Student Puzzle Editor Anirban DasGupta returns to consideration of statistical problems in this issue. The problem falls in the class of irregular problems. He says, “Certainly all of you have seen inference problems about uniform distributions with one or more unknown endpoints. That is one of the simplest irregular inference problems.”

A tiger moves around a circular home territory of an unknown radius $\rho $, the circle being centered at a known point $(x_0, y_0)$. Paw prints of the tiger have been detected at points $(x_j, y_j), j = 1, 2, \cdots , n$.

(a) Write a model for the problem.

(b) Find the MLE $\hat{\rho}$ of $\rho $ under your model.

(c) Find, if it is possible, the exact distribution of the MLE in part (b).

(d) Find the asymptotic distribution of this MLE, i.e., find sequences $a_n, b_n$ and a non-degenerate distribution $G$ such that the distribution of $a_n\,(\hat{\rho} – b_n)$ converges in law to $G$.

Student members of IMS are invited to submit solutions to (with subject “Student Puzzle Corner”). Send your solution by May 1.


Solution to Puzzle 37

Guest Puzzler Stanislav Volkov explains his problem about a speeding random walk:

We will show that $\sum_n \mathbb{P}(S_n=0)<\infty$. Then, by the Borel-Cantelli lemma $S_n=0$ only for finitely many $n$s a.s. Next, note that if $\mathbb{P}(S_n=a\text{ i.o.})>0$ for some $a\in\mathbb{Z}_+$, then $\mathbb{P}(S_n=a-1\text{ i.o.})>0$ as well, since
\mathbb{P}(S_{n+2}=a-1\mid S_n=a)&=\mathbb{P}(X_{n+1}=n+1,X_{n+2}=-(n+2)\mid S_n=a)
\\ &=\mathbb{P}(X_{n+1}=n+1,X_{n+2}=-(n+2))=\frac14.
By induction we get a contradiction with the fact that $\mathbb{P}(S_n=0\text{ i.o.})=0$. The argument for $a\in\mathbb{Z}_-$ is similar, so we conclude that $\mathbb{P}(S_n=a\text{ i.o.})=0$ for all $a\in\mathbb{Z}$ and thus $S_n$ is transient in the sense that $|S_n|\to\infty $ a.s.

Now it remains to estimate $\mathbb{P}(S_n=0)=A_n/2^{n}$ where $A_n$ is the number of ways to put plus and minus signs in the sequence $1,2,3,\dots,n$ so that the result will be zero;
\pm 1\pm2 \pm\dots\pm n=0.
From the parity argument it follows that $A_n=0$ unless $n\mod 4=0$ or $1$; for $n=0,1,2,\dots$ we can also easily compute $ A=(1,0,0,2,2,0,0,8,14,0,0,\dots)$. An internet search, e.g., indicates that this is a sequence A063865, see~\cite{A}. The asymptotic formula for $A_n$ was given in~\cite{SUL}: $A_n\sim \sqrt{6/\pi}\cdot 2^n\cdot n^{-3/2}$. Hence
\sum_n \mathbb{P}(S_n=0)\sim \sum_n \frac{\sqrt{6/\pi}}{ n^{3/2}}<\infty $$ as required.