Domain: Programming Languages, Compilers & Formal Methods

In the secretary problem, an adversary chooses $n$ values $x_1 \ge x_2 \ge \dots \ge x_n$ in $[0,1]$ that arrive online at independent uniform random times in $[0,1]$. An online algorithm must irrevocably select exactly one value upon seeing it (selecting $0$ if none is chosen). The regret of an algorithm on instance $\vec{x}$ is $x_1 - \mathbb{E}[\text{selected value}]$, and the worst-case regret is its supremum over all $\vec{x}$.

A pricing-curve algorithm is defined by a measurable threshold function $f:[0,1] \to [0,1]$ and selects the first arriving value $v$ at time $t$ with $v \ge f(t)$, or $0$ if none arrives.

Prove rigorously that the pricing-curve algorithm with $f(t)=1-t$ has worst-case regret at most $1/4$, and that every pricing-curve algorithm (any $f$) has worst-case regret at least $1/4$. The proofs must be fully self-contained, use only elementary probability, Lebesgue measure, and calculus, and hold for all $n \ge 1$.

Grading rubric:
criterion: Does the proof explicitly consider the harder variant in which $t_1 \sim U[0,1]$ is random but the adversary chooses all arrival times $t_2, \dots, t_n$ in advance?
weight: 10
criterion: Does the proof remove every $x_i < f(t_i)$ for $i \ge 2$ and state that this does not change the algorithm’s behavior or regret?
weight: 10
criterion: Does the proof define $j = \arg\min_{i \ge 2}$?
weight: 10
criterion: Does the proof reduce the instance to keeping only $x_1$ and $x_j$ while preserving behavior and regret?
weight: 10
criterion: Does the proof state that the algorithm selects $x_1$ if and only if $t_1 \in [1-x_1, t_j]$?
weight: 10
criterion: Does the proof write the exact regret expression $(x_1 - x_j)(1 - (t_j - (1 - x_1)))$?
weight: 10
criterion: Does the proof invoke the inequality $x_j \ge 1 - t_j$ to bound the regret expression?
weight: 10
criterion: Does the proof claim that the quadratic $(a - b)(1 - a + b)$ for $a \ge b \in [0,1]$ is at most $1/4$?
weight: 10
criterion: Does the lower-bound proof define $\alpha = \mu(\{t \in [0,1] : f(t) < 0.5\})$?
weight: 10
criterion: Does the lower-bound proof separately handle the case $\alpha > 1/2$ by arguing that, as $n \to \infty$, the algorithm accepts some value by time $0.5$ with probability approaching $1$ when $t_1 \ge 0.5$?
weight: 10
criterion: Does the lower-bound proof correctly handle the $\alpha \le 1/2$ case by setting $\alpha = \mu(\{t : f(t) < 0.5\})$?
weight: 10

Answer:
Upper bound ($f(t)=1-t$): Consider the harder variant where $t_1 \sim U[0,1]$ but $t_2, \dots, t_n$ are chosen adversarially in advance (any bound here implies the original). Remove every $x_i < f(t_i)$ for $i \ge 2$ (unchanged behavior). Let $j = \arg\min_{i \ge 2} t_i$ (add $x_2=0$ at $t_2=1$ if needed), then keep only $x_1, x_j$ (unchanged). The algorithm picks $x_1$ iff $t_1 \in [1-x_1, t_j]$, else $x_j$. Regret equals $(x_1 - x_j)(1 - (t_j - (1 - x_1)))$. Since $x_j \ge 1 - t_j$, this is at most $(x_1 - x_j)(1 - x_1 + x_j) \le 1/4$ (maximum of quadratic $(a-b)(1-a+b)$ on $a \ge b \in [0,1]$ is $1/4$).

Lower bound (any $f$): Let $\alpha = \mu(\{t : f(t) < 0.5\})$.
 * If $\alpha \le 1/2$, take $n=1$, $x_1=0.5$: expected value $\le \alpha \cdot 0.5 \le 0.25$, regret $\ge 0.25$.
 * If $\alpha > 1/2$, take $x_1=1$, $x_i=0.5$ for $i \ge 2$. With probability $1/2$, $t_1 \ge 0.5$. Also $\mu(\{t \in [0, 0.5] : f(t) \ge 0.5\}) \ge \alpha - 1/2 > 0$. As $n \to \infty$, probability of accepting some value by time $0.5$ approaches $1$. Thus regret approaches at least $0.5(1 - 0.5) = 0.25$.

In both cases worst-case regret $\ge 1/4$.