Student Puzzle Editor Anirban DasGupta poses another two problems, and says, “For the first time in our problem corner, we will consider a problem that is hard to solve analytically, and we will ask you to write an algorithm to solve the problem as accurately as you can by simulation. The second one is a traditional problem.” Send us your solution, to either or both. The deadline is May 1, 2023.

Puzzle 44.1.

As you were making coffee in your kitchen on waking up, you noticed that a huge centipede has curled itself up on your kitchen floor and its head and tail are joined together. Petrified by it, you took a coffee mug with a circular mouth and tried to cover the centipede. Assume that the centipede is eight inches long. A coffee mug of what internal diameter can cover the centipede completely? Give the smallest diameter suggested by your algorithm. Send your algorithm with your answer.

Puzzle 44.2.

Suppose $X_1, X_2, \cdots$ are i.i.d. standard Cauchy. For $n \geq 2$, denote by $T_n$ the proportion of pairs $i < j \leq n$ such that $X_i + X_j \geq 0$.

(i) Does $T_n$ converge in probability to some constant $c$? If so, to what?

(ii) Does there exist a sequence $\{a_n\}$ and a nontrivial distribution $G$ such that $a_n\,(T_n – c) \stackrel {\mathcal{L}} \Rightarrow G$? If so, identify $\{a_n\}$ and $G$.

## Solution to Puzzle 43

Thank you to Alberto Bordino (University of Warwick), Soham Bonnerjee (University of Chicago), and Abhinandan Dalal (University of Pennsylvania) for sending their solutions. Anirban DasGupta explains:

Puzzle 43.1

If you take a large value of $n$, and plot the pairs $(k, \bar{X}_k)$ for $k = 1, 2, \cdots , n$ for various sets of simulated standard Cauchy variables, you should see that the paths oscillate and that they revisit neighborhoods of a given level $b$ within the range of the path. We cannot show many such plots here due to reasons of space. We can frame this rigorously. We can show by a simple calculation that given any $b$, and $0 < \epsilon \leq 1$, $\sum_n\,\frac{1}{n}\, P(|\bar{X}_n-b| \leq \epsilon) = \infty$. All we need to use is that for each $n, \bar{X}_n$ is distributed as a standard Cauchy. This is sufficient to guarantee that with probability one, $b$ is an accumulation point for the sequence $\{\bar{X}_n\}$. If we choose $b = 0$, then we can conclude that with probability one, $\limsup\,|\bar{X}_n – e^n| = \infty$. You will notice that $e^n$ is nothing special; we can choose basically any sequence going to $\infty$.

Puzzle 43.2

We will use the standard result that a unique Bayes rule with a finite Bayes risk is admissible. The uniqueness of a Bayes rule follows from strict convexity of $L(\theta , a) = (\theta – a)^2$ in $a$; here $\theta$ stands for $N$. Consider now the proper prior with mass function $\frac{c}{N^4}, N = 1, 2, \cdots$. Denoting the digamma function by $\psi (x)$, the posterior mean $E(N\,|X)$ equals $\delta _B(X) = \frac{-4\,\psi ^{(3)}\,(X)} {\psi ^{(4)}\,(X)}$. $\delta _B(X)$ has a quadratically bounded risk function and so a finite Bayes risk for the prior $\frac{c}{N^4}$. Thus $\frac{-4\,\psi ^{(3)}\,(X)}{\psi ^{(4)}\,(X)}$ is one admissible estimator of $N$.