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let $x_1,x_2, \ldots, x_n$ be a random sample from distribution under function distribution:

$$F(x)= \left( \frac{x}{\theta} \right)^\beta, \quad 0 \leq x \lt \theta.$$

Where $β$ is unknown but $θ$ is known. find a shortest confidence interval in level $(1-\alpha)$ for $\beta^2$.

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I give you a solution, but this is surely not the solution wanted by your teacher (as this looks like homework?).

I guess $F(x) = \left( {x\over\theta} \right)^\beta$ is the cumulative distribution function (CDF).

Let $Y = \max_i X_i$. Its CDF is $$G(y) = P(Y < y) = P\left(\cap_i X_i < y\right) =\prod_i P(X_i < y) = F(y)^n =\left( {y\over\theta} \right)^{n\beta}.$$

Let $q_1$ and $q_2$ be the quantiles of order $\alpha\over2$ and ${1-\alpha\over 2}$. $G(q_1) = {\alpha \over 2}$ leads to $n\beta (\log q_1 - \log \theta) = \log\left({\alpha\over 2}\right)$ hence $\log(q_1) = \log(\theta) + {1\over n\beta} \log\left({\alpha\over 2}\right)$, and $G(q_2) = 1-{\alpha\over 2}$ leads to $\log(q_2) = \log(\theta) + {1\over n\beta} \log\left(1-{\alpha\over2}\right)$.

From $P(\log q_1 < \log Y < \log q_2) = 1-\alpha$ you get $$P\left(\log\left({\alpha\over 2}\right) < n\beta(\log Y-\log\theta) < \log\left(1-{\alpha\over2}\right)\right) = 1-\alpha,$$ hence (as $\log Y < \log \theta$) $$ P\left({ \log\left({\alpha\over 2}\right) \over n (\log Y -\log\theta)} > \beta > {\log\left(1-{\alpha\over2}\right) \over n (\log Y -\log\theta)} \right) = 1-\alpha.$$

To get the shortest CI replace $q_1$ and $q_2$ by quantiles $\xi$ and $1-\alpha+\xi$ and seek $\xi$ that minimizes the length of the interval. To get a CI on $\beta^2$ take the square (everything is positive, so $x\mapsto x^2$ is monotonic).

PS If you get some other anwser by an other way, please share

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