3.1. Results of [Reference Samuel-Cahn26]

It is shown in [Reference Samuel-Cahn26, Theorem 2] that for the class ${\mathscr T}_{\rm s}$ of single-threshold rules without randomization we have

(21) \begin{equation} \lim_{n\to\infty} \mathcal{R}^*_n({\mathscr T}_{\rm s};\, \mathcal{F}_{[0,\infty)}) = \frac{1}{2}.\end{equation}

In the proof of (21) [Reference Samuel-Cahn26] considers a discrete least favorable distribution $F_*$ having three atoms at 0, $a\in (0,1)$ , and 1 with respective probabilities $1-(b+c)/n$ , $c/n$ , and $b/n$ ; here, $b>0$ , $c>0$ , and $b+c<n$ . Thus, the distribution $F_*$ is

\begin{align*} F_*(x) = \begin{cases} 1 - \dfrac{b+c}{n}, & 0\leq x < a, \\ 1 - \dfrac{b}{n}, & a\leq x< 1, \\ 1, & x\geq 1. \end{cases} \end{align*}

In this example a straightforward calculation yields

\begin{equation*} M_n(F_*) = a\bigg[\bigg(1-\frac{b}{n}\bigg)^n-\bigg(1-\frac{b+c}{n}\bigg)^n\bigg] + 1-\bigg(1-\frac{b}{n}\bigg)^n.\end{equation*}

For the stopping rule $\tau_0(\theta)$ in (8) there are three possible thresholds: $\theta=0$ , $\theta=a$ , and $\theta=1$ . In view of (14) with $p=0$ , the corresponding values are given by

\begin{align*} \mathbb E X_{\tau_0(0)} & = \bigg[1-\bigg(1-\frac{b+c}{n}\bigg)^{n-1}\bigg]\frac{ac+b}{c+b} + \bigg(1-\frac{b+c}{n}\bigg)^{n-1}\frac{ac+b}{n}, \\ \mathbb E X_{\tau_0(a)} & = 1 - \bigg(1-\frac{b}{n}\bigg)^{n-1} + \bigg(1-\frac{b}{n}\bigg)^{n-1}\frac{ac+b}{n}, \\ \mathbb E X_{\tau_0(1)} & = \mathbb E(X_1) = \frac{ac+b}{n}.\end{align*}

Now, put $a= a_n= 1/n$ , $b=b_n= 1/n$ , and $c=c_n=\sqrt{n}$ , and let $F_n$ stand for the distribution $F_*$ with parameters $a_n$ , $b_n$ , and $c_n$ . Then, asymptotically, $M_n(F_n)\sim{2}/{n}$ , $\mathbb E X_{\tau_0(0)}\sim{1}/{n}$ , $\mathbb E X_{\tau_0(a_n)}\sim{1}/{n}$ , and $\mathbb E X_{\tau_0(1)}\sim{1}/{n^{3/2}}$ , where $v_n\sim w_n$ means that $\lim_{n\to\infty}(v_n/w_n)=1$ . Thus, the limit $\frac12$ is achieved asymptotically for $\mathcal{R}_n({\mathscr T}_{\rm s};\, F_n)$ .

It is instructive to calculate the competitive ratio of the best stopping rule from ${\mathscr T}_{\rm s,r}$ (the best rule with randomization) on the sequence of distributions $\{F_n\}$ defined above. Consider the stopping rule $\tau_p(\theta)$ defined in (11) with

\begin{align*} \theta=a_n=\frac{1}{n}, \qquad p=p_n= \frac{F_n(a_n)-(1-{1}/{n})}{F_n(a_n)-F_n(a_n-)}=\frac{1-b_n}{c_n}=\frac{1-{1}/{n}}{\sqrt{n}}. \end{align*}

With this choice of parameters, in view of (13),

\begin{align*} \mathbb E X_{\tau_{p_n}(a_n)} & = \bigg[1-\bigg(1-\frac{1}{n}\bigg)^n\bigg]\bigg[a_n+ n\int_{a_n}^\infty (1-F_n(x))\,\mathrm{d} x\bigg] \\ & \quad + \bigg(1-\frac{1}{n}\bigg)^{n-1}\bigg[\int_{[0,a_n]}x\,\mathrm{d} F_n(x) - p_na_n\Delta(a_n)\bigg] \\ & = \bigg[1-\bigg(1-\frac{1}{n}\bigg)^n\bigg]\big[a_n + b_n(1-a_n)\big] + \bigg(1-\frac{1}{n}\bigg)^{n-1}\bigg[\frac{a_n c_n}{n} - \frac{a_n(1-b_n)}{n}\bigg] \\ & = \bigg[1-\bigg(1-\frac{1}{n}\bigg)^n\bigg]\bigg(\frac{2}{n}- \frac{1}{n^2}\bigg) + \bigg(1-\frac{1}{n}\bigg)^{n-1}\frac{1}{n\sqrt{n}\,}\bigg[1-\frac{1}{\sqrt{n}\,}+\frac{1}{n\sqrt{n}\,}\bigg].\end{align*}

Since $M_n(F_n)\sim 2/n$ , on the sequence of distributions $\{F_n\}$ we obtain

\begin{align*} \lim_{n\to\infty}\frac{\mathbb E X_{\tau_{p_n}(a_n)}}{M_n(F_n)} = 1-\frac{1}{{\mathrm{e}}}. \end{align*}

Thus, $\lim_{n\to\infty}\mathcal{R}_n^*({\mathscr T}_{\rm s, r};\, \mathcal{F}_{[0,\infty)}) = 1-{1}/{{\mathrm{e}}}$ , i.e. the prophet inequality (16) is asymptotically sharp for the single-threshold stopping rules with randomization, and it is achieved on the same sequence of worst-case distributions as the prophet inequality (21).

3.2. Online auctions literature

Recently there has been renewed interest in prophet inequalities due to relations between optimal stopping theory and the design of online auction mechanisms. In one such scenario, a seller has a single item to sell to n customers, where the ith customer, $i=1, \ldots, n$ , has a random valuation $X_i$ for the item. The valuations of different customers are assumed to be independent. The customers arrive sequentially, and they are presented with a price (perhaps, customer-dependent). The customer purchases the item if his/her valuation exceeds the quoted price. The goal of the seller is to design a pricing policy to maximize the revenue. This setting is often referred to as a posted-price auction.

The prophet secretary model in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9, Reference Esfandiari, Hajiaghayi, Liaghat and Monemizade10] assumes that the valuations $X_1, \ldots, X_n$ are independent random variables with distributions $F_1, \ldots, F_n$ , and customers arrive sequentially in a random order. This implies that the realized sequence of valuations is $X_{\pi_1}, \ldots, X_{\pi_n}$ , where $\pi=(\pi_1, \ldots, \pi_n)$ is a random permutation of $\{1, \ldots, n\}$ , independent of $X_1, \ldots, X_n$ . Various pricing policies were studied in this setting and corresponding prophet inequalities were derived. We refer to [Reference Correa, Foncea, Hoeksma, Oosterwijk and Vredeveld5] for a review of results in this area.

It was shown in [Reference Correa, Foncea, Hoeksma, Oosterwijk and Vredeveld6] that, for continuous distribution functions, $\mathcal{R}_n({\mathscr T}_{\rm s};\, F) \geq 1-{1}/{{\mathrm{e}}}$ for all $F\in \mathcal{C}_{[0,\infty)}$ . A similar result was obtained in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9] for the class of all distributions $\mathcal{F}_{[0,\infty)}$ , but that paper considered stopping rules with random breaking of ties when the observed random variable $X_t$ is exactly equal to a chosen threshold. Theorem 21 in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9] claims that the prophet inequality $\mathcal{R}_n^*({\mathscr T}_{\rm s};\, \mathcal{F}_{[0,\infty)})\geq 1-{\mathrm{e}}^{-1}$ is asymptotically sharp in the i.i.d. setting. In the proof of this statement the authors first consider the distribution $F_*$ having two atoms at $n/({\mathrm{e}}-1)$ and $({\mathrm{e}}-2)/({\mathrm{e}}-1)$ with respective probabilities $1/n^2$ and $1-1/n^2$ . For this distribution there are two possible choices of the threshold: $\theta_1=0$ and $\theta_2= ({\mathrm{e}}-2)/({\mathrm{e}}-1)$ . Then [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9] showed that the best single-threshold stopping rule achieves the competitive ratio at most $0.58$ . Furthermore, the authors argued that this bound can be improved by randomization to $1-{\mathrm{e}}^{-1}$ .

For the distribution $F_*$ we have

\begin{equation*} M_n(F_*) = \frac{{\mathrm{e}}-2}{{\mathrm{e}}-1}\bigg(1-\frac{1}{n^2}\bigg)^n + \frac{n}{{\mathrm{e}}-1}\bigg[1 - \bigg(1-\frac{1}{n^2}\bigg)^n\bigg].\end{equation*}

Using (14) for the single-threshold rule with no randomization ( $p=0$ ), we obtain

\begin{align*} \mathbb E X_{\tau_0(\theta_1)} & = \frac{{\mathrm{e}}-2}{{\mathrm{e}}-1}\bigg(1-\frac{1}{n^2}\bigg) + \frac{1}{n({\mathrm{e}}-1)}, \\ \mathbb E X_{\tau_0(\theta_2)} & = \frac{n}{{\mathrm{e}}-1}\bigg[1-\bigg(1-\frac{1}{n^2}\bigg)^{n-1}\bigg] + \bigg[\frac{{\mathrm{e}}-2}{{\mathrm{e}}-1}\bigg(1-\frac{1}{n^2}\bigg) + \frac{1}{n({\mathrm{e}}-1)}\bigg] \bigg(1-\frac{1}{n^2}\bigg)^{n-1}.\end{align*}

These formulas imply that

\begin{align*} \lim_{n\to\infty} M_n(F_*)=1, \qquad \lim_{n\to\infty} \mathbb E X_{\tau_0(\theta_1)} = \frac{{\mathrm{e}}-2}{{\mathrm{e}}-1}, \qquad \lim_{n\to\infty} \mathbb E X_{\tau_0(\theta_2)} = 1, \end{align*}

so that for the specified distribution $\lim_{n\to\infty} \mathcal{R}_n({\mathscr T}_{\rm s};\, F_*)=1$ . Thus, for the distribution $F_*$ given in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9], the single-threshold stopping rules with no randomization (from ${\mathscr T}_{\rm s}$ ) achieve a factor of one as $n\to\infty$ . In fact, for any distribution with two atoms the single-threshold rules achieve the factor one; this was pointed out in [Reference Samuel-Cahn26, Remark 1].

The difference between the above calculation and the conclusion in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9] stems from the fact that, in spite of the close connection between optimal stopping and online auctions, the algorithms considered in the latter area (see, e.g., [Reference Correa, Foncea, Hoeksma, Oosterwijk and Vredeveld6, Reference Ehsani, Hajiaghayi, Kesselheim and Singla9, Reference Esfandiari, Hajiaghayi, Liaghat and Monemizade10], etc.) are not completely equivalent to those in optimal stopping problems (see, e.g., [Reference Assaf, Goldstein and Samuel-Cahn1–Reference Chow, Robbins and Siegmund4, Reference Hill and Kertz15, Reference Hill and Kertz17, Reference Samuel-Cahn26], etc.). In particular, the decision rules in [Reference Correa, Foncea, Hoeksma, Oosterwijk and Vredeveld6, Reference Ehsani, Hajiaghayi, Kesselheim and Singla9, Reference Esfandiari, Hajiaghayi, Liaghat and Monemizade10] (which are fully in line with various other papers in this strand of literature) are left undefined on the event where the random variables $X_1, \ldots, X_n$ do not exceed the corresponding thresholds (see, e.g., Algorithm Prophet Secretary in [Reference Esfandiari, Hajiaghayi, Liaghat and Monemizade10, p. 1687] and [Reference Correa, Foncea, Hoeksma, Oosterwijk and Vredeveld6, Algorithm 1]). Strictly speaking, the considered decision rules are not stopping times because they do not take values in the stopping set $\{1, \ldots, n\}$ . Effectively, the implication of this fact is that if $X_1, \ldots, X_n$ are below the respective thresholds then the reward obtained is equal to zero. This formally corresponds to computation of the reward of a single-threshold rule without the last term on the right-hand side of (13). This distinction affects statements about sharpness of prophet inequalities derived in the online auctions literature, and the comparison across literatures should take this into consideration. In contrast, stopping times are naturally defined to be equal to the terminal value $X_n$ of the sequence at the end of the horizon n on the event when all observations are less than the corresponding thresholds, i.e. the last observation is selected. All this implies that sharpness of the inequality (16) for single-threshold stopping times does not follow from the results in [Reference Ehsani, Hajiaghayi, Kesselheim and Singla9].

In the next section we demonstrate that derivation of sharp non-asymptotic prophet inequalities for the stopping rules from the class ${\mathscr T}_{\rm s,r}$ is equivalent to the solution of an infinite two-person zero-sum game on the unit square with particular payoff kernels. The value of this game provides a sharp constant in the prophet inequality, while the optimal solution yields the corresponding least favorable distribution and optimal single-threshold stopping rule.

5.1. Ratio-type prophet inequalities

To compute the sharp constants in ratio-type prophet inequalities we solve the game (31) as follows. Let $r_i^{{\top}}$ denote the ith row of matrix $R_N$ defined in (29), i.e. $R_N=[r_{1}^{{\!\top}};\,r_{2}^{{\!\top}};\,\ldots;\,r_{N-1}^{{\!\top}}]$ . Then the problem (31) is equivalent to

\begin{align*} \min_\mu \max_{i=1,\ldots,N-1} r_i^{{\top}}\mu \quad {\rm such\ that}\quad \mu^{{\!\top}}\textbf{1}=1,\quad \mu \geq 0,\end{align*}

which is directly cast as the linear optimization problem

(R) \begin{equation} \min_{\mu,t} t \quad {\rm such\ that}\quad r_i^{{\top}}\mu \leq t, \ i=1, \ldots, N-1, \quad \mu^{{\!\top}}\textbf{1} = 1, \quad \mu\geq 0.\end{equation}

If $(\mu^*, t^*)$ is the optimal solution of (R) then the value of the matrix game (31) is $\mathcal{R}_{n, N}^*=t^*$ , and a least favorable distribution has at most N atoms with $(i/N)$ -quantiles $\{u^*_i, i=1, \ldots, N\}$ which are determined by the relationship

\begin{align*}u_1^*=0, \qquad u_{i+1}^*=\sum_{j=1}^i\frac{\mu_j^*}{1-({j}/{N})^n}, \quad i=1,\ldots, N-1.\end{align*}

Then the optimal stopping rule is associated with the threshold $\theta^*=u_{i^*+1}$ , where $i^*$ satisfies $r_{i_*}^{{\!\top}} \mu^*=t^*$ .

We solve problem (R) for different values of horizon $n\in\{10$ , 25, 50, 100, 300, 500, 1000, 2000, $3000\}$ with fixed discretization parameter $N=13\,500$ using the Mosek optimization solver [25]. This choice of N is dictated by computer power limitations (the computations were performed on a laptop with 32 GB RAM and an 11th-generation i7 processor). The results are reported in the second and third columns of Table 1. The second column presents the values $\mathcal{R}_{n, N}^*$ , while the third column gives the lower bound $\underline{\mathcal{R}}_{n, N}^*$ , which is the maximum of the lower bound of Theorem 4 and $1- (1-{1}/{n})^n$ , given by the prophet inequality (16):

\begin{equation*} \underline{\mathcal{R}}_{n, N}^* = \max\big\{1-(1-{1}/{n})^n,\,\mathcal{R}_{n, N}^* - \Delta_R\big\}.\end{equation*}

Table 1.

Optimal values $\mathcal{R}^*_{n, N}$ and $\mathcal{A}_{n, N}^*$ of problems (R) and (A), and bounds on $\mathcal{R}^*_n({\mathscr T}_{\rm s, r};\,\mathcal{F}_{[0,\infty)})$ and $\mathcal{A}_{n}^*({\mathscr T}_{\rm s,r};\, \mathcal{F}_{[0,1]})$ for different values of n, where $N=13\,500$ for problem (R) and $N=13\,000$ for problem (A).

Table 1 Long description

The table presents data for different values of n, specifically 10, 25, 50, 75, 100, 200, 500, 1000, 2000, and 3000. It includes columns for R sub n, N, R sub n, N asterisk, A sub n, N asterisk, and A sub n, N asterisk plus delta A. Each row lists the corresponding values for these columns. Row 1: n, 10; R sub n, N, 0.6698; R sub n, N asterisk, 0.669; A sub n, N asterisk, 0.1395; A sub n, N asterisk plus delta A, 0.140. Row 2: n, 25; R sub n, N, 0.6540; R sub n, N asterisk, 0.651; A sub n, N asterisk, 0.1572; A sub n, N asterisk plus delta A, 0.158. Row 3: n, 50; R sub n, N, 0.6458; R sub n, N asterisk, 0.641; A sub n, N asterisk, 0.1644; A sub n, N asterisk plus delta A, 0.166. Row 4: n, 75; R sub n, N, 0.6427; R sub n, N asterisk, 0.636; A sub n, N asterisk, 0.1671; A sub n, N asterisk plus delta A, 0.170. Row 5: n, 100; R sub n, N, 0.6411; R sub n, N asterisk, 0.634; A sub n, N asterisk, 0.1699; A sub n, N asterisk plus delta A, 0.172. Row 6: n, 200; R sub n, N, 0.6392; R sub n, N asterisk, 0.633; A sub n, N asterisk, 0.1708; A sub n, N asterisk plus delta A, 0.178. Row 7: n, 500; R sub n, N, 0.6409; R sub n, N asterisk, 0.632; A sub n, N asterisk, 0.1723; A sub n, N asterisk plus delta A, 0.191. Row 8: n, 1000; R sub n, N, 0.6468; R sub n, N asterisk, 0.632; A sub n, N asterisk, 0.1729; A sub n, N asterisk plus delta A, 0.211. Row 9: n, 2000; R sub n, N, 0.6595; R sub n, N asterisk, 0.632; A sub n, N asterisk, 0.1733; A sub n, N asterisk plus delta A, 0.250. Row 10: n, 3000; R sub n, N, 0.6726; R sub n, N asterisk, 0.632; A sub n, N asterisk, 0.1731; A sub n, N asterisk plus delta A, 0.288.

The accuracy of the presented bounds deteriorates as n grows; computation of more accurate bounds requires selecting much bigger values of N. For instance, in order to compute the sharp constant in the prophet inequality for $n=100$ with accuracy $10^{-3}$ , we need $N > 1.25\times 10^5$ , which leads to a linear program with non-sparse matrices of several hundred thousand dimensions.

From the numbers presented in the second and third columns of Table 1, some very useful conclusions on sharp constants can be reliably drawn. For instance, if $n=25$ then $0.651 \leq \mathcal{R}^*_{25}\big({\mathscr T}_{\rm s, r};\, \mathcal{F}_{[0,\infty)}\big) \leq 0.654$ .

5.2. Difference-type prophet inequalities

The sharp constants for the difference-type prophet inequality are obtained as solutions of the finite matrix game (32) with payoff matrix $A_N$ defined in (30). If $a_i^{{\top}}$ denotes the ith row of matrix $A_N$ , i.e. $A_N=[a_{1}^{{\!\top}};\, a_{2}^{{\!\top}};\,\ldots;\,a_{N-1}^{{\!\top}}]$ , then (32) is equivalent to the linear program

(A) \begin{equation} \max_{\mu,t} t \quad {\rm such\ that} \quad a_i^{{\top}}\,\mu \geq t,\ i=1, \ldots, N-1, \quad \mu^{{\!\top}}\textbf{1}= 1, \quad \mu\geq 0.\end{equation}

If $(\mu^*, t^*)$ is the optimal solution of (A) then $\mathcal{A}_{n, N}^*=t^*$ , and the least favorable distribution $F_*$ is the discrete distribution with $i/N$ -quantiles $u^*_i$ determined by $u_1^*=0$ , $u^*_{i+1} = \sum_{j=1}^i \mu_j^*$ , $i=1,\ldots, N-1$ .

Problem (A) is solved for $n\in\{10$ , 25, 50, 100, 300, 500, 1000, 2000, $3000\}$ and $N=13\,000$ . The last two columns in Table 1 contain the optimal value $\mathcal{A}_{n, N}^*$ and $\mathcal{A}_{n, N}^*+ \Delta_A$ , which are the lower and upper bounds on the sharp constant $\mathcal{A}_n^*\big({\mathscr T}_{\rm s, r};\, \mathcal{F}_{[0,1]}\big)$ in the difference-type prophet inequality. To the best of our knowledge, the existing literature does not contain prophet inequalities for single-threshold stopping rules in the i.i.d. setting. However, the numbers in the last two columns of Table 1 can be compared to the bounds on the values $\mathcal{A}_n^*({\mathscr T}_{\rm all};\, \mathcal{F}_{[0,1]})$ reported in [Reference Hill and Kertz16] for the class of all possible stopping rules: $\mathcal{A}_n^*\big({\mathscr T}_{\rm all};\, \mathcal{F}_{[0,1]}\big) \leq b_n$ with $b_{10\,000}\approx 0.111$ . The inferiority of the optimal single-threshold stopping rule in comparison with the optimal stopping rule is rather moderate; for example, for $n=200$ the regret increases from $0.11$ to $0.17$ .

6.1. Difference-type prophet inequality for distributions with bounded variance

Let $\mathcal{F}_\sigma$ be the class of distribution functions on $[0,\infty)$ with variance bounded by a constant $\sigma^2<\infty$ ,

\begin{align*} \mathcal{F}_\sigma \,:\!=\, \bigg\{F\in\mathcal{F}_{[0,\infty)}\colon \operatorname{var}(X_i) = \int_{(0,1)}\int_{(0,1)}[x\wedge y-xy]\,\mathrm{d} F^{\leftarrow}(x)\,\mathrm{d} F^{\leftarrow}(y) \leq \sigma^2\bigg\}.\end{align*}

If $F\in \mathcal{F}_\sigma \cap \mathcal{D}_N$ then the condition $\operatorname{var}(X_i)\leq \sigma^2$ is equivalent to

\begin{align*} \sum_{i=1}^{N-1}\sum_{j=1}^{N-1}\bigg(\frac{i}{N}\wedge\frac{j}{N} - \frac{ij}{N^2}\bigg) (u_{i+1}-u_{i})(u_{j+1}-u_{j}) = v^{{\top}}Qv,\end{align*}

where $v=[v_1;\ldots;\,v_{N-1}]$ with $v_i=u_{i+1}-u_{i}$ , and $Q_{ij}= ({i \wedge j})/{N} - {ij}/{N^2}$ , $i,j=1, \ldots, N-1$ .

Problem (35) associated with the ratio-type prophet inequality is scale invariant in the sense that its optimal solution is defined up to a scale parameter. Therefore the optimal value does not change when an upper bound on the variance is added. The same is true when any one-sided linear constraint on v is imposed. Therefore we discuss only the difference-type prophet inequality for the family $\mathcal{F}_\sigma\cap \mathcal{D}_N$ .

In this situation, in problem (36) the constraint $\textbf{1}^{{\!\top}}v=1$ should be replaced by the quadratic constraint $v^{{\top}}Qv\leq \sigma^2$ :

\begin{equation*} \max_v\min_{i=1,\ldots,N-1}[d^{{\top}}v - b_i^{{\top}}v] \quad {\rm such\ that} \quad v^{{\top}}Qv \leq \sigma^2, \quad v\geq 0.\end{equation*}

The optimal value of this problem is $\mathcal{A}_n^*({\mathscr T}_{\rm s};\, \mathcal{F}_\sigma\cap\mathcal{D}_N) = \kappa_n\sigma$ , where $\kappa_n$ is given by

(37) \begin{equation} \kappa_n \,:\!=\, \max_{z,t}\big\{t\colon(\textbf{1}d^{{\top}} - B)z - t\textbf{1} \geq 0,\, \|Q^{1/2}z\|_2 \leq 1,\,z\geq 0,\,t\geq 0\big\},\end{equation}

where $B=\{b_{i,j}\}$ , $i,j=1, \ldots, N-1$ . Table 2 displays values of $\kappa_n$ for $n\in\{10$ , 25, 50, 75, 100, 250, 500, 750, $1000\}$ obtained by solving the optimization problem in (37) for $N=7000$ ; we used the CVX package for specifying and solving convex programs [Reference Grant and Boyd12] together with the Mosek optimization solver.

Table 2.

The values of $\kappa_n$ in (37) as a function of n.

Table 2 Long description

A table with two columns and ten rows. The first column is labeled n and the second column is labeled kappa sub n. The values in the table are as follows: Row 1: n, 10; kappa sub n, 0.594. Row 2: n, 25; kappa sub n, 0.957. Row 3: n, 50; kappa sub n, 1.361. Row 4: n, 75; kappa sub n, 1.670. Row 5: n, 100; kappa sub n, 1.930. Row 6: n, 250; kappa sub n, 3.056. Row 7: n, 500; kappa sub n, 4.319. Row 8: n, 750; kappa sub n, 5.280. Row 9: n, 1000; kappa sub n, 6.082.

It can be seen that the sequence $\{\kappa_n\}$ increases with n. The growth of $\{\kappa_n\}$ can be compared with the results of [Reference Kennedy and Kertz20], which considered the setting of independent random variables, $F^{(n)}=\prod_{i=1}^nF_i$ , with the marginal distributions $F_i$ having bounded variance, ${\rm var}\{X_i\}\leq \sigma^2$ , $i=1, \ldots, n$ . They established an upper bound on the worst-case regret of the optimal stopping rules:

\begin{align*} \mathcal{A}_n^*\big({\mathscr T}_{\rm all};\, \mathcal{F}_\sigma^{(n)}\big) = \sup\nolimits_{F^{(n)}\in\mathcal{F}_\sigma^{(n)}}\mathcal{A}_n\big({\mathscr T}_{\rm all};\, F^{(n)}\big) \leq c_n \sigma\sqrt{n-1},\end{align*}

where $c_n\leq \frac12$ , $\liminf_n c_n\geq \sqrt{\ln 2- \frac12}\approx 0.439$ . Since the setting of the i.i.d. random variables is a specific case, we also have $\mathcal{A}^*_n({\mathscr T}_{\rm all};\, \mathcal{F}_\sigma)\leq c_n \sigma \sqrt{n-1}$ . Note that for large N, the value $\mathcal{A}^*_n({\mathscr T}_{\rm s, r};\, \mathcal{F}_\sigma\cap \mathcal{D}_N)$ approximates $\mathcal{A}^*_n({\mathscr T}_{\rm s, r};\, \mathcal{F}_\sigma)$ . For the data in Table 2 we have $\max_{n}\{\kappa_n/c_n\sqrt{n-1}\} \leq 0.451$ , where the maximum is taken over $n\in\{10$ , 25, 50, 75, 100, 250, 500, 750, $1000\}$ . Thus, the upper bound of [Reference Kennedy and Kertz20] is at least twice as high as the actual worst-case regret of the optimal single-threshold stopping rule on the class $\mathcal{F}_\sigma\cap \mathcal{D}_N$ with the values of n and N indicated above. Note, however, that [Reference Kennedy and Kertz20] considered a more general setting of independent random variables, and did not make statements on the sharpness of the derived prophet inequality.

6.2. Ratio-type prophet inequality for Pareto-like distributions

Let $1<p_1<p_0$ be real numbers and consider the family of distributions

\begin{align*} \mathcal{F}(p_0,p_1) = \big\{F\in\mathcal{F}_{[1,\infty)}\colon 1-x^{-p_1} \leq F(x) \leq 1-x^{-p_0} \text{ for all } x\in [1,\infty)\big\}.\end{align*}

If $F\in \mathcal{F}(p_0,p_1)$ then $(1-t)^{-1/p_0}\leq F^\leftarrow(t)\leq (1-t)^{-1/p_1}$ , and for $F\in \mathcal{F} (p_0,p_1)\cap \mathcal{D}_N$ we have

(38) \begin{equation} q_{1,i}\,:\!=\,\bigg(\frac{N}{N-i}\bigg)^{1/p_0} \leq u_i \leq \bigg(\frac{N}{N-i}\bigg)^{1/p_1} \,=\!:\, q_{0,i}, \quad i=1,\ldots, N-1.\end{equation}

The condition in the definition of $\mathcal{F}(p_0,p_1)$ imposes restrictions on the distribution tail. Larger values of $p_1$ result in lighter distribution tails, and it is expected that in such a situation the worst-case competitive ratio of the single-threshold stopping rules will be closer to 1.

Table 3.

The optimal values $\mathcal{R}_n^*(p_0,p_1)$ of (39) as a function of n for $p_1=5$ , $p_0=20$ , and $N=7000$ .

Table 3 Long description

A table with 12 rows and 12 columns. The first row contains the header ‘n’ followed by numerical values 10, 25, 50, 75, 100, 250, 500, 750, and 1000. The first column contains the header ‘R*_R,N(p0, p1)’ followed by numerical values 0.897, 0.865, 0.846, 0.837, 0.831, 0.815, 0.806, 0.802, and 0.799. The table presents the optimal values of R*_R,N(p0, p1) for different values of n and p. The values decrease as n increases.

The optimization problem associated with the family $\mathcal{F}(p_0,p_1)\cap \mathcal{D}_N$ takes the form

(39) \begin{equation} \min_v\max_{i=1,\ldots,N-1}\frac{b_i^{{\top}}v}{d^{{\top}}v} \quad {\rm such\ that} \quad q_1 \leq Qv \leq q_0, \quad v\geq 0,\end{equation}

where Q is the lower triangular matrix with all entries equal to one, and $q_0=[q_{0,1};\ldots;\, q_{0, N-1}]$ and $q_1=[q_{1,1};\ldots;\, q_{1, N-1}]$ ; see (38). The optimal values $\mathcal{R}^*_{n, N}(p_0,p_1)\,:\!=\,\mathcal{R}_n^*({\mathscr T}_{\rm s};\, \mathcal{F}(p_0,p_1)\cap \mathcal{D}_N)$ of (39) for $p_1=5$ , $p_0=20$ , $n\in \{10$ , 25, 50, 75, 100, 250, 500, 750, $1000\}$ computed with $N=7000$ are presented in Table 3. These values can be compared with values of $\mathcal{R}_{n, N}^*$ in Table 1. As expected, $\mathcal{R}^*_{n, N}(p_0,p_1)>\mathcal{R}_{n, N}^*$ because the family of distributions considered is narrower. Note also that the values of $\mathcal{R}^*_{n, N}(p_0,p_1)$ decrease as n increases.

7.1. Proof of Theorem 1

Proof. Let $\theta \geq 0$ be a fixed real number, $p\in [0,1]$ , and consider the stopping rule $\tau_p(\theta)$ defined in (11). In the subsequent proof, by convention we set $\prod_{j=1}^0 = 1$ .

We have

\begin{align*} \mathbb E X_{\tau_p(\theta)} & = \mathbb E \sum_{t=1}^{n-1} X_t \big[1_{(X_t>\theta)} + 1_{(X_t=\theta, \xi_t=1)}\big] \prod_{j=1}^{t-1}\big[1_{(X_j<\theta)} + 1_{(X_j=\theta, \xi_j=0)}\big] \\ & \quad + \mathbb E X_n \prod_{j=1}^{n-1}\big[1_{(X_j<\theta)} + 1_{(X_j=\theta, \xi_j=0)}\big] \\ & = \big[\mathbb E X_t 1_{(X_t>\theta)} + \theta p \Delta(\theta)\big] \sum_{t=1}^{n-1}\big[F(\theta\!-\!)+ (1-p)\Delta(\theta)\big]^{t-1} \\ & \quad + \mathbb E X_n\big[F(\theta-)+ (1-p)\Delta(\theta) \big]^{n-1} \\ & = \big[\mathbb E X_t 1_{(X_t>\theta)} + \theta p \Delta(\theta)\big]\frac{1-F_p^{n-1}(\theta)}{1-F_p(\theta)} + F_p^{n-1}(\theta)\mathbb E X_n \\ & = \bigg[\theta(1-F_p(\theta)) + \int_\theta^\infty[1-F(x)]\,\mathrm{d} x\bigg]\frac{1-F_p^{n-1}(\theta)}{1-F_p(\theta)} + F_p^{n-1}(\theta) \mathbb E X_n, \end{align*}

and (14) follows because $\int_{\theta}^\infty[1-F(x)]\,\mathrm{d} x=\int_{\theta}^\infty[1-F_p(x)]\,\mathrm{d} x$ . Thus, (14) is proved.

Furthermore,

\begin{align*} \mathbb E X_{\tau_p(\theta)} & = \mathbb E \sum_{t=1}^{n}X_t\big[1_{(X_t>\theta)} + 1_{(X_t=\theta, \xi_t=1)}\big] \prod_{j=1}^{t-1}\big[1_{(X_j<\theta)} + 1_{(X_j=\theta, \xi_j=0)}\big] \\ & \quad + \mathbb E X_n \prod_{j=1}^{n}\big[1_{(X_j<\theta)} + 1_{(X_j=\theta, \xi_j=0)}\big] \\ & = \big[\mathbb E X_t 1_{(X_t>\theta)} + \theta p \Delta(\theta)\big] \sum_{t=1}^n[F(\theta-) + (1-p)\Delta(\theta)]^{t-1} \\ & \quad + \big[\mathbb E X_n 1_{(X_n<\theta)} + \theta(1-p)\Delta(\theta)\big] [F(\theta-) + (1-p)\Delta(\theta)]^{n-1} \\ & = \bigg\{\theta(1-F_p(\theta)) + \int_\theta^\infty[1-F(x)]\,\mathrm{d} x\bigg\} \frac{1-F_p^{n}(\theta)}{1-F_p(\theta)} \\ & \quad + F_p^{n-1}(\theta)\bigg\{\int_{[0,\theta]}x\,\mathrm{d} F(x) - p\theta\Delta(\theta)\bigg\}. \end{align*}

This completes the proof of (13).

7.2. Proof of Theorem 2

Proof. Let $F\in \mathcal{F}_{[0,\infty)}$ be a fixed distribution, and for fixed $\theta\geq 0$ and $p\in [0,1]$ let $F_p(\theta)$ be defined in (12). For any $x\in [0,1]$ there exists a pair $(\theta_x, p_x)\in [0,\infty]\times [0,1]$ such that $F_{p_x}(\theta_x)=x$ . Indeed, by the definition of $F_p(\theta)$ ,

1. (i) if $\theta_x\,:\!=\, F^{\leftarrow}(x)$ and $F(\theta_x)=x$ then, for any $p_x\in [0,1]$ , $F_{p_x}(\theta_x)=x$ ;

2. (ii) if $\theta_x\,:\!=\,F^{\leftarrow}(x)$ and $F(\theta_x)>x$ then, for $p_x\,:\!=\, (F(\theta_x)-x)/(F(\theta_x)-F(\theta_x-))$ , $F_{p_x}(\theta_x)=x$ .

Then, according to (14), the reward of $\tau_{p_x}(\theta_x)$ is given by

(40) \begin{align} V_n(\tau_{p_x}(\theta_x);\, F) & = (1-x^{n-1})\bigg[F^\leftarrow(x)+\frac{1}{1-x}\int_{F^\leftarrow(x)}^\infty[1-F(t)]\,\mathrm{d} t\bigg] + x^{n-1}\int_0^\infty[1-F(x)]\,\mathrm{d} x \nonumber \\ & = (1-x^{n-1})\bigg[F^\leftarrow(x)+\frac{1}{1-x}\int_{x}^1(1-y)\,\mathrm{d} F^\leftarrow(y)\bigg] + x^{n-1}\int_{0}^1(1-y)\,\mathrm{d} F^\leftarrow(y) \nonumber \\ & = \int_{0}^1\bigg[(1-x^{n-1})\min\bigg\{1,\frac{1-y}{1-x}\bigg\} + x^{n-1}(1-y)\bigg]\,\mathrm{d} F^\leftarrow(y). \end{align}

Furthermore, $M_n(F)= \mathbb E \max_{1\leq t\leq n} X_t = \int_{0}^{1} (1-y^n)\,\mathrm{d} F^\leftarrow(y)$ . Therefore the constant in the sharp ratio-type prophet inequality is equal to the optimal value of the optimization problem

(41) \begin{equation} \mathcal{R}^*_n({\mathscr T}_{\rm s,r};\, \mathcal{F}_{[0,\infty)}) = \inf_{F^\leftarrow}\sup\limits_{0\leq x\leq 1} \frac{\int_{0}^{1}\big[(1-x^{n-1})\min\{1,({1-y})/({1-x})\big\} + x^{n-1}(1-y)\big]\,\mathrm{d} F^\leftarrow(y)} {\int_{0}^{1}(1-y^n)\,\mathrm{d} F^\leftarrow(y)}, \end{equation}

where the infimum is taken over all quantile functions of probability distributions on $[0,\infty)$ . The problem is equivalent to

(42) \begin{equation} \begin{aligned} \textstyle\inf_{F^\leftarrow}\sup_{0\leq x\leq 1} \quad & \int_0^1\bigg[(1-x^{n-1})\min\bigg\{1,\frac{1-y}{1-x}\bigg\} + x^{n-1}(1-y)\bigg]\,\mathrm{d} F^\leftarrow(y) \\ {\rm such\ that} \quad & \int_0^1(1-y^n)\,\mathrm{d} F^\leftarrow(y)=1, \end{aligned} \end{equation}

Let $\mu$ be the right-continuous function such that $\mathrm{d} \mu (y)\,:\!=\,(1-y^n)\,\mathrm{d} F^{\leftarrow}(y)$ for all $y\in [0,1]$ . Then $\mu$ is a probability distribution on [0,1]. Considering the randomized choice of $x\in [0,1]$ according to a distribution $\lambda$ on [0,1], we observe that the optimal values of (41) and (25) are equal. This proves the first statement of the theorem.

As for the game-theoretic representation for the difference-type prophet inequality, we observe that

\begin{align*} \sup\limits_{F^\leftarrow}\inf_{0\leq x\leq 1}[M_n(F)-V_n(\tau_{p_x}(\theta_x);\, F)] = \sup\limits_{F^\leftarrow}\inf_{0\leq x\leq 1}\int_0^1 A(x, y)\,\mathrm{d} F^\leftarrow (y), \end{align*}

where the supremum is taken over all quantile functions of distributions on [0,1]. This constraint can be written in the form $F^{\leftarrow}(1)= \int_0^1 \mathrm{d} F^{\leftarrow}(t) \leq 1$ . Defining the right-continuous function $\mu$ such that $\mathrm{d} \mu(y)= \mathrm{d} F^{\leftarrow}(y)$ and using the same reasoning as above, we come to (26).

7.3. Proof of Theorem 3

Proof. Assume that $F\in \mathcal{D}_N$ , i.e. $F(x)=({1}/{N})\sum_{i=1}^N I(u_i\leq x)$ with some $0=u_0 \leq u_1\leq \cdots \leq u_N$ for $x\geq 0$ . For such distributions the set of all possible thresholds of stopping rules is restricted to the $(i/N)$ -quantiles $\{u_i, i=0,\ldots,N\}$ of F. Define $v_0\,:\!=\, u_1$ , $v_j\,:\!=\,u_{j+1}-u_{j}$ , $j=1,\ldots,N-1$ . The quantile function $F^{\leftarrow}(u)$ of $F\in \mathcal{D}_N$ is given by $F^{\leftarrow}(u)=\sum_{j=0}^{N-1}v_jI\{{j}/{N} < u\leq({j+1})/{N}\}$ . Therefore, it follows from (40) that the reward of the stopping rule $\tau_{p_i}(u_i)$ with threshold $u_i$ , $i=0, \ldots, N-1$ , and randomization probability $p_i\,:\!=\, (F(u_i)-i/N)/(F(u_i)-F(u_i-))$ is

(43) \begin{align} V_n(\tau_{p_i}(u_i);\, F) & = \sum_{j=0}^{N-1}\bigg[\bigg(1-\bigg(\frac{i}{N}\bigg)^{n-1}\bigg) \min\bigg\{1,\frac{1-{j}/{N}}{1-{i}/{N}}\bigg\} + \bigg(\frac{i}{N}\bigg)^{n-1}\bigg(1-\frac{j}{N}\bigg)\bigg]v_j \nonumber \\ & = \sum_{j=0}^{N-1}R\bigg(\frac{i}{N},\frac{j}{N}\bigg)v_j, \end{align}

where $R(\cdot, \cdot)$ is given by (22). It follows from (43) that $V_n(\tau_{p_i}(u_i);\, F)= b_i^{{\top}}v$ , $i=0, \ldots, N-1$ , where $b_i=[b_{i,0};\ldots;\,b_{i,(N-1)}]$ , $i=0, \ldots, N-1$ , are vectors with entries

\begin{align*} b_{i,j} = \begin{cases} 1-\bigg(\dfrac{i}{N}\bigg)^{n-1}\dfrac{j}{N}, & 0 \leq j\leq i\leq N-1, \\[4mm] \dfrac{1-({i}/{N})^n}{1-{i}/{N}}\bigg(1-\frac{j}{N}\bigg), & 0 \leq i < j\leq N-1. \end{cases} \end{align*}

Moreover, for $F\in\mathcal{D}_N$ we have $M_n(F)=\mathbb E\max_{1\leq t\leq n}X_t=\sum_{j=0}^{N-1}[1-({j}/{N})^n]v_j \,=\!:\, d^{{\top}}v$ , where $d\in {\mathbb R}^{N}$ , $d_j= 1-({j}/{N})^n$ , $j=0, \ldots, N-1$ . Then (42) implies that

(44) \begin{equation} \mathcal{R}^*_n({\mathscr T}_{\rm s,r};\, \mathcal{D}_N) = \max_\lambda\min_v\big\{\lambda^{{\!\top}}Bv\colon\lambda^{{\!\top}}\textbf{1}=1,\,d^{{\top}}v=1,\,\lambda\geq 0,\,v\geq 0\big\}, \end{equation}

where $B=[b_0^{{\!\top}};\,b_1^{{\!\top}};\ldots;\,b_{N-1}^{{\!\top}}]$ , and $\lambda$ stands for the probability vector that defines the single-threshold rule: the threshold $u_i$ is selected with the probability $\lambda_i$ , $i=0, \ldots, N-1$ . Observing that

\begin{align*} \bigg[\frac{b_0^{{\!\top}}}{d_0}; \frac{b_1^{{\!\top}}}{d_1};\cdots;\frac{b_{N-1}^{{\!\top}}}{d_{N-1}}\bigg] = \bigg\{R\bigg(\frac{i}{N},\frac{j}{N}\bigg)\colon i,j=0,\ldots,N-1\bigg\} \,=\!:\, \tilde{R}_N, \end{align*}

we obtain that (44) is equivalent to

(45) \begin{equation} \mathcal{R}^*_n({\mathscr T}_{\rm s,r};\, \mathcal{D}_N) = \max_\lambda\min_\mu\big\{\lambda^{{\!\top}}\tilde{R}_N\mu\colon\lambda^{{\!\top}}\textbf{1}=1,\, \mu^{{\!\top}}\textbf{1}=1,\,\lambda\geq 0,\,\mu\geq 0\big\}. \end{equation}

To complete the proof that $\mathcal{R}^*_n({\mathscr T}_{\rm s,r};\, \mathcal{D}_N)=\mathcal{R}_{n, N}^*$ , we need to show that the optimal value of the above game with $N\times N$ matrix $\tilde{R}_N=\{R({i}/{N},{j}/{N})\colon i,j=0,\ldots,N-1\}$ coincides with the optimal value of the game with $(N-1)\times (N-1)$ matrix $R_N=\{R({i}/{N},{j}/{N})\colon i,j=1,\ldots,N-1\}$ . This fact is a consequence of the following dominance relationships between the rows and columns of matrix $\tilde{R}_N$ :

1. (i) The pure $\lambda$ -strategy $[1;\,0;\ldots;\,0]$ is dominated by the strategy $[0;\,1;\,0\ldots;\,0]$ . Indeed,

   \begin{align*} R(0,{j}/{N}) & = \frac{1-{j}/{N}}{1-({j}/{N})^n} \quad \text{for all}\ j=0,\ldots,N-1, \\ R({1}/{N},0) & = R({1}/{N},{1}/{N})=1, \\ R({1}/{N},{j}/{N}) & = \frac{1-({1}/{N})^n}{1-{1}/{N}}\frac{1-{j}/{N}}{1-({j}/{N})^n} \quad \text{for all}\ j=2,\ldots,N-1, \end{align*}
   so that $R(0,0)=R({1}/{N},0)$ and $R(0,{j}/{N})<R(1/N,{j}/{N})$ for all $j=1,\ldots,N-1$ . Thus, the zeroth row of the matrix $\tilde{R}_N$ may be eliminated.

2. (ii) The pure $\mu$ -strategy $[1;0;\ldots;\,0]$ is dominated by the strategy $[0;1;0;\ldots;\,0]$ . Indeed,

   \begin{align*} R({i}/{N}, 0) & = 1 \quad \text{for all}\ i=0, \ldots, N-1, \\ R(0,{1}/{N}) & = \frac{1-{1}/{N}}{1-({1}/{N})^n}, \\ R({i}/{N},{1}/{N}) & = \frac{1-({i}/{N})^{n-1}({1}/{N})}{1-({1}/{N})^{n}} \quad \text{for all}\ i=2,\ldots,N-1. \end{align*}
   Thus, $R({i}/{N},0)>R({i}/{N},{1}/{N})$ for all $i=0,\ldots, N-1$ , i.e. the zeroth column of matrix $\tilde{R}_N$ can be eliminated.

Facts (i) and (ii) together with (45) and the definition of $\mathcal{R}^*_{n, N}$ in (31) imply that $\mathcal{R}^*_n ({\mathscr T}_{\rm s,r};\, \mathcal{D}_N)=\mathcal{R}_{n, N}^*$ .

To prove that $\mathcal{A}_{n}^*({\mathscr T}_{\rm s,r};\, \mathcal{D}_N)= \mathcal{A}_{n, N}^*$ we note that with the introduced notation for any $F\in \mathcal{D}_N$ the regret of the stopping rule $\tau_{p_i}(u_i)$ associated with the threshold $u_i$ is

\begin{align*} \mathcal{A}_n(\tau_{p_i}(u_i);\, F) = M_n(F)-V_n(\tau_{p_i}(u_i);\, F) = d^{{\top}}v - b_i^{{\top}}v, \quad i=0, \ldots, N-1. \end{align*}

For $F\in \mathcal{F}_{[0,1]}$ , $v^{{\top}}\textbf{1} \leq 1$ ; therefore

\begin{align*} \mathcal{A}_n^*({\mathscr T}_{\rm s,r};\, \mathcal{D}_N) & = \max_{v}\min_{i=1,\ldots,N-1}\{d^{{\top}}v - b_i^{{\top}}v\colon v^{{\top}}\textbf{1} \leq 1,\,v\geq 0\} \\ & = \max_{v}\min_{\mu}\big\{\mu^{{\!\top}}(\textbf{1}d^{{\top}}-B)v\colon \mu^{{\!\top}}\textbf{1}=1,\, v^{{\top}}\textbf{1} = 1,\,v\geq 0, \,\mu\geq 0\big\}. \end{align*}

Observe that $\textbf{1}d^{{\top}}-B = \{A({i}/{N},{j}/{N})\colon i,j=0,\ldots,N-1\} \,=\!:\, \tilde{A}_N$ . Then the statement of the theorem follows from the fact that, similarly to above, the zeroth row and zeroth column of $\tilde{A}_N$ can be eliminated by dominance considerations.

7.4. Proof of Theorem 4

We begin with a simple lemma.

Lemma 1.

\begin{align*} |A(x, y) - A(x^\prime, y)| & \leq (n-1)|x-x^\prime| \quad {for\, all}\ x,x^\prime,y \in [0,1], \\ |A(x, y) - A(x, y^\prime)| & \leq (n-1)|y-y^\prime| \quad {for\, all}\ x,y,y^\prime \in [0,1]. \end{align*}

Proof. We note that A(x, y) is continuous on $[0,1]^2$ . We have, for any $y\in [0,1]$ ,

\begin{align*} \bigg|\frac{\partial A(x,y)}{\partial x}\bigg| & = (n-1)yx^{n-2} \leq n-1 \quad \text{for all}\ (x,y) \ \colon x > y, \\ \bigg|\frac{\partial A(x,y)}{\partial x}\bigg| & = (1-y)\sum_{k=1}^{n-1}kx^{k-1} \\ & = (1-y)\frac{\sum_{j=0}^{n-2}(j+1)x^{\,j}}{\sum_{j=0}^{n-2}x^{\,j}}\sum_{j=0}^{n-2}x^{\,j} \\ & \leq \bigg(\frac{1-y}{1-x}\bigg)\max_{j=0,\ldots,n-2}\{(j+1)\} \leq n-1 \quad \text{for all}\ (x,y) \ \colon x < y. \end{align*}

Therefore, for any fixed $y\in [0,1]$ , if $x > y$ and $x^\prime > y$ then $|A(x, y) - A(x^\prime, y)|\leq (n-1)|x-x^\prime|$ , and the same inequality holds for any fixed $y\in [0,1]$ and $x < y$ , $x^\prime < y$ . Thus, for any $y\in [0,1]$ , $|A(x, y) - A(x^\prime, y)|\leq (n-1)|x-x^\prime|$ for all $x, x^\prime\in [0,1]$ . Similarly, for any $x\in [0,1]$ , $|{\partial A(x,y)}/{\partial y}| = |x^{n-1}-n y^{n-1}|\leq n-1$ for all $(x,y)\colon x>y$ . For $x < y$ we have

\begin{align*} \frac{\partial A(x,y)}{\partial y} & = \sum_{k=1}^{n-1}ky^{k-1} - \sum_{k=1}^{n-1}(k+1)y^k + \sum_{k=1}^{n-1}x^k \leq \sum_{k=1}^{n-1}ky^{k-1}(1-y) \\ & \leq \frac{\sum_{k=1}^{n-1}ky^{k-1}}{\sum_{k=1}^{n-1}y^{k-1}}(1-y)\sum_{k=1}^{n-1}y^{k-1} \leq \max_{j=0,\ldots,n-2}\{(j+2)\} \leq n-1. \end{align*}

On the other hand, the above expression implies that

\begin{align*} \frac{\partial A(x,y)}{\partial y} = 1-ny^{n-1} + \sum_{k=1}^{n-1} x^k \geq 1-n. \end{align*}

Thus, $|\partial A(x,y)/\partial y|\leq n-1$ , which implies that $|A(x, y) - A(x, y^\prime)|\leq (n-1) |y-y^\prime|$ for all $(x,y)\colon x<y$ . This completes the proof of the lemma.

Proof of Theorem 4(i). For brevity we write $\mathcal{A}^*_n=\mathcal{A}_n^*({\mathscr T}_{\rm s,r};\, \mathcal{F}_{[0,1]})$ , and recall that $\mathcal{A}_n^*=$ $\sup_\mu \inf_\lambda \bar{A}(\lambda, \mu)$ , where $\bar{A}(\lambda,\mu)$ is defined in (24). Consider the finite matrix game (32) associated with $A_N$ whose value is $\mathcal{A}^*_{n, N}$ . Assume that $\lambda^{N}=(\lambda_1^N,\ldots,\lambda^N_{N-1})$ and $\mu^N=(\mu_1^N,\ldots,\mu_{N-1}^N)$ are the optimal mixed strategies in this game, i.e. $\bar{A}(\lambda^N, \mu^N)=\mathcal{A}_{n, N}^*$ .

Let $\delta_y$ denote the degenerate distribution at $y\in [0,1]$ . First, we note that $\max_y \bar{A}(\lambda^N, \delta_y) \geq \mathcal{A}_n^*$ . Therefore, there exists a point y, say $y_*$ , such that $\bar{A}(\lambda^N, \delta_{y_*}) \geq \mathcal{A}_n^*$ . By Lemma 1, there exists an index $j_*\in \{0, 1, \ldots, N-1\}$ such that

\begin{equation*} \big|\bar{A}\big(\lambda^N, \delta_{y_*}\big) - \bar{A}\big(\lambda^N, \delta_{j_*/N}\big)\big| \leq \sum_{i=1}^{N-1}\lambda_i^N\bigg|A\bigg(\frac{i}{N},y_*\bigg)-A\bigg(\frac{i}{N},\frac{j_*}{N}\bigg)\bigg| \leq (n-1)\bigg|y_*-\frac{j_*}{N}\bigg| \leq \frac{n-1}{2N}. \end{equation*}

Therefore

\begin{align*} \mathcal{A}_n^* \leq \bar{A}(\lambda^N,\delta_{y_*}) \leq \bar{A}(\lambda^N, \delta_{j_*/N}) + \frac{n-1}{2N} \leq \mathcal{A}^*_{n, N}+ \frac{n-1}{2N}, \end{align*}

which yields the upper bound in (33).

Furthermore, note that if $\delta_x$ is the degenerate distribution at $x\in [0,1]$ then $\min_x \bar{A}(\delta_x, \mu^N)\leq \mathcal{A}_n^*$ . Therefore there exists $x_*$ such that $\bar{A}(\delta_{x_*}, \mu^N)\leq \mathcal{A}_n^*$ . By Lemma 1, there exists an index $i_*\in \{1, \ldots, N-1\}$ such that

\begin{align*} \big|\bar{A}\big(\delta_{x_*}, \mu^N\big) - \bar{A}\big(\delta_{i_*/N}, \mu^N\big)\big| \leq \sum_{j=1}^{N-1}\mu_j^N\bigg|A\bigg(x_*,\frac{j}{N}\bigg) - A\bigg(\frac{i_*}{N},\frac{j}{N}\bigg)\bigg| \leq (n-1)\bigg|x_*-\frac{i_*}{N}\bigg| \leq \frac{n-1}{2N}. \end{align*}

Therefore

\begin{align*} \mathcal{A}_n^* \geq \bar{A}(\delta_{x_*}, \mu^N) \geq \bar{A}(\delta_{i_*/N},\mu^N) - \frac{n-1}{2N} \geq \mathcal{A}_{n, N}^* - \frac{n-1}{2N}, \end{align*}

which completes the proof of statement (i).

We now require another auxiliary lemma.

Lemma 2. Let $\mathcal{R}_n^*\,:\!=\,\inf_\mu \sup_\lambda \bar{R}(\lambda, \mu)$ be the value of the game on the unit square with payoff kernel R(x,y) (see (22) and (24)), and let $\lambda_*$ and $\mu_*$ be the optimal strategies. Let

(46) \begin{equation} c_n\,:\!=\,-\ln \bigg\{1-(1-{\mathrm{e}}^{-1})^2+\frac{1}{n-1}\bigg\}. \end{equation}

If $n\geq 4$ then the interval $[1- c_n/n, 1]$ does not belong to the support of $\lambda_*$ .

Proof. Let $x_0=1- c/n$ for some $c \in (0,c_n)$ , and assume to the contrary that $x_0\in {\rm supp}(\lambda_*)$ . Under this assumption we have $\bar{R}(\delta_{x_0}, \mu^*)=\mathcal{R}_n^*$ ; see, e.g., [Reference Karlin19, Lemma 2.2.1]. Now let $y_0=1-{1}/{n}$ ; since $x_0>y_0$ , it follows from (22) that

\begin{align*} \bar{R}(\delta_{x_0}, \delta_{y_0}) = R(x_0, y_0) & = \frac{1-(1-{c}/{n})^{n-1}(1-{1}/{n})}{1-(1-{1}/{n})^n} \\ & \leq \frac{1}{1-{\mathrm{e}}^{-1}}\bigg[1-\bigg(1-\frac{c}{n}\bigg)^n + \bigg(1-\frac{c}{n}\bigg)^n\frac{1-c}{n-c}\bigg]. \end{align*}

Because ${\mathrm{e}}^{-c} \geq (1-{c}/{n})^n \geq {\mathrm{e}}^{-c}(1-{c^2}/({2(n-1)}))$ we obtain

\begin{align*} \bar{R}(\delta_{x_0}, \delta_{y_0}) & \leq \frac{1}{1-{\mathrm{e}}^{-1}}\bigg[1- {\mathrm{e}}^{-c}\bigg(1-\frac{c^2}{2(n-1)}-\frac{1}{n-c}\bigg)\bigg] \\ & \leq \frac{1}{1-{\mathrm{e}}^{-1}}\bigg[1-{\mathrm{e}}^{-c} + \frac{{\mathrm{e}}^{-c}(c^2+2)}{2(n-1)}\bigg]. \end{align*}

For $c\in (0,1)$ ${\mathrm{e}}^{-c}(c^2+2)\leq 2$ ; therefore

\begin{align*} \bar{R}(\delta_{x_0}, \delta_{y_0}) \leq \frac{1}{1-{\mathrm{e}}^{-1}}\bigg[1-{\mathrm{e}}^{-c}+\frac{1}{n-1}\bigg]. \end{align*}

This inequality shows that for any pair of numbers n and $c\in (0,1)$ such that

(47) \begin{equation} 1-{\mathrm{e}}^{-c}+ \frac{1}{n-1} < (1-{\mathrm{e}}^{-1})^2 \end{equation}

we obtain $\mathcal{R}^*_n = \bar{R}(\delta_{x_0},\mu_*) \leq \bar{R}(\delta_{x_0},\delta_{y_0}) = R(x_0,y_0) < 1-{\mathrm{e}}^{-1}$ . In particular, if $n\geq 4$ and

\begin{align*} c_n= -\ln\bigg\{1-(1-{\mathrm{e}}^{-1})^2+\frac{1}{n-1}\bigg\} \approx -\ln\bigg\{0.6004 + \frac{1}{n-1}\bigg\} \end{align*}

then (47) holds for all $c\in (0,c_n)$ . This, however, stands in contradiction to inequality (16). Therefore $x_0\not\in {\rm supp}(\lambda_*)$ , and the proof is complete.

The next result is an analogue of Lemma 1 for the function R(x, y).

Lemma 3. For any $\varepsilon\in(0,1)$ ,

(48) \begin{align} |R(x, y) - R(x^\prime, y)| & \leq \frac{n-1}{1-(1-\varepsilon)^n}|x-x^\prime| \ {for\, all}\ x,x^\prime \in [0,1-\varepsilon] \ {and\, all}\ y\in [0,1], \end{align}

(49) \begin{align} |R(x, y) - R(x, y^\prime)| & \leq \frac{n-1}{1-(1-\varepsilon)^n}|y-y^\prime| \ {for\, all}\ x\in [0,1-\varepsilon] \ {and\, all}\ y, y^\prime \in [0,1]. \\[6pt] \nonumber\end{align}

Proof. Let $y \in [0,1]$ be fixed. By (22), if $0\leq x < y\leq 1$ then

\begin{equation*} \frac{\partial R(x,y)}{\partial x} = \frac{1-y}{1-y^n}\sum_{j=1}^{n-1}jx^{j-1} = \frac{\sum_{j=0}^{n-2}(j+1)x^{\,j}}{\sum_{j=0}^{n-2} x^{\,j}}\,\frac{\sum_{j=0}^{n-2}x^{\,j}}{\sum_{j=0}^{n-1}y^j} \leq \max_{0 \leq j\leq n-2}\{(j+1)\} \leq n-1, \end{equation*}

and for $0\leq y < x\leq 1-\varepsilon$ ,

\begin{equation*} \bigg|\frac{\partial R(x,y)}{\partial x}\bigg| = \frac{(n-1)x^{n-2} y}{1-y^n} \leq \frac{n-1}{1-(1-\varepsilon)^n}. \end{equation*}

In view of these inequalities, for all $x, x^\prime$ such that $0\leq x < y < x^{\prime}\leq 1-\varepsilon$ we have

\begin{equation*} |R(x, y)- R(x^\prime, y)|\leq |R(x, y) - R(y,y)| + |R(y,y)- R(x^\prime, y)| \leq \frac{n-1}{1-(1-\varepsilon)^n} |x-x^\prime|. \end{equation*}

The inequalities imply (48).

Similarly, if $x\in [0,1-\varepsilon]$ is fixed and $x\leq y\leq 1$ , then

\begin{align*} \bigg|\frac{\partial R(x,y)}{\partial y}\bigg| = \frac{1-x^n}{1-x}\frac{\sum_{j=0}^{n-2}(j+1)y^j}{\big(\sum_{j=0}^{n-1} y^j\big)^2} \leq \frac{\sum_{j=0}^{n-1} x^{\,j}}{\sum_{j=0}^{n-1} y^j}\frac{\sum_{j=0}^{n-2} (j+1)y^j}{\sum_{j=0}^{n-1} y^j} \leq \max_{0 \leq j\leq n-2} \{(j+1) \}\leq n-1, \end{align*}

and for $y < x\leq 1-\varepsilon$ ,

\begin{equation*} \frac{\partial R(x,y)}{\partial y} = -\frac{x^{n-1}}{1-y^n} + \frac{(1-x^{n-1}y)ny^{n-1}}{(1-y^n)^2}\leq \frac{(n-1)x^{n-1}}{1-y^n} \leq \frac{n-1}{1-(1-\varepsilon)^n}; \end{equation*}

the same inequality holds for $|\partial R(x,y)/\partial y|$ . Combining these inequalities we obtain (49).

Now, using Lemmas 2 and 3 we complete the proof of statement (ii) of Theorem 4.

Proof of Theorem 4(ii). The proof goes along the lines of the proof of statement (i), with the following minor changes: the payoff function A(x, y) is replaced by R(x, y), and the infinite game is considered on the rectangle $[0, 1- c_n/n]\times [0,1]$ . Lemma 2 ensures that the optimal values of the game on this set coincide with those on the unit square.

Recall that $\mathcal{R}_n^*=\inf_\mu \sup_\lambda \bar{R}(\lambda, \mu)$ , and $\mathcal{R}_{n, N}^*= \bar{R}(\lambda^N, \mu^N)$ , where $\lambda^N=(\lambda_1^N, \ldots, \lambda_{N-1}^N)$ and $\mu^N=(\mu_1^N, \ldots, \mu_{N-1}^N)$ are optimal mixing strategies in the finite matrix game (31). It follows from Lemma 2 that $\lambda^N_i=0$ for $\lceil (1-c_n)N\rceil\leq i\leq N-1$ , where $c_n$ is given in (46). The definitions imply that $\min_y \bar{R}(\lambda^N, \delta_y)\leq \mathcal{R}_n^*$ , i.e. there exists $y_*\in [0,1]$ such that $\bar{R}(\lambda^N, \delta_{y_*})\leq \mathcal{R}_n^*$ . By Lemma 3 applied with $\varepsilon=c_n/n$ , for some $j_*$ ,

\begin{align*} \big|\bar{R}(\lambda^N, \delta_{y_*}) - \bar{R}(\lambda^N, \delta_{j_*/N})\big| \leq \frac{n-1}{1-(1-{c_n}/{n})^n}\bigg|y_*-\frac{j_*}{N}\bigg| \leq \frac{n-1}{2N\big(1-(1-{c_n}/{n})^n\big)}. \end{align*}

Therefore

\begin{align*} \mathcal{R}_n^* \geq \bar{R}(\lambda^N, \delta_{y_*}) \geq \bar{R}(\lambda^N, \delta_{j_*/n}) - \frac{n-1}{2N(1-(1-{c_n}/{n})^n)} \geq \mathcal{R}_{n, N}^* - \frac{n-1}{2N(1-(1-{c_n}/{n})^n)}. \end{align*}

Then the lower bound in (34) follows by substitution of (46). The upper bound on $\mathcal{R}_n^*$ is proved similarly.