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?
1 Answer
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 $c$.
