I need some help with the following problem:
I want to estimate $n$ of $X_i \sim U(1, n)$ with a $90\%$ confidence level. What is given is the sample size with $10$ and the maximum of the sample with $100$. $X_1,…,X_{10}$ are iid.
When calculating the MLE for this distribution I got $max\{X_1,…,X_{10}\}$, which is $100$. Now I want to know $Pr[U_1 \le 100 \le U_2] \ge 0.9$. I don't know how to calculate the confidence interval. Using the central limit theorem would be wrong since the estimator is not a sum of random variables, would it?
Best Answer
What you need is a random variable depending on $n$ whose distribution does not depend on $n$. Also this random variable should only depend on the data through a sufficient statistic, which in this case is $X_{(10)}$.
One possible idea is to consider that $(X_i - 1) / (n - 1) \sim$ uniform$(0, 1)$ and so $- \log[(X_i - 1) / (n - 1)] \sim$ exponential$(1)$. Now $- \log[(X_{(10)} - 1) / (n - 1)]$ is the minimum of a random sample of $10$ unit exponential random variables, and therefore is itself exponential with rate $\lambda = 10$. So using the $0.9$ quantile of this distribution,
$$ \begin{align} P \left (0 \leq - \log \left [ \frac{X_{(10)} - 1}{n - 1} \right ] \leq 0.23 \right ) = 0.9. \end{align} $$
This event is the same as $X_{(10)} \leq n \leq (X_{(10)} - 1)e^{0.23} + 1$, and so these are the endpoints of a $90\%$ confidence interval for $n$.