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

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?

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!

• 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? – Alex Turner May 14 at 0:38
• Works I think, and several other solutions! – mich95 May 14 at 22:34