# Finding a confidence interval for shifted exponential distribution

Let $$X_1,\ldots, X_n$$ are i.i.d. random variables such that: $$f(x;\sigma ,\theta)=\frac{1}{\sigma}e^{\frac{-(x-\theta)}{\sigma}}, x\gt \theta$$ where $$\sigma \gt 0$$ and $$\theta \in R$$ .

a) if $$\theta$$ is known, find a $$100(1-\alpha)%$$% confidence interval for $$\sigma$$. (Hint: use $$\sum_{i=0}^n (X_i -\theta)$$ or a modification)

b) if $$\theta$$ is unknown, find a $$100(1-\alpha)%$$% confidence interval for $$\sigma$$. (Hint: use $$\sum_{i=0}^n (X_i -X_{(n)})$$ or a modification)

a) Since $$\theta$$ is a location parameter then $$X_i - \theta \sim Exp(1/\sigma)$$ then $$Y=\sum_{i=0}^n (X_i - \theta) \sim Gamma(n,1/ \sigma)$$, so $$\frac{\sigma}{2}Y\sim Gamma (n,1/2) \sim {\chi^2}_{2n}$$ and solving $$P(a\le\frac{\sigma}{2}Y\le b)=1-\alpha$$ I can find the condidence interval. Is it right?

b)I found that $$X_{(1)}-\theta\sim Exp (n/\sigma)$$ I was thinking sum $$X_i - \theta$$ and $$X_{(1)}-\theta$$ but due to $$X_{(1)}$$ and $$X_i$$ are not independent I couldn't compute the distribution. Any ideas?

• Should be $x>\theta$ in the pdf instead of $0<x<\theta$. – StubbornAtom May 16 at 6:24

## 1 Answer

For part (a), note that $$X_i-\theta$$ are i.i.d $$\mathsf{Exp}$$ with mean $$\sigma$$. In other words, $$\frac{2}{\sigma}(X_i-\theta)$$ are i.i.d $$\mathsf{Exp}$$ with mean $$2$$, i.e. a $$\chi^2_2$$ variable. Therefore the correct pivot is $$T_1(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_i-\theta)\sim \chi^2_{2n}$$

You can get a $$100(1-\alpha)\%$$ confidence interval for $$\sigma$$ starting from $$P_{\sigma}\left[\chi^2_{1-\alpha/2,2n}

where $$\chi^2_{\alpha,2n}$$ is such that $$P(\chi^2_{2n}>\chi^2_{\alpha,2n})=\alpha$$.

For part (b), I think a suitable pivot is (see this post for justification) $$T_2(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=2}^n (X_i-X_{(1)})\sim \chi^2_{2n-2}$$

• Care to explain the downvote? – StubbornAtom May 18 at 3:57