Let $(X_1, \ldots, X_n)$ be an i.i.d. random sample. Determine asymptotic confidence Let $(X_1, \ldots, X_n)$ be an i.i.d. random sample. Determine asymptotic confidence intervals at level $\gamma \in (0, 1)$ for the parameter $\theta$, if
$(a)$ $X_1 \sim U(0, \theta), \quad \theta > 0$
$(b)$ $X_1 \sim \text{Exp}(\theta), \quad \theta > 0$
$X_1 \sim U(0, \theta)$}
The probability density function of the uniform distribution is $f(x;\theta) = \frac{1}{\theta}$ for $0 \leq x \leq \theta$.
The maximum likelihood estimator (MLE) for $\theta$ is the maximum observed value, i.e., $\hat{\theta} = \max(X_1, X_2, \ldots, X_n)$.
The standard error can be estimated using the Fisher information:
$[ \text{SE}(\hat{\theta}) = \frac{\hat{\theta}}{\sqrt{3n}} ]$
where $n$ is the sample size.
The asymptotic confidence interval is then:
$[ \left(\hat{\theta} - z_{\frac{\alpha}{2}} \cdot \frac{\hat{\theta}}{\sqrt{3n}}, \ \hat{\theta} + z_{\frac{\alpha}{2}} \cdot \frac{\hat{\theta}}{\sqrt{3n}} \right) ]$
$(b)$ $X_1 \sim \text{Exp}(\theta)$}
For the exponential distribution, the probability density function is $f(x;\theta) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$ for $x \geq 0$.
The MLE for $\theta$ is the sample mean, i.e., $\hat{\theta} = \frac{1}{n}\sum_{i=1}^{n}X_i$.
The standard error can be estimated using the Fisher information:
$[ \text{SE}(\hat{\theta}) = \frac{\hat{\theta}}{\sqrt{n}} ]$
The asymptotic confidence interval is then:
$ [ \left(\hat{\theta} - z_{\frac{\alpha}{2}} \cdot \frac{\hat{\theta}}{\sqrt{n}}, \ \hat{\theta} + z_{\frac{\alpha}{2}} \cdot \frac{\hat{\theta}}{\sqrt{n}} \right) ] $
In both cases, $z_{\frac{\alpha}{2}}$ is the critical value from the standard normal distribution corresponding to the confidence level $1 - \alpha$.
Is my solution correct in this way? I'm pretty unsure about it