Obtain a $(1-\alpha) 100$% confidence interval for $\theta$ using the moment estimator 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!
 A: The method of moments estimator of $\theta$ is $$\hat\theta(X_1,\ldots,X_n)=\frac{\overline X}{1-\overline X}$$
Assuming $n$ is large enough, asymptotic distribution of $\overline X$ by CLT is $$\sqrt n\left(\overline X-\frac{\theta}{\theta+1}\right)\stackrel{a}\sim N\left(0,\sigma^2(\theta)\right)$$
, where $\sigma^2(\theta)=\frac{\theta}{(\theta+1)^2(\theta+2)}$.
Let $g(x)=\frac{x}{1-x}$ be defined on $(0,1)$, so that $g'$ exists and $g'(\theta)\ne 0$.
Then by Delta method, $$\sqrt{n}\left(g(\overline X)-g\left(\frac{\theta}{\theta+1}\right)\right)\stackrel{a}\sim N\left(0,\sigma^2(\theta)[g'(\theta)]^2\right)$$
Or, $$\sqrt{n}\left(\hat\theta-\theta\right)\stackrel{a}\sim N\left(0,\sigma^2(\theta)[g'(\theta)]^2\right)$$
Hence a suitable pivot is $$\frac{\sqrt{n}\left(\hat\theta-\theta\right)}{\sqrt{\sigma^2(\theta)(g'(\theta))^2}}\stackrel{a}\sim N\left(0,1\right)$$
From here, one can find a Wald-type asymptotic confidence interval for $\theta$ as mentioned in the answer by @BruceET. Finding the exact distribution of $\overline X$ or $\hat\theta$ for the confidence interval is almost too complicated.
A: 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 

