Anirban DasGupta is offering another three problems in the December 2025 IMS Bulletin issue.  He says, “You can send a solution to one, two, or ideally all three! We hope you will enjoy thinking about these problems.”

Puzzle 59.1
Suppose $X, Y, Z$ are iid standard normal. Find, without doing any calculations, the distribution of $\frac{X+YZ}{\sqrt{1+Z^2}}$.

Puzzle 59.2
Suppose $X_1, X_2, \cdots $ are iid $U[0,1]$. For $n \geq 2$, let $X_{n-1:n}$ be the second largest observation among $X_1, \cdots , X_n$. Find sequences $a_n, b_n$ and a nondegenerate distribution $G$ such that $\frac{X_{n-1:n} – a_n}{b_n}$ converges in distribution to a random variable with distribution $G$.

Puzzle 59.3
Consider the standard linear model $\bf{Y} = \bf{X}\,\bf{\beta} + \bf{\epsilon}$. Find an element $u$ in the column space of $X$ such that $||u-\bf{Y}||^2 < ||v – \bf{Y}||^2$ for all $v$ different from $u$ in the column space of $X$.

Solution to Puzzle 58

Congratulations to Samprit Chakraborty (ISI Kolkata), who sent detailed solutions to the problems from first principles, and also to Kavya Sharma (Birla Institute of Technology, Mesra), who sent a correct answer to the first problem. Anirban DasGupta explains:

Puzzle 58.1
By an easy application of the iterated expectation formula, the unconditional mean and variance of $Y$ are $\frac{n}{2}$ and $\frac{3}{4}\,n$. Therefore, immediately, from Markov’s inequality, $P(|Y-\frac{n}{2}| > n^{3/4})\to 0$.

Puzzle 58.2
Using the characterization that if $\bf{X} \sim N_p(\bf{0}, \Sigma)$, where $\Sigma $ is positive definite, then $\bf{X}’A\bf{X}$ has a chi-square distribution if and only if $A\Sigma\,A = A$ (see, e.g., Rao (1973)), a simple calculation shows that one of $a$ and $b$ must be $\frac{1}{4}$ and the other zero. For the second part of the problem, using the characterization for independence of a linear form and a quadratic form of a Gaussian vector (again, see Rao, 1973), it follows that for this part there is no nontrivial solution, other than the trivial solution $a = b = 0$.

Puzzle 58.3
The calculation of the expected number of triangles is trivial, and the expected value is ${n \choose 3}\,p^3 \to \frac{c^3}{6}$. The variance calculation is longer. By using the standard indicator variable method, one gets that the variance of the number of triangles is $c^3 + o(1)$. Now by using the second moment theorem for convergence to Poisson of sums of indicators (see e.g., Barbour, Holst and Janson, or Galambos and Simonelli, or Kingman), one obtains that the number of triangles has a limiting Poisson distribution with mean $\frac{c^3}{6}$.

A note of thanks for Puzzle 56
Thanks very much to Bernard Silverman for computing the numerical value of the expected overshoot for the first few values of $k$ in Puzzle 56. They are very, very close to 0.5.