Anirban DasGupta says, “We are staying with our contest model introduced in the previous puzzles. Each correct answer receives 3 points, each incorrect answer receives -2 points, and each item left unanswered receives -1 point. The top three scorers will be recognized. You can answer just one of the two problems, 53I and 53II, but it will be great if you attempt both.”

Student IMS members: send your solutions to bulletin@imstat.org by December 1, 2024.

Puzzle 53.1

Suppose $X$ has a Poisson distribution with mean one. Prove that the distribution of $X$ is determined by its moments, i.e., if $Y$ is any real valued random variable such that $E(Y^n)=E(X^n)$ for all $n=1,2,\cdots$, then $Y$ has a Poisson distribution with mean one.

Puzzle 53.2

And now the contest problem. For each question, just say {\it True} or {\it False}, without the need to provide a proof. But answers with some explanations are especially welcome. Here are the items.

(a) If $X$ is a nonnegative random variable, and has a finite MGF everywhere, then $E(X^X)<\mathrm{\infty }$.

(b) For estimating the variance of a normal distribution with an unknown mean, the MLE of the variance is inadmissible under squared error loss function.

(c) Suppose $X_1,X_2,\cdots$ are iid Poisson with mean one. Let $S_n=\sum _{i=1}^n{X}_i,n\ge 1$. Let $N$ be the first $n$ for which $S_n>1$. Then $E(S_N–2)\le 1$.

(d) Suppose $X$ is uniformly distributed in the $p$-dimensional disk $\{x:||x|{|}^2={\rho }^2\}$. Treat $\rho$ as an unknown positive parameter. Then the bias of the MLE of $\rho$ converges to zero when $p\to \mathrm{\infty }$.

(e) Consider a $2\times 2$ random matrix in which all four entries are iid standard normal. Then the probability that the determinant of this matrix exceeds 1 is an irrational number.

Solution to Puzzle 52

Marco Dalla Pria sent a complete set of correct answers to Puzzle 52

Congratulations to Marco Dalla Pria, PhD student in Modeling and Data Science at the University of Turin, Italy (pictured here), for his extremely well done solutions. Puzzle editor Anirban DasGupta explains:

Puzzle 52.1

By a standard indicator variable argument, $E(T(\sigma ))=\frac{n(n-1)}{4}$.

Puzzle 52.2

(a) Obviously FALSE.

(b) TRUE. One can take $Y$ to be such that $Y$ and $\frac{1}{Y}$ have the same distribution, e.g., a standard Cauchy.

(c) Since chromatic polynomials are monic, the sum of its roots (by Vieta’s theorem) is the negative of the coefficient of $x^{n-1}$. It is easily seen that there are graphs for which this coefficient depends on $n$. So FALSE.

(d) TRUE; use the fact that $E(|X-c|)$ is minimized at any median and then use the Cauchy–Schwarz inequality.

(e) TRUE; this is a consequence of a well known theorem due to Poincaré.