This is a question in a practice exam handed out by my professor in preparation for the final.
Suppose that $\{X_i\}_{i=1}^{n}$, are i.i.d. with common density
$f(X; \theta) = \theta X^{\theta−1}I(0 < x < 1)$ where $\theta \in (0, \infty)$
Assume that n is large. Derive an approximate 95% confidence interval for $\theta$ based on the m.l.e estimate.
$\textbf{Step 1:}$ Find MLE.
$$\prod_{i=1}^{n}f(X_i;\theta)=\theta^{n} \prod_{i=1}^{n}f(X_i;\theta)X^{\theta−1}I(0 < x < 1)$$
I derived the MLE to be $\hat{\theta}=\frac{n}{-\sum_{i=1}^{n}ln(X_i)}$. Define $S_i=-ln(X_i)$ then $\hat{\theta}=\frac{1}{\bar{S}}$
$\textbf{Step 2:}$ Find Distribution of MLE.
$$S_i\sim exp(\theta)\Rightarrow\sum_{i=1}^{n}S_i\sim gam(n,\theta)\Rightarrow\frac{1}{\sum_{i=1}^{n}S_i}\sim InvGama(n,\theta)$$
$\textbf{Step 3:}$ Find Asymptotic distribution of MLE.
By CLT $$\sqrt{n}(\bar{S}-\frac{1}{\theta})\stackrel{\text{d}}{\rightarrow}N(0,\frac{1}{\theta^2})$$
Consider:
$$h(t)=\frac{1}{t} \hspace{4mm} \text{and} \hspace{4mm} h^{'}(t)=-\frac{1}{t^2}$$
Since $h^{'}(\frac{1}{\theta})=-\theta^2\neq0$ by assumption $\theta\in(0,\infty)$
By the delta method:
$$\sqrt{n}(h(\bar{S})-h\bigg(\frac{1}{\theta}\bigg))\stackrel{\text{d}}{\rightarrow}N(0,\frac{1}{\theta^2}*\bigg[h^{'}\bigg(\frac{1}{\theta}\bigg)\bigg]^2)=N(0,\theta^2)$$
Then for large enough n,
$$P\bigg(Z_{\frac{\alpha}{2}}<\frac{\sqrt{n}(\frac{1}{\bar{S}}-\theta)}{\theta}<Z_{1-\frac{\alpha}{2}}\bigg)\approx 1-\alpha$$
Then the confidence interval is($Z$ represent the quantiles of a standard normal.
$$\bigg(\frac{\sqrt{n}}{(Z_{1-\frac{\alpha}{2}}+\sqrt{n})\bar{S}},\frac{\sqrt{n}}{(Z_{\frac{\alpha}{2}}+\sqrt{n})\bar{S}}\bigg)$$
Is my solution correct? Is there an easier way to approach this problem?