Obtain a $(1-\alpha) 100$% confidence interval for $\theta$ using the moment estimator

Question

Suppose $x_1,..x_n$ is a random sample from a distribution with probability density $f(x|\theta )=\theta x^{\theta -1} 0<x<1 \ and \ \theta >0.$

Find the moment estimator of $\theta$ and obtain a $(1-\alpha )100%$ confidence interval for $\theta$ using the moment estimator.

I know I can find the moment estimator by calculating $EX$ and I find the moment estimator as $\theta=\mu/(1-\mu)$. However, I have no idea how to get the confidence interval by using this moment estimator. Could someone help me out? Thank you so much!

Answer 1

Comment: Briefly, I can think of three approaches. Maybe you or someone on this site can fill in details of (1) or (2).

1) Hope that $n$ is large enough to use the Central Limit Theorem. Find $\sigma = SD(\hat \theta)$ in terms of $\theta$, and estimate it with $\hat \theta.$ Use $\hat \theta \pm 1.96\hat \sigma$ as a large-sample 95% CI for $\theta.$

Note: This is pretty much in the spirit of the traditional Wald CI for binomial proportion $\theta = P(Success),$ for which the MME is $\hat \theta = X/n$. Then the CI is $\hat \theta \pm 1.96\sqrt{\hat \theta(1-\hat \theta)/n},$ where $V(\hat \theta) = \theta(1-\theta)/n.$ (Such intervals are called 'asymptotic'. For example, this CI for the binomial success probability does not have the intended 95% coverage unless $n$ is quite large.)

2) Find the exact distribution of $\bar X$ from the beta distribution of the $X_i$, and from there the exact distribution of $\hat \theta/\theta.$ Take $L$ and $U$ that cut 2.5% from the lower and upper tails of this distribution respectively. Then $P(L < \hat \theta/\theta < U) = .95$ and a 95% CI for $\theta$ is $(\hat \theta/U, \hat \theta/L).$

3) For a particular $n$ and $\hat \theta$, do a parametric bootstrap CI of $\theta$.

I have not tried (2) and so am not sure it is feasible. Here is an example of the bootstrap, in a case where $\hat \theta = 3.5$ based on a sample of size $n = 100.$ The resulting 95% CI for $\theta$ is $(2.67, 4.12).$

 m = 10^4;  th.hat.obs = 3.5;  n = 100
 x = rbeta(m*n, th.hat.obs, 1)
 DTA = matrix(x, nrow=m)  # 10000 x 100 matrix, each row a 're-sample'
 a = rowMeans(DTA);  th.hat.boot = a/(1-a)
 v.boot = th.hat.boot - th.hat.obs
 th.hat.obs - quantile(v.boot, c(.975, .025))
 ##    97.5%     2.5% 
 ## 2.666264 4.118063