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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?

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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!

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  • $\begingroup$ If I do: $P(X_{(n)}/\theta \le a) = \alpha /2$ and $P(X_{(n)}/\theta \ge b)= \alpha /2$ I get $a=(\alpha /2)^{1/nk}$ and $b= (1 - \alpha /2)^{1/nk}$ and the confidence interval would be $(X_{(n)}/(1- \alpha /2)^{1/nk}, X_{(n)}/( \alpha /2)^{1/nk})$ Is this correct? $\endgroup$ – Alex Turner May 14 at 0:38
  • $\begingroup$ Works I think, and several other solutions! $\endgroup$ – mich95 May 14 at 22:34

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