Find $1-\alpha$ confidence interval for $\theta$ based on $X_1,...,X_n \sim \frac{2x}{\theta^2}$ (iid)

Question

Find $1-\alpha$ confidence interval for $\theta$ based on $X_1,...,X_n \sim \frac{2x}{\theta^2}1(0<x<\theta)$ (iid)

I am unfamiliar with confidence intervals. Does $1-\alpha$ confidence interval mean $(1-\alpha)100$% mean the random variable will fall into an interval for $\theta$?

Am I to calculate $\displaystyle \int_a^b \frac{2x}{\theta^2}dx = 1-\alpha$ or do I need the likelihood function here and to estimate for $\theta$ prior?

Answer 1

Let $S = \sum X_i$ with $\mathbb{E}(S) = \dfrac{2n \theta}{3}$ and $\operatorname{Var}(S) = \dfrac{5n\theta^2}{18}$.

So we have $\mathbb{P}(|S - 2n\theta/3| \ge \epsilon n) \le \dfrac{5\theta^2}{18\epsilon^2n} \iff \mathbb{P}(\dfrac{2\theta}{3} -\epsilon < \dfrac{1}{n}S < \dfrac{2\theta}{3} + \epsilon) \ge 1- \dfrac{5\theta^2}{18\epsilon^2n} = 1 - \alpha$.

Now express $\epsilon$ in terms of $n,\theta, \alpha$. And represent inequality inside of probability measure as $\mathbb{P}(f(S,n,\alpha) \le \theta \le g(S,n,\alpha)) \ge 1 - \alpha$