The freshmen at East China Normal University has just received their student ID. Let the last three digits of a student ID be ABC, then A is the class he is in, whereas BC is his number in the class.

For example, a freshman whose student ID ends with 153 is the 53rd student in Class One. There are $\theta$ classes in total, and one class has $\phi$ students. Both the classes and students within a class are numbered from one. (Actually, the student numbers in each class are not exactly the same in real life, but to simplify the problem let's just pretend they are.)

One of the freshmen is wondering how many classmate he has, and as a senior statistics major I must help him! I requested other curious freshmen to share their student IDs in a private online group so that we might be able to find something from the data.

Formally, the samples are $X_i = (Y_i, Z_i)$ where $Y_i$ takes $1, 2, \dots, \theta$ and $Z_i$ takes $1, 2, \cdots, \phi$, both with equal probability. Furthermore, $Y_i$ and $Z_i$ are independent. Since the student IDs are assigned randomly, we are basically doing a simple random sampling without replacement.

An obvious answer is to apply the method of moments by doubling the average, i.e. $(2\bar{Y}, 2\bar{Z})$, but I wonder if that's the optimal approach. I have learned that the maximum of the samples is a sufficient statistic if the samples were independent, and the UMVUE can be constructed accordingly. However, "without replacement" means they are not, so this claim probably won't stand.

Another issue is that both $\theta$ and $\phi$ are unknown. I'm tempted to estimate them separately because of the independence of these two random variables, but I'm not 100% sure if that is correct given "without replacement".

To sum up, I want to find the UMVUE of $(\theta, \phi)$. If it doesn't exist, the MLE or something better (defined in terms of mean squared error) than $(2\bar{Y}, 2\bar{Z})$ will be considered as an answer.

  • $\begingroup$ Dudu it's not from a textbook! It's a real-world question I'm wondering. Plus, I have shown my attack at it in the last but one paragraph, didn't I? $\endgroup$
    – nalzok
    Aug 10, 2019 at 9:20
  • $\begingroup$ This question is more real-world and contains at least as much effort as the magic money tree problem. I don't understand why are you putting it on hold while that one is well-received. $\endgroup$
    – nalzok
    Aug 10, 2019 at 9:35
  • 2
    $\begingroup$ It's an exercise, quite a well-known one. But on re-reading, I think you have indicated where you're stuck , so I've re-opened it. Sorry. $\endgroup$ Aug 10, 2019 at 10:00
  • $\begingroup$ @Scortchi Thanks sir! $\endgroup$
    – nalzok
    Aug 10, 2019 at 10:04
  • 1
    $\begingroup$ This is similar to the famous German tank problem. $\endgroup$
    – BruceET
    Aug 11, 2019 at 1:25


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