# On Zero-Sum Optimal Stopping Games

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## Abstract

On a filtered probability space \((\Omega ,\mathcal {F},P,\mathbb {F}=(\mathcal {F}_t)_{t=0,\ldots ,T})\), we consider stopping games \(\overline{V}:=\inf _{{\varvec{\rho }}\in \mathbb {T}^{ii}}\sup _{\tau \in \mathcal {T}}\mathbb {E}[U({\varvec{\rho }}(\tau ),\tau )]\) and \(\underline{V}:=\sup _{{\varvec{\tau }}\in \mathbb {T}^i}\inf _{\rho \in \mathcal {T}}\mathbb {E}[U(\rho ,{\varvec{\tau }}(\rho ))]\) in discrete time, where *U*(*s*, *t*) is \(\mathcal {F}_{s\vee t}\)-measurable instead of \(\mathcal {F}_{s\wedge t}\)-measurable as is assumed in the literature on Dynkin games, \(\mathcal {T}\) is the set of stopping times, and \(\mathbb {T}^i\) and \(\mathbb {T}^{ii}\) are sets of mappings from \(\mathcal {T}\) to \(\mathcal {T}\) satisfying certain non-anticipativity conditions. We will see in an example that there is no room for stopping strategies in classical Dynkin games unlike the new stopping game we are introducing. We convert the problems into an alternative Dynkin game, and show that \(\overline{V}=\underline{V}=V\), where *V* is the value of the Dynkin game. We also get optimal \({\varvec{\rho }}\in \mathbb {T}^{ii}\) and \({\varvec{\tau }}\in \mathbb {T}^i\) for \(\overline{V}\) and \(\underline{V}\) respectively.

## 1 Introduction

*U*(

*s*,

*t*) is \(\mathcal {F}_{s \vee t}\)-measurable.

^{1}Note that

*U*in (1.3) is \(\mathbb {F}_{s\wedge t}\)-measurable, and thus Dynkin game ends at the minimum of \(\rho \) and \(\tau \). However, for a general

*U*, it may be possible that \(\overline{A}>\underline{A}\) even for some very natural choices of

*U*. For example, consider

*U*, the problem in (1.2) becomes deterministic, and it is easy to see that \(\overline{A}=\lceil T/2 \rceil >0=\underline{A}\). As opposed to the Dynkin game in which

*U*is given by (1.3), we can see that the game with

*U*given by (1.4) has not ended when only one of the players has stopped, i.e., the payoff will be further affected by the player who stops later. Intuitively, the failure of the equality \(\overline{A}=\underline{A}\) of the two game values in (1.2) is due to the fact that the minimizer in the first game (represented by \(\overline{A}\)) is weaker than the minimizer in the second game (represented by \(\underline{A}\)). The opposite holds for the maximizer. In other words, the inner player in either of the games knows the outer player’s prefixed strategy but not vice versa.

### Example 1.1

Let \(T=1\) and \(U(s,t)=1_{\{s\ne t\}},\ t=0,1\). Then there are only two elements, \({\varvec{\rho }}^0\) and \({\varvec{\rho }}^1\), in \(\mathbb {T}^i\), with \({\varvec{\rho }}^0(0)={\varvec{\rho }}^0(1)=0\) and \({\varvec{\rho }}^1(0)={\varvec{\rho }}^1(1)=1\). It can be shown that \(\overline{B}=1\) and \(\underline{B}=0\).

^{2}the outer players here have more power compared to the case in (1.5).

### Example 1.2

Let \(T=1\) and \(U(s,t)=1_{\{s\ne t\}},\ t=0,1\). Then in this case \(\mathbb {T}^{ii}\) is the set of all the maps from \(\mathcal {T}\) to \(\mathcal {T}\). By letting \({\varvec{\rho }}(0)=0\) and \({\varvec{\rho }}(1)=1\), we have that \(\overline{C}=0\). By Letting \({\varvec{\tau }}(0)=1\) and \({\varvec{\tau }}(1)=0\), we have that \(\underline{C}=1\).

### Remark 1.1

In the continuous-time case, we have also have (1.6) and (1.9) in general (see Remark 2.1, in [2]). But in order to solve this problem the latter paper assumes that the pay-off is right continuous along stopping times in the sense of expectation as in [8], which is the most relaxed assumption in the continuous optimal stopping time literature (see also [6, 7]), and a result the difference between the two types of non-anticipativity conditions disappear, i.e. \(\overline{B}=\underline{B}\) and \(\overline{C}=\underline{C}\). Hence the discrete-time case is interesting, because no such technical condition is needed and one is able to observe the structure of the problem more clearly.

Our new zero-sum stopping game, which can be considered as a **two stage** Dynkin game, also captures the “game nature”— the interaction between players (i.e., players can adjust their own strategies according to the other’s behavior). For example, unlike in \(\overline{A}\) in (1.2) where \(\rho \) does not depend on \(\tau \) (but \(\tau \) depends on \(\rho \)), in \(\overline{V}\), \({\varvec{\rho }}\) depends on \(\tau \) by the definition of \(\mathbb {T}^{ii}\) (and of course \(\tau \) still depends on \({\varvec{\rho }}\) as \(\tau \) is the inner player).

Since [3] Dynkin games have been studied extensively, and we refer to the survey paper [5] and the references therein. The stopping game we introduce here is more suitable for handing conflict since in general players take turns in playing the game and the game does not end when one of the players act. In other words, stopping games are not always *duels*, which is what the usual Dynkin game models. In Sects. 3.2 and 3.3, we will provide two applications of our game, including a robust utility maximization problem involving two American options, and an example of competing companies choosing times to enter the market.

We can also relate our paper to the following interesting phenomenon observed in stochastic differential games: In a zero-sum game with one player doing the inside optimization using an open loop strategy and the other player doing outside optimization using an Elliott–Kalton (non-anticipative) strategy, it has been observed (see for example [1], Theorem 3.11) that the infimum and supremum can not be exchanged. For the game to have a value when one needs to change the strength of the players appropriately: the outside player should have closed loop strategies and the inside player should have open loop controls. (This was only proved analytically using a viscosity comparison). We are able to observe this in an optimal stopping problem for the first time. Moreover, we prove it directly using probabilistic techniques only.

The rest of the paper is organized as follows. In the next section, we introduce the setup and the main result. We provide three examples in Sect. 3. In Sect. 4, we give the proof of the main result. Finally we give some insight for the corresponding problems in continuous time in Sect. 5.

## 2 The Setup and the Main Result

Let \((\Omega ,\mathcal {F},P)\) be a probability space, and \(\mathbb {F}=(\mathcal {F}_t)_{t=0,\ldots ,T}\) be the filtration enlarged by *P*-null sets, where \(T\in \mathbb {N}\) is the time horizon. Let \(U:\{0,\ldots ,T\}\times \{0,\ldots ,T\}\times \Omega \mapsto \mathbb {R}\), such that \(U(s,t,\cdot )\in \mathcal {F}_{s\vee t}\). For simplicity, we assume that *U* is bounded. Denote \(\mathbb {E}_t[\cdot ]\) for \(\mathbb {E}[\cdot |\mathcal {F}_t]\). We shall often omit to write “almost surely” (when a property holds outside a *P*-null set). Let \(\mathcal {T}_t\) be the set of \(\mathbb {F}\)-stopping times taking values in \(\{t\cdots ,T\}\), and \(\mathcal {T}:=\mathcal {T}_0\). We define the stopping strategies of Type I and Type II as follows:

### Definition 2.1

### Remark 2.1

We can treat \(\mathcal {T}\) as a subset of \(\mathbb {T}^i\) and \(\mathbb {T}^{ii}\) (i.e., each \(\tau \in \mathcal {T}\) can be treated as the map with only one value \(\tau \)). Hence we have \(\mathcal {T}\subset \mathbb {T}^i\subset \mathbb {T}^{ii}\).

### Theorem 2.1

### Remark 2.2

As the inner player in \(\overline{V}\), \(\tau \) depends on \({\varvec{\rho }}\). Therefore, as a good reaction to \({\varvec{\rho }}\), \(\tau ^*(\cdot )\) defined in (2.8) is a map from \(\mathbb {T}^{ii}\) to \(\mathcal {T}\) instead of a stopping time. (To convince oneself, one may think of \(\inf _x\sup _y f(x,y)=\inf _x f(x,y^*(x))\)).

### Corollary 2.1

### Proof

## 3 Examples

In this section we provide three examples that fall within the setup of Sect. 2. The first example shows that in the classical Dynkin game one does not need to use non-anticipative stopping strategies. The second example is a relevant problem from mathematical finance in which our results can be applied. This problem is on determining the optimal exercise strategy when one trades two different American options in different directions. In the third example we consider two competing companies making an entry decision into a particular market.

### 3.1 Dynkin Game Using Non-anticipative Stopping Strategies

*U*, the \({\varvec{\rho }}^*\) and \({\varvec{\tau }}^{**}\) defined in (2.7) and (2.9) can w.l.o.g. be written as

### Remark 3.1

In this example we let \({\varvec{\rho }}\in \mathbb {T}^{ii}\) and \({\varvec{\tau }}\in \mathbb {T}^i\). The same conclusion holds if we let \({\varvec{\rho }}\in \mathbb {T}^i\) and \({\varvec{\tau }}\in \mathbb {T}^{ii}\) instead.

### 3.2 A Robust Utility Maximization Problem

*f*and

*g*are adapted to \(\mathbb {F}\). Consider

*f*and shorts an American option

*g*, and the goal is to choose an optimal stopping strategy to maximize the utility according to the stopping behavior of the holder of

*g*. Here we assume that the maturities of

*f*and

*g*are the same (i.e.,

*T*). This is without loss of generality. Indeed for instance, if the maturity of

*f*is \(\hat{t}<T\), then we can define \(f(t)=f(\hat{t})\) for \(t=\hat{t}+1,\ldots ,T\).

### 3.3 Time to Enter the Market

There are two companies choosing when to enter a specific market. These two companies will produce the same kind of product. The one that enters the market first can start collecting profit earlier, while the one that enters second can use the other’s experience (e.g., marketing strategies, technologies) to reduce its own cost. Hence, each company’s entering time will affect the profit and market share no matter if it enters the market first or second. Then a natural and simple model for this set-up would be given by our game defined in (2.2).

## 4 Proof of Theorem 2.1

We will only prove the results for \(\overline{V}\), since the proofs for \(\underline{V}\) are similar.

### Lemma 4.1

For any \(\sigma \in \mathcal {T}\), \(\rho _u(\sigma )\in \mathcal {T}\) and \(\tau _u(\sigma )\in \mathcal {T}\).

### Proof

### Lemma 4.2

\({\varvec{\rho }}^*\) defined in (2.7) is in \(\mathbb {T}^{ii}\) and \(\tau ^*\) defined in (2.8) is a map from \(\mathbb {T}^{ii}\) to \(\mathcal {T}\).

### Proof

It remains to show that \({\varvec{\rho }}^*\) satisfies the non-anticipative condition of Type II in (2.1). Take \(\tau _1,\tau _2\in \mathcal {T}\). If \({\varvec{\rho }}^*(\tau _1)<\tau _1\wedge \tau _2\le \tau _1\), then \(\tau _1>\rho _d\) and thus \({\varvec{\rho }}^*(\tau _1)=\rho _d<\tau _1\wedge \tau _2\le \tau _2\), which implies \({\varvec{\rho }}^*(\tau _2)=\rho _d={\varvec{\rho }}^*(\tau _1)<\tau _1\wedge \tau _2\). If \({\varvec{\rho }}^*(\tau _1)\ge \tau _1\wedge \tau _2\), then if \({\varvec{\rho }}^*(\tau _2)<\tau _1\wedge \tau _2\) we can use the previous argument to get that \({\varvec{\rho }}^*(\tau _1)={\varvec{\rho }}^*(\tau _2)<\tau _1\wedge \tau _2\) which is a contradiction, and thus \({\varvec{\rho }}^*(\tau _2)\ge \tau _1\wedge \tau _2\). \(\square \)

### Lemma 4.3

### Proof

### Lemma 4.4

### Proof

## 5 Some Insight Into the Continuous-Time Version

We need to make sure that \(V^1\) and \(V^2\) defined in (2.3) and (2.4) have RCLL modifications.

On an intuitive level, the optimizers (or choose to be \(\epsilon \)-optimizers in continuous time) \(\rho _u(\cdot )\) and \(\tau _u(\cdot )\) are maps from \(\mathcal {T}\) to \(\mathcal {T}\). Yet this may not be easy to prove in continuous time, as opposed to the argument in Lemma 4.2.

*U*in (

*s*,

*t*) (maybe also in \(\omega \)). On the other hand, with such continuity, there will essentially be no difference between using stopping strategies of Type I and using stopping strategies of Type II, as opposed to the discrete-time case (see Examples 1.1 and 1.2).

## Footnotes

- 1.
For example, let \(f_t=g_t=t\). Then \(U(s,t)=s\wedge t\). By setting \(\rho \equiv 0\) we can see that \(\overline{A}=\underline{A}=0\).

- 2.
To wit, let \({\varvec{\rho }}\in \mathbb {T}^i\) and take \(\sigma _1,\sigma _2\in \mathcal {T}\). If \({\varvec{\rho }}(\sigma _1)\wedge {\varvec{\rho }}(\sigma _2)>\sigma _1\wedge \sigma _2\), then (1.7) is satisfied. Now assume \({\varvec{\rho }}(\sigma _1)={\varvec{\rho }}(\sigma _2)\le \sigma _1\wedge \sigma _2\). If the strict inequality holds, then the first part of (1.7) is satisfied. Otherwise, \({\varvec{\rho }}(\sigma _1)={\varvec{\rho }}(\sigma _2)=\sigma _1\wedge \sigma _2\), which implies the second part of (1.7).

## Notes

### Acknowledgements

This research was supported in part by the National Science Foundation under grant DMS-1613170.

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