Student members of IMS are invited to submit solutions to bulletin@imstat.org (subject “Student Puzzle Corner”). If correct, we’ll publish your name (and photo, if there’s space), and the answer, in the next issue. The Puzzle Editor is Anirban DasGupta. His decision is final.
Guest Student Puzzle editor Stanislav Volkov sets another problem.
Let ξ be a non-negative random variable with finite expectation, and c > 0 be a constant. Alice plays the following game. She draws a value from the distribution ξ and then she can either stop or draw again, paying a cost of c. She can play as many rounds as she wants, paying c every time, and her payoff will be the last drawn value when she decides to stop, minus the costs she paid. Alice’s goal is to maximise the expected net payoff of the game.
(a) Assuming the optimal strategy by Alice, what is her expected payoff? Solve the problem when ξ has a continuous distribution (you can make additional assumptions if needed).
(b) Find the explicit formula when ξ ~ U[0, 1].
(c) Find the explicit formula when ξ has a uniform discrete distribution with atoms on {1, 2, …, n}.
Solution to Puzzle 59
Hats off to student members Martín Alcalde Navarro (University of Zaragoza, Spain) and Samprit Chakraborty (Indian Statistical Institute, Kolkata), who both sent perfect solutions to all three of the Puzzle 59 problems. Well done, too, to Kavya Sharma (Birla Institute of Technology, Mesra) and Priyansh Pratap Singh (Birla Institute of Technology, Mesra) who sent correct solutions to two and one of the problems, respectively.
Puzzle 59 was first published in the December 2025 issue. Anirban DasGupta explains:
Puzzle 59.1
By conditioning on $Z$, one has that the conditional distribution of $\frac{X+YZ}{\sqrt{1+Z^2}}$ given $Z = z$ is $N(0,1)$, and thus free of $z$. So that is also the unconditional distribution.
Puzzle 59.2
$1-X_{n-1:n}$ has the same distribution as $\frac{Z_1 + Z_2}{Z_1 + Z_2 + \cdots + Z_{n+1}}$, where the $Z_i$ are i.i.d. standard exponential.
By the Weak Law of Large Numbers, $\frac{Z_1 + Z_2 + \cdots + Z_{n+1}}{n}$ converges in probability to $1$.
So, by Slutsky’s theorem, $n\,(1-X_{n-1:n})$ converges in distribution to the sum of two i.i.d. standard exponentials, which has the density $x\,e^{-x}$ with support $(0, \infty )$.
Puzzle 59.3
The unique $u$ minimizing $||u-\bf{Y}||^2$ over the column space of $X$ is by definition $\bf{X}\,\hat{\bf{\beta}}$, where $\hat{\bf{\beta}}$ is the unique LSE of $\bf{\beta}$ when the LSE is unique. The unique $u$ is $\bf{X}\,(\bf{X}’\,\bf{X})^{-1}\,\bf{X}’\,\bf{Y}$ when $\bf{X}’\,\bf{X}$ is nonsingular.