Finding confidence interval for $\frac{kx^{k-1}}{\theta^k}$

Question

Let $X_1,\ldots, X_n$ are i.i.d. random variables such that: $$f(x;\theta)=\frac{kx^{k-1}}{\theta^k}, x\in (0,\theta)$$ where $\theta \gt 0 $ and $k$ is a positive integer. Find a $100(1-\alpha)% $% confidence interval for $\theta$.

I couldn't find a pivot yet, any ideas?

Answer 1

The cdf of $X_1, \ldots,X_n$ is $F$ defined by $F(x)=(x/\theta)^{k}$ for $0 \le x \le \theta$. For $x<0$, $F(x)=0$ and for $x>\theta$, $F(x)=1$. Let $X_{(n)}=\max(X_1, \ldots,X_n)$. For $0 \le y \le \theta$. $\Pr(X_{(n)} \le y)=F(x)^{n}=(x/\theta)^{nk}$. Therefore, for all $0 \le y \le 1$, $\Pr(X_{(n)}/ \theta \le y)= y^{nk}$. Consider the interval $(0,b=X_{(n)}/\alpha^{1/nk})$.

$\Pr(\theta \le b)=\Pr(1/ \theta \ge 1/b)=\Pr(X_{(n)}/\theta \ge \alpha^{1/nk})=1-\alpha$. There are other confidence intervals!