Deadline: November 1, 2022. Student members of IMS are invited to submit solutions to bulletin@imstat.org (with subject “Student Puzzle Corner”). The names of student members who submit correct solutions to either or both of these puzzles, and the answer, will be published in the issue following the deadline. The Puzzle Editor is Anirban DasGupta. His decision is final.

Student Puzzle Editor Anirban DasGupta says, “This time we pose two standard, but fun, problems in traditional statistics and probability. You are most welcome to send answers to one or both, we will still mention your name. The first is a problem on discovery. I hope you have some fun thinking about it. The second involves randomizing the number of total objects in a multinomial experiment.”

Puzzle 41.1:

Suppose X ~ U[0, f(θ)] for some suitable function f, and θ is a real parameter taking values in the entire real line. Discover a prior distribution on θ and a function f such that the posterior mean of θ given X = x is identically equal to zero for all x.

Puzzle 41.2:
(This problem can be solved analytically. If you are unable to find an analytic solution, send us a careful simulation-based answer. If the simulation-based answer is satisfactory, we will still mention your name.)

N balls are distributed, one by one, completely at random, independently into one of 300 urns, where N has a Poisson distribution with mean 1000. Find explicitly the most likely number of urns that will remain empty

## Solution to Puzzle 40

Problem Corner Editor Anirban DasGupta explains the previous puzzle:

The posterior distribution is uniform on the support $\mathcal{S}$, where $\mathcal{S}$ depends on whether the responses to the two questions were Yes-Yes(YY), Yes-No(YN), No-Yes(NY) or No-No(NN).

When the responses are YY, $\mathcal{S} = \{11, 13, 17, 31, 37, 71, 73, 79, 97\}$. The median of a CDF $F$ being $\inf\{x: F(x) \geq \frac{1}{2}\}$, in this case the posterior median is 37.

When the responses are YN, $\mathcal{S} = \{19, 23, 29, 41, 43, 47, 53, 59, 61, 67, 83, 89\}$. In this case the posterior median is 47.

When the responses are NY, $\mathcal{S} = \{14, 16, 32, 34, 35, 38, 74, 76, 91, 92, 95, 98\}$, and the posterior median is 38.

In the case that the responses are NN, the posterior median is 57.

The unknown number being an integer, the posterior median is easier to interpret as it is by definition an integer, while the posterior mean will usually not be an integer.