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A random variable is uniformly distributed over $(0,\theta)$. The maximum of a random sample of $n$, call $y_n$ is sufficient for $\theta$ and it is also the maximum likelihood estimator. Show also that a $100\gamma\%$ confidence interval for $\theta$ is $(y_n, y_n /(1 − \gamma )^{1/n})$.

Could anyone tell me how to deal with this problem? Do I have to use the central limit theorem?

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1 Answer

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You need to show that$$ \Pr\left(y_n<\theta<\frac{y_n}{(1-\gamma)^{1/n}}\right) = \gamma. $$Since $y_n$ is necessarily always less than $\theta$, this probability is the same as\begin{align} & \Pr\left(\theta<\frac{y_n}{(1-\gamma)^{1/n}}\right) \\ = {} & \Pr\left( \theta(1-\gamma)^{1/n} < y_n\right) \\ = {} & 1-\Pr\left( y_n < \theta(1-\gamma)^{1/n} \right). \end{align}

Notice that\begin{align} & \Pr(y_n < c) = \Pr(\text{All $n$ observations}<c) \\ = {} & \Big( \Pr(\text{A single observation}<c) \Big)^n = \left( \frac c\theta \right)^n. \end{align}Apply this last sequence of equalities with $\theta(1-\gamma)^{1/n}$ in place of $c$.

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