Let $X_1, X_2, \dots , X_n$ be a random sample of size $n$ from the uniform distribution $U(2\theta)$ whose pdf is $f(x;\theta)=(2\theta)^{-1}$ , if $0<x<2\theta$ , zero otherwise.
i) We can see that the MLE estimator is $\hat{\theta}_{MLE}=\frac{1}{2}\underset{i}{\max} \ x_i$
ii) Also we can find that $Z_n = n (\theta-\hat{\theta}) \overset{d}{\to} \exp(\theta^{-1})$
I would like to construct a confidence interval with approximate $(1 − \alpha)$-confidence level based on $\hat{\theta}_{MLE}$.
$\textbf{I wondered if both of these approaches are correct or if one is not, what is the problem?}$
$\textbf{A}$:
We would like to have $\mathbb{P}[a<\hat{\theta}<b] = 1-\alpha$, where $a$ and $b$ are constants.
$$1-\alpha = \mathbb{P}[a<\hat{\theta}<b] = F_{\hat{\theta}}(b)-F_{\hat{\theta}}(a) \ \ \ \ \ \ \ , \text{where} \ F_{\hat{\theta}}(x)=(\frac{x}{\theta})^n $$ We can find $a$ and $b$ by solving: $$1-\frac{\alpha}{2}= F_{\hat{\theta}}(b) =(\frac{b}{\theta})^n \implies b = (1-\frac{\alpha}{2})^{\frac{1}{n}}\theta $$ $$\frac{\alpha}{2}=F_{\hat{\theta}}(a) = (\frac{a}{\theta})^n \implies a = (\frac{\alpha}{2})^{\frac{1}{n}}\theta$$ A confidence interval with $1-\alpha$ confidence level is
$$\text{CI}_\theta(1-\alpha) \equiv \Bigg[ \hat{\theta}(1-\frac{\alpha}{2})^{\frac{1}{n}} \ , \ \hat{\theta}(\frac{\alpha}{2})^{\frac{1}{n}} \Bigg]$$
$\textbf{B}$
We would like to have $\mathbb{P}[a<\theta-\hat{\theta}<b] = 1-\alpha$, where $a$ and $b$ are constants. $$1-\alpha=\mathbb{P}[a<\theta-\hat{\theta}<b] = \mathbb{P}[na<n(\theta-\hat{\theta})<nb]$$ So give (ii): $$na = \theta \log(1-\alpha/2)$$ $$nb = \theta \log(\alpha/2)$$ So ;
$$\text{CI}_\theta(1-\alpha) \equiv \Bigg[ \hat{\theta}-\frac{\hat{\theta}}{n}\log(1-\frac{\alpha}{2}) \ , \ \hat{\theta}-\frac{\hat{\theta}}{n}\log(\frac{\alpha}{2}) \Bigg]$$
Update: I got my answer from this Does invertible monotone transformation of a confidence interval give you a confidence interval (at the same level) in the transformed space? So both give a confidence interval with the same confidence level.
