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
    $\begingroup$ The central limit theorem won't help because it's about the distribution of a mean or a sum, not of a maximum. $\endgroup$ – Michael Hardy Sep 3 '12 at 12:26
  • $\begingroup$ I'm getting $(y_n,y_n/(1-\gamma)^{1/n})$. I'm guessing what you wrote was intended to be that. $\endgroup$ – Michael Hardy Sep 3 '12 at 12:32
  • $\begingroup$ yes, that was a typo. I apologize for the mistake. $\endgroup$ – adamG Sep 3 '12 at 12:42
  • $\begingroup$ In general asymptotic theory doesn't help here because the question requires an exact result. But there is asymptotic theory for the maximum. It is called Gnedenko's theorem and can be applied to the uniform distribution. $\endgroup$ – Michael R. Chernick Sep 3 '12 at 16:03

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 $c$.

  • $\begingroup$ Thanks Michael, I didn't think about using the given interval and work the reverse way. How would you reason to choose this confidence interval if it was not specified in advance? Could you give an intuitive answer? $\endgroup$ – adamG Sep 3 '12 at 13:03
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    $\begingroup$ This confidence interval has $100\gamma%$ within the interval, and $100(1-\gamma)%$ to the right of the interval. One could also construct a confidence interval with $(100(1-\gamma)/2)%$ in each tail. Or even $$100(1-\gamma)%$ to the left of the interval, with the right endpoint equal to $\infty$. The problem so far doesn't completely specify these things. But one could speak of a unique "confidence distribution", and find that. The second line of displayed $\TeX$ in my answer above is the first step in doing that. $\endgroup$ – Michael Hardy Sep 3 '12 at 13:18

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