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?
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 $$P_{\sigma}\left[\chi^2_{1-\alpha/2,2n}&lt;T_1&lt;\chi^2_{\alpha/2,2n}\right]=1-\alpha\quad\forall\,\sigma\,,$$

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

For part (b), use the  pivot (see this answer for proof) $$T_2(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_{(i)}-X_{(1)})=\frac{2}{\sigma}\sum_{i=1}^n(X_i-X_{(1)})\sim \chi^2_{2n-2}$$
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 $$P_{\sigma}\left[\chi^2_{1-\alpha/2,2n}&lt;T_1&lt;\chi^2_{\alpha/2,2n}\right]=1-\alpha\quad\forall\,\sigma\,,$$

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

For part (b), use the  pivot (see this answer for proof) $$T_2(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_{(i)}-X_{(1)})=\frac{2}{\sigma}\sum_{i=1}^n(X_i-X_{(1)})\sim \chi^2_{2n-2}$$