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

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

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