If $X\sim\text{Beta}(\theta,1)$, obtain the confidence interval of $100(1-\alpha)\%$ based on the asymptotic distribution of the score function

Question

Let $X_{1},X_{2},\ldots,X_{n}$ be a random sample whose distribution is given by $\text{Beta}(\theta,1)$. Obtain the approximate confidence interval of $100(1-\alpha)\%$ based on the asymptotic distribution of the score function.

MY ATTMEPT

In the first place, observe that the probability density function of $\text{Beta}(\theta,1)$ is given by $f(x|\theta) = \theta x^{\theta-1}$. Consequently, the Fischer information of the parameter $\theta$ is given by

\begin{align*} & f(x|\theta) = \theta x^{\theta-1} \Rightarrow \ln f(x|\theta) = \ln(\theta) + (\theta - 1)\ln(x) \Rightarrow\\\\ & \frac{\partial\ln f(x|\theta)}{\partial\theta} = \frac{1}{\theta} + \ln(x) \Rightarrow \frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}} = -\frac{1}{\theta^{2}} \Rightarrow\\\\ & -\textbf{E}\left(\frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}}\right) = \frac{1}{\theta(1+\theta)} \Rightarrow I_{F}(\theta) = \frac{1}{\theta(1+\theta)} \end{align*}

However, I am little bit lost here. I know the expectation of the score function equals zero and its variance equals $I_{F}(\theta)$. I also know that the distribution of the score function is asymptotic normal, but I am unsure about which pivotal quantity should we use. Precisely speaking, should it be \begin{align*} \text{Pivot}(\textbf{X},\theta) \stackrel{\displaystyle?}{=} \frac{1}{\sqrt{nI_{F}(\theta)}}\displaystyle\sum_{i=1}^{n}U(X_{i},\theta),\quad\text{where}\quad U(X_{i},\theta) = \frac{\partial\ln f(x_{i}|\theta)}{\partial\theta} \end{align*}

If so, can someone tell me the explicit expression of the confidence interval? Any help is appreciated. Thanks in advance.

Answer 1

To expand on my comment, you could have worked directly with the score for the combined sample $\mathbf X$.

From the joint pdf of $\mathbf X$, it follows that the score function is $$\frac{\partial}{\partial\theta}\ln f(\mathbf X\mid\theta)=\sum_{i=1}^n \ln X_i+\frac{n}{\theta}$$

So the asymptotic distribution of the score is equivalent to the limiting distribution of $\sum\limits_{i=1}^n \ln X_i$.

By CLT, $$\frac{\sum_{i=1}^n \ln X_i+\frac{n}{\theta}}{\sqrt{\frac{n}{\theta^2}}}\stackrel{L}\longrightarrow N(0,1)$$

Note that the Fisher information in $\mathbf X$ is actually $$I(\theta)=-n E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f(X_1\mid\theta)\right]=\frac{n}{\theta^2}$$

So the pivot you are looking for is eventually $$\frac{1}{\sqrt n}\left(\theta\sum_{i=1}^n \ln X_i+n\right)\stackrel{L}\longrightarrow N(0,1)$$

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Exact confidence intervals for $\theta$ are also availabe, a suitable pivot being

$$-2\theta\sum_{i=1}^n \ln X_i\sim \chi^2_{2n}$$