Recursive Inspection Games

Transcription

1 Recursive Inspection Gaes Bernhard von Stengel February 7, 2016 arxiv: v2 [cs.gt] 7 Feb 2016 Abstract We consider a sequential inspection gae where an inspector uses a liited nuber of inspections over a larger nuber of tie periods to detect a violation (an illegal act of an inspectee. Copared with earlier odels, we allow varying rewards to the inspectee for successful violations. As one possible exaple, the ost valuable reward ay be the copletion of a sequence of thefts of nuclear aterial needed to build a nuclear bob. The inspectee can observe the inspector, but the inspector can only deterine if a violation happens during a stage where he inspects, which terinates the gae; otherwise the gae continues. Under reasonable assuptions for the payoffs, the inspector s strategy is independent of the nuber of successful violations. This allows to apply a recursive description of the gae, even though this norally assues fully infored players after each stage. The resulting recursive equation in three variables for the equilibriu payoff of the gae, which generalizes several other known equations of this kind, is solved explicitly in ters of sus of binoial coefficients. We also extend this approach to non-zero-su gaes and, siilar to Maschler (1966, inspector leadership where the inspector coits to (the sae randoized inspection schedule, but the inspectee acts legally (rather than ixes as in the siultaneous gae as long as inspections reain. Keywords: inspection gae, ultistage gae, recursive gae, Stackelberg leadership, binoial coefficients MSC2010 subject classification: 91A05, 91A20 OR/MS subject classification: Gaes/group decisions: Noncooperative; Military: Search/surveillance JEL subject classification: C72 Published in: Matheatics of Operations Research (2016, Departent of Matheatics, London School of Econoics, London WC2A 2AE, United Kingdo. Eail: stengel@nash.lse.ac.uk 1

2 1 Introduction Inspection gaes odel situations where an inspector with liited resources verifies by eans of inspections that an inspectee adheres to legal obligations, which the inspectee has an incentive to violate. Inspection gaes have been applied to ars control and disaraent, tax auditing, fare evasion, environental pollution, and hoeland security; for a survey see Avenhaus, von Stengel, and Zair [5], Avenhaus and Canty [2], and other recent works such as [9] or [8]. This paper presents a generalization of a classical sequential inspection gae by Dresher [10]. In Dresher s gae, the inspector has to distribute a given nuber of inspections over a larger nuber of inspection periods to detect a violation that the inspectee, who can count the inspector s visits, perfors in at ost one of these periods. In our extension of this gae, the inspectee ay violate ore than once, and collect a possibly different reward for each successful violation; for exaple, a violation ay be the diversion of a certain aount of nuclear aterial in a tie period, with the highest reward to the inspectee once he has diverted enough aterial to build a nuclear bob. As in Dresher s gae, the gae ends if a violation is discovered by the inspector who inspects at the sae tie. This is in line with an application to ars control, and ay also apply in other contexts where an identified violator of legal rules becoes subject to uch tighter surveillance. A central aspect of our odel and its analysis, and the reason for its choice of paraeters, is that the inspector s ixed strategy in equilibriu does not depend on whether a successful violation took place during a tie period without an inspection, about which the inspector is norally not infored. As we will explain in 3, the gae can therefore, despite this lack of inforation, be described recursively by a sequence of 2 2 gaes for each stage. As long as there are reaining inspections (but fewer than the nuber of reaining tie periods and intended violations, the inspector and inspectee randoize at each stage whether to inspect and to violate. For the payoffs and ixed strategy probabilities in equilibriu we give explicit solutions in ters of the gae paraeters. Our analysis starts with a zero-su gae, which is then extended to non-zero-su payoffs. Furtherore, if, as in Maschler [19], the inspector can coit to his ixed equilibriu strategy, the inspectee will act legally as long as inspections reain. This coitent power, known as inspector leadership, increases the inspector s payoff. The ain precursor to this work is [26], which, however, only considers the extra paraeter of a varying nuber of intended violations, not different rewards for the. Inspection gaes with two paraeters (tie periods and inspections were considered by Dresher [10], Thoas and Nisgav [24], and Baston and Bostock [6]. Maschler [19] introduced non-zero-su gaes and inspector leadership. Höpfinger [16] and Avenhaus and von Stengel [4] extended Dresher s odel to non-zero-su payoffs. Rinderle [21] studied the case that inspections ay have probabilities of false alars and non-detection of a violation. Avenhaus and Canty [1] considered sequential inspections where tieliness of detection atters. Gaes with a third paraeter of intended violations were considered by Kuhn [18], Sakaguchi [22, 23], and Ferguson and Melolidakis [13]. In these odels, the gae continues even after a detected violation, unlike in our odel. In addition, the 2

3 inspector is fully infored after each stage whether a violation took place or not even when he did not inspect. As already noted by Kuhn [18, p. 174], this full inforation is iplicit in a recursive description. In 2, we describe the inspection gae and its paraeters. The recursive equation for the value of this zero-su gae is solved explicitly. The equilibriu strategy of the inspector depends only on the nuber of reaining tie periods and inspections. Section 3 discusses the key property that the strategy of the inspector does not depend on the inspectee s intended violations and their rewards, which allows to apply the solution also when the inspector has no inforation about undiscovered violations. We also show that our odel is as general as possible to achieve this property. In 4, we show how to extend the solution relatively easily to the non-zero-su gae where a detected violation incurs a negative cost to both players copared to the case of legal action and no inspection. The inspector leadership gae is studied in 5. In 6, we discuss possible extensions of our odel, and general aspects of the recursive gaes we consider, in particular coputational advantages copared to gaes in extensive for. 2 Zero-su inspection gae with ultiple violations We consider a two-player gae Γ(n,,k, were n,,k are three nonnegative integer paraeters. The gae is played over n discrete tie periods. The nuber is the nuber of inspections available to the inspector (the first player. The nuber k is the axiu nuber of intended violations of the inspectee (the second player. In each tie period, the inspector can use one of his inspections (if > 0 or not, and siultaneously (if k > 0 the inspectee chooses between legal action and violation. The gae has also a real-valued penalty paraeter b and nonnegative reward paraeters r k,r k 1,...,r 1 that deterine the payoffs, as follows. In this section, we assue that the payoffs are zero-su. Let v(n,,k be the value of the gae Γ(n,,k, as the equilibriu payoff to the inspector. If n = 0, then the gae is over and v(n,,k = 0. More generally, if n, then the inspector can inspect in every reaining tie period, where we assue that the inspectee acts legally throughout, with v(n,,k = 0 if n. (1 If n > 0, the gae Γ(n,,k is described recursively. Suppose first that n > > 0, so that the inspector decides whether to use one of his inspections or not, and k > 0, so the inspectee decides whether to act legally or to violate. The recursive description of Γ(n,,k, with value v(n,,k, is given by the following payoffs to the inspector, which are the costs to the inspectee, at the first tie period. inspectee inspector legal action violation inspection v(n 1, 1,k b r k (2 no inspection v(n 1,,k v(n 1,,k 1 r k 3

4 Of the four possible cobinations of the player s actions in the first period, one of the terinates the gae, naely when the inspector inspects and the inspectee perfors a violation, which we assue is caught with certainty. In all other cases, the gae continues. If the inspectee acts legally and the inspector inspects, then the gae continues as the gae Γ(n 1, 1,k. If the inspectee acts legally and the inspector does not inspect, then the gae continues as the gae Γ(n 1,,k. If the inspectee violates and the inspector did not inspect, then the gae continues as the gae Γ(n 1,,k 1, where in addition the inspectee collects the reward r k which he can keep even if he is caught in a later tie period. The corresponding botto-right cell in (2 has therefore payoff entry v(n 1,,k 1 r k to the inspector. If the gae terinates because the inspectee is caught, we assue that inspectee has to pay the penalty b r k, which is proportional to his reward r k if the violation had been successful, ultiplied by the penalty factor b. We assue only that b > 1, to allow for the possibility (in particular when payoffs are no longer zero-su, discussed in 4 that even a caught violation is less preferred by the inspector than legal action (with reference payoff zero. A single successful (uncaught violation with payoff r k should however still be worse for the inspector than a caught violation with payoff b r k, hence the requireent that b > 1. This condition holds obviously when b > 0 where a caught violation creates an actual penalty to the inspectee that is worse than legal action. The nonnegative rewards to the inspectee r k,r k 1,...,r 1 are nubered in that order to identify the fro the gae paraeter k in Γ(n,,k as the gae progresses. That is, r k is the reward for the first successful violation, r k 1 for the second, and so on until the reward r 1 for the kth and last violation if the gae has not ended earlier. The inspectee can perfor at ost one violation per tie period. Hence, if there are no inspections left ( = 0, then the inspectee can violate in each of the reaining n tie periods up to k ties in total, that is, in{k,n} v(n,0,k = i=1 r k+1 i. (3 We allow soe rewards to be zero. If all reaining rewards are zero, then this gives the sae payoffs as when the inspectee only acts legally fro now on, so this ay instead be represented by a saller k. However, the ter v(n,0,k in (3 ay be zero even if soe reaining rewards are nonzero. This case ay arise in the course of the gae after soe tie periods without violations so that n < k, for exaple when n = 1, k = 2, r 2 = 0, r 1 = 1, so we allow for this possibility. The gae Γ(n,,k is copletely described by the base cases (1 and (3 (both of which iply v(0,, k = 0 and the recursive description (2. This description of the gae assues that both players are fully infored about the other player s action after each tie period, and thus know in which of the four cells in (2 the gae continues. We call this the gae with full inforation and will weaken this assuption in 3. The following ain theore gives an explicit forula for the gae value v(n,,k and the optial inspection strategy. A large part of the proof is to show that (2 has a circular preference structure and hence a ixed equilibriu. The ost iportant, but very direct 4

5 part of the proof (fro (28 onwards is that the explicit representation (6 holds. We discuss a possible derivation of the ter t(n,,k in 5 after Theore 5. Theore 1 Let n,,k be nonnegative integers, b > 1, and r k,r k 1,,r 1 0. Define ( n s(n, = b i (4 i and t(n,,k = i=0 k i=1 ( n i r k+1 i. (5 Then the zero-su gae Γ(n,,k defined by (2 for n > > 0 and k > 0 and by (1 and (3 otherwise has value v(n,,k = t(n,,k. (6 s(n, For n > > 0 and k > 0, the gae (2 has a copletely ixed equilibriu where the inspector inspects with probability p and the inspectee violates with probability q, where p = s(n 1, 1 s(n,, 1 p = s(n 1, s(n,, (7 and q = v(n 1,,k v(n 1, 1,k v(n 1,,k v(n 1, 1,k + b r k v(n 1,,k 1 + r k. (8 This equilibriu is unique, unless r k+1 i = 0 for 1 i in{k,n }, in which case all entries in (2 are zero and the players can play arbitrarily. Proof. Proof. We first consider soe properties of s(n, as defined in (4. Clearly, and Furtherore, b s(n 1, 1 = s(n,0 = 1, s(n,n = (1 + b n (9 1 i=0 ( n 1 i b i = s(n 1, ( n 1. (10 s(n, = s(n 1, 1 + s(n 1, (0 < < n (11 which holds because ( ( n n s(n, = b i = b + i=0 i 0 ( n 1 = b ( n i i=1 i=1 b i + ( ( ( n 1 n 1 b i + i i 1 ( n 1 b 1 i 1 i=0 i (12 = s(n 1, + s(n 1, 1. 5

6 This eans s(n, is uniquely defined inductively by (9 and (11. Recall that ( x 0 = 1 for any x [12, p. 50]. The following alternative representation s(n, = i=0 ( n 1 i i (1 + b i (13 holds because it also fulfills (9 and (11, which is shown siilarly to (12. Because b > 1, we have s(n, > 0 for 0 n by (13 (or by (9 and (11. The ain assertion to prove is the explicit representation (6 for v(n,, k. Clearly, v(n,,0 = 0, and (3 and (1 hold because in (5, ( n i = 0 if i > n. Hence, we can assue that n > > 0 and k > 0 where the recursive description (2 applies. By induction on n, we can assue as inductive hypothesis that A = v(n 1, 1,k, C = v(n 1,,k, D = v(n 1,,k 1 r k (14 are given using (6. These nubers and B = b r k define the gae (2 as 1 q q p A B 1 p C D (15 which also shows the probabilities p, 1 p and 1 q, q of playing the rows and coluns. To coplete the induction, we will show that this gae has value v(n,,k as in (6. If r k+1 i = 0 for 1 i in{k,n }, then by (5 and (6 A = B = C = D = 0, so this is the all-zero gae with value zero in agreeent with (6, and arbitrary equilibriu strategies of the players. So assue that this is not the case, so that and hence A < 0. t(n 1, 1,k > 0 (16 Intuitively, (2 has a ixed equilibriu because the inspector prefers not to inspect if the inspectee acts legally and to inspect if he violates, and the inspectee prefers to act legally if inspected and to violate otherwise. In (15, this holds if It is easy to see (and well known that then with p = A < C, B > D, A < B, C > D. (17 C D B A +C D, 1 p = B A B A +C D, (18 B D 1 q = C A + B D, q = C A C A + B D, (19 the gae has value v(n,,k, where v(n,,k = p A + (1 p C = p B + (1 p D, (20 v(n,,k = (1 q A + q B = (1 q C + q D, (21 6

7 where p in (18 is uniquely deterined by (20 and q in (19 is uniquely deterined by (21, and p, 1 p, q, and 1 q are all positive by (17. For (17, we first show A < C. By (6, this is equivalent to or, by (16, to s(n 1, s(n 1, 1 > t(n 1, 1,k s(n 1, 1 t(n 1,,k t(n 1, 1,k = r k < t(n 1,,k s(n 1, ( n 2 ( + n 3 ( rk n 1 k r1 ( r n 2 ( k + n 3 ( rk n 1 k. (22 r1 Assue that k n, otherwise replace k by n because ( n 1 i ( 1 = n 1 i = 0 for i > n. We show that the right expression in (22 is largest when r k > 0 and r k 1 = = r 1 = 0. Naely, for general nonnegative ρ 1,...,ρ k, not all zero, positive h 1,...,h k, and any g 1,...,g k so that we have g 1 h 1 g 2 h 2 g k h k, (23 g 1 h 1 ρ 1g ρ k g k ρ 1 h ρ k h k (24 which is seen by induction as follows. By oitting the ters where ρ i = 0, we can assue ρ i > 0 for all i. For k = 1, (24 is true. For k > 1, let G = ρ 1 g 1 +ρ 2 g 2 and H = ρ 1 h 1 +ρ 2 h 2. Then g 1 G h 1 H = ρ 1g 1 + ρ 2 g 2 g 2 (25 ρ 1 h 1 + ρ 2 h 2 h 2 because the left inequality in (25 is equivalent to g 1 (ρ 1 h 1 + ρ 2 h 2 h 1 (ρ 1 g 1 + ρ 2 g 2 and thus to g 1 ρ 2 h 2 h 1 ρ 2 g 2 which holds by (23; the right inequality in (25 is shown siilarly. This shows (24 for k = 2, and for k > 2 using the inductive hypothesis G H g 3 g k G h 3 h k H G + ρ 3g ρ k g k. H + ρ 3 h ρ k h k With ρ i = r k+1 i, g i = ( n 1 i, hi = ( n 1 i 1 for 1 i k we have g i h i = n i and thus (23 and (24, so (22 holds if ( n 2 s(n 1, s(n 1, 1 > = n 1 (26 ( n 2 1 which we now show. By (13, the following are equivalent: (1 + b + 1 i=0 i=0 ( n 1 i 1 i s(n 1, > n 1 s(n 1, 1, ( n 1 i (1 + b i > n 1 1( n 1 i i (1 + b i, i=0 1 i ( n n 1 i n 1 i (1 + bi > (1 + b i, 1 i 7 1 i=0

8 which is true because 0 < < n and thus n and thus A < C. i > n 1 for 0 i <. This shows (26 The reaining inequalities in (17 are seen as follows. Because b > 1 and r k 0, we have B = b r k 0 r k v(n 1,,k 1 r k = D, with inequality possible only if r k = 0 and t(n 1,,k 1 = 0, which because t(n 1,,k 1 = r k 1 ( n 2 + r k 2 ( n 3 ( n k + + r 1 eans r k+1 i = 0 for 1 i in{k,n } which we have excluded. So B > D. Suppose p given by (7 fulfills (20, which iplies (18. By (11, the real nuber p defined in (7 is indeed a probability. Also, p > 0 and 1 p > 0, so that by (18 either C < D and B < A or C > D and B > A. The forer can be excluded because it would iply B < A < C < D < B. This proves (17. So it reains to show (20, that is, (27 v(n,,k = p v(n 1, 1,k + (1 p v(n 1,,k, (28 v(n,,k = p b r k + (1 p (v(n 1,,k 1 r k. (29 After ultiplication with s(n,, (28 and (29 are by (6 and (7 equivalent to t(n,,k = t(n 1, 1,k t(n 1,,k, (30 t(n,,k = s(n 1, 1 b r k + ( t(n 1,,k 1 s(n 1, r k. (31 Equation (30 holds because, by (5, t(n 1, 1,k +t(n 1,,k = = k i=1 k i=1 Equation (31 holds because, by (10 and (27, ( n 1 i k ( n 1 i r k+1 i + 1 r k+1 i i=1 ( n i r k+1 i = t(n,,k. (32 s(n 1, 1 b r k + ( t(n 1,,k 1 s(n 1, r k = (s(n 1, ( n 1 rk + ( t(n 1,,k 1 s(n 1, r k = ( n 1 rk t(n 1,,k 1 = t(n,,k. (33 This shows (28 and (29, which copletes the induction on n. The inspectee s violation probability q in (8 is just given by (19. By Theore 1, the gae in (2 has a unique ixed equilibriu (unless all payoffs are zero. This uniqueness applies recursively to all stages of Γ(n,,k if the players use behavior strategies. The sae probabilities for their actions could result fro ixed strategies that correlate these actions, which we do not consider because behavior strategies suffice [17]. 8

9 A special case of Theore 1 has been shown in [26], naely when r i = 1 for k i 1. In that case, t(n,,k in (5 can be written as t(n,,k = k i=1 (see Feller [12, p. 63, equation (12.6]. ( n i = ( n + 1 ( n k + 1 Dresher [10] actually considered two special cases of this gae for b = 1, naely k = 1 and k = n, where (34 siplifies to ( ( n 1 n t(n,,1 = and t(n,,n =. ( The corresponding expressions (6 were stated and proved by Dresher [10], and, apparently independently, by Sakaguchi [23]. (34 3 Discussion of the odel and interpretation of the ain theore In this section, we discuss the ain Theore 1, in particular the fact that the inspector s equilibriu strategy depends only on the nuber of tie periods and inspections. Consequently, the sae strategy also applies to a new gae Γ (n,,k where the inspector is not infored about violations at previous tie periods when he did not inspect, which we call the gae without full inforation. In a basic for, this assuption is iplicit in the odels by Dresher [10]. The recursive definition of Γ(n,,k as in (2 allows to copute the gae value even without an explicit forula as stated in (6. If a gae as in (15 fulfills the inequalities (17 so that the gae has a ixed equilibriu, then the equilibriu probabilities (18 BC AD and (19 give the value of the gae as B A+C D. Sakaguchi [23] recursively coputes the gae value v(n,,k in this way for different entries in (2. As entioned, the recursive description (2 assues that, in particular, the inspector knows whether the inspectee chose legal action or violation even after a tie period where the inspector did not inspect. In practice, it ay be rather questionable how the inspector would obtain this knowledge. In the gaes studied by Dresher [10], it actually does not atter whether the inspector has this knowledge or not. In Dresher s first gae, the inspectee has only a single intended violation, corresponding to k = 1 in our odel (and, throughout, r i = 1 for all i. Then the lower-right entry in (2 given by v(n 1,,0 r 1 is equal to 1. In that case, because the inspectee has successfully violated once and will not violate further, the gae is effectively over because the inspectee acts legally fro then on and will not be caught. Then any action of the inspector is optial, and so the inspector can act as if the violation is still to take place. That is, if the inspector does know whether he is in the gae Γ(n 1,,1 or Γ(n 1,,0 (the latter with added payoff 1 due to the uncaught violation, then 9

10 he can always act as if he is in the gae Γ(n 1,,1 because that is the only situation where his strategy atters. Therefore, the recursive description is justified. In the second gae described by Dresher [10], the inspectee tries to violate as often as possible. This corresponds to our gae Γ(n,,n because the inspectee can only violate once per tie period and will therefore not violate ore than n ties because otherwise he would be caught with certainty. Then the botto-left and botto-right entries in (2 are v(n 1,,n and v(n 1,,n 1 1, respectively. However, the lower-left gae Γ(n 1,,n (where the inspectee has issed out to violate during an uninspected tie period is equivalent to the gae Γ(n 1,,n 1, that is, again a gae with a axial nuber of intended violations. The botto-right gae is the sae, except for the added 1 to the inspector s payoff, so again it does not atter whether the inspector knows if the inspectee violated or not. Dresher [10] gave explicit values for these two gaes as in (35. However, Dresher did not copute the optial inspection probabilities, because he would otherwise ost likely have noted that they are the sae in the two gaes Γ(n,,1 and Γ(n,,n. These inspection probabilities are given by (7. A key aspect of our odel is that they hold, independently of k, in the gae Γ(n,,k with the nuber k of intended violations as a new paraeter. Because of this independence of k, the equilibriu strategy of the inspector, and the gae value v(n,,k, apply also to the gae Γ (n,,k without full inforation where the inspector does not know if a violation occurred or not in an uninspected tie period. Naely, by induction the inspection strategy is the sae in the two gaes Γ (n 1,,k and Γ (n 1,,k 1 which correspond to the two botto cells in (2, the latter with an additional loss of r k to the inspector, as long as the inspectee has still an incentive to violate; if that is not the case, as in the gae Γ (n 1,,0 which has value zero, then any inspection strategy is optial and so the inspector should act as if there are still violations to take place because only then his action atters, as in Dresher s first gae. Forally, the gae Γ (n,,k without full inforation is not described recursively. However, it can be odelled as an extensive for gae with inforation sets [17] that represent the inspector s lack of inforation. If we then change the gae to the gae with full inforation, then these inforation sets are cut, which transfors Γ (n,,k into the recursively described gae Γ(n,,k in (2. Because the inspector s behavior strategy in Γ(n,,k is the sae at all inforation sets obtained fro cutting an inforation set h, say, in the original gae Γ (n,,k, it can also be defined uniquely as the behavior at h and thus defines a behavior strategy for Γ (n,,k. In particular, the value of Γ (n,,k stays the sae at v(n,, k. This (straightforward anipulation of inforation sets is described in detail in [26]. In suary: Corollary 1 The equilibriu payoff and the equilibriu strategies for the inspection gae with full inforation described in Theore 1 also apply in the gae Γ (n,,k without full inforation where the inspector is not infored about the action of the inspectee after a tie period without inspection. In the gae Γ (n,,k without full inforation, the inspectee has typically additional equilibriu behavior strategies copared to Γ(n,,k. As an exaple, let n,,k = 3,1,2 10

11 and b = r 2 = r 1 = 1. Then the botto cells of (2 both correspond to the gae Γ(2,1,1, with added payoff 1 in the botto-right cell. In Γ(2,1,1, which is (15 with A,B,C,D = 1,1,0,1, the optial strategies are p = q = 1/3, with v(2,1,1 = 1/3. At the first stage in Γ(3,1,2, they are p = 1/4 and q = 5/12, which also applies to Γ (3,1,2. However, in the gae Γ (3,1,2, the inspector does not know if the inspectee violated in the first tie period or not, which gives the inspectee additional optial behavior strategies for the second tie period. For exaple, the inspectee can violate with probability 4/7 if he acted legally in the first period and violate with probability zero if he violated in the first period. Another such coordinated different behavior in the second tie period would be to violate with probability zero following legal action in the first period and to violate with probability 4/5 following a violation in the first period. We next discuss the rewards to the inspectee r k,...,r 1 for successful violations, and the corresponding scaled penalty b r k to the inspectee in (2. These paraeters are new copared to von Stengel [26], who proved Theore 1 with r i = 1 for k i 1. With general nonnegative rewards r i, it sees that one can dispense with the paraeter k and siply assue that only the first k rewards r i are nonzero if the inspectee intends only k violations. In one respect this is a different gae than when the inspectee will not carry out ore than k violations, because when all rewards are zero, the inspectee can violate and be caught without penalty, which just terinates the gae; one ay argue that this is an acceptable gae outcoe that just has to be interpreted appropriately. The ain reason for the paraeter k in the recursive description of the gae is to identify the next reward to the inspectee after a successful violation when the gae continues in the botto-right cell in (2. The nuber of intended violations serves as a counter for the rewards, which we have therefore nubered in the order r k,r k 1,...,r 1. Such a counter is needed for the recursive description in one way or another. The payoff b r k to the inspector for a caught violation ay see strange in the gae Γ (n,,k where the inspector is not infored about k. However, we think it is justifiable to ake the stakes of a violation proportional to r k even if the inspector does not know r k, because the inspectee knows what is at stake. We have chosen this payoff as b r k because otherwise the optial inspection strategy would not be independent of k as it is according to the solution (7. In fact, the next theore states that the payoffs in (2 are as general as possible so that this independence holds. For siplicity, we assue that the arginal gain r j to the inspectee for the next of j reaining violations is always positive, and that the gae has a circular preference structure. Theore 2 Suppose that n,,k are the nuber of tie periods, inspections, and intended violations in a zero-su inspection gae where the inspectee can violate at ost once per tie period, where his overall payoff depends only on (and is strictly increasing in the total nuber of successful violations, and whether he is ever caught (in which case the gae terinates or not. Consider this gae with full inforation and value v(n,,k. Then the ost general for of this gae fulfills (1 and (3, and for n > > 0 and k > 0 is the gae p v(n 1, 1,k f (k (36 1 p v(n 1,,k v(n 1,,k 1 r k 11

12 siilar to (2, where we assue that (36 has a unique copletely ixed equilibriu. Here f (k is the arginal penalty and r k is the arginal gain to the inspectee for the first of k reaining violations, r k > 0. Then the probability p of inspection is independent of k (so that it can be applied to the gae without full inforation if and only if there is soe b > 1 so that f (k = b r k for all k, as in (2. Proof. Proof. Consider the gae with full inforation. At the beginning of the gae, we can assue k n because the inspectee will not perfor ore than n violations because he would otherwise be caught with certainty. For 1 i n, let r k i+1 be the arginal gain to the inspectee for the ith successful violation, which by assuption is strictly positive. Suppose that over the n tie periods, the inspectee perfors i successful violations, 0 i k, and thus gains r k + r k r k i+1. This is his payoff (and loss to the inspector, in copletely general for, if he is not caught. If the inspectee is caught when attepting the (i + 1st violation, then he pays the penalty f (k i, which is subtracted fro this su (this penalty ay include, for exaple, repaying all previous gains; the inspector s payoff is then r k r k 1 r k i+1 + f (k i. Then v(n,,k = 0 when k = 0 or n as in (1 (for legal action throughout, and v(n,0,k given by (3. For n > > 0 and k > 0, the gae with value v(n,,k is given by (36, which is therefore the general for of an inspection gae under the stated assuptions. If k = 1, there is only one paraeter f (1 so that we can set b = f (1/r 1 and the inspector s strategy can be applied to the gae without full inforation; this is essentially the first gae by Dresher [10]. Hence, we can assue k 2. Let j 2 and suppose that the inspectee has perfored k j successful violations (and therefore, so far, gained r k + r k r j+1, that the inspector has perfored 1 inspections, and that n 3 tie periods have passed. The successful violations and the inspections have to take place in different tie periods, which is possible because k j + 1 n = n 3, and this occurs with positive probability because of the ixed equilibriu at every stage of the gae. Then at this stage there are three tie periods, one inspection, and j intended violations reaining, and the reaining gae has value v(3,1, j and is of the for In the botto left cell of (37, v(2,1, j is the value of v(2,0, j f ( j. (37 v(2,1, j v(2,1, j 1 r j v(1,0, j f ( j, that is, of v(1,1, j v(1,1, j 1 r j p r j f ( j. (38 1 p 0 r j In the botto right cell of (37, the inspectee collects a reward of r j, and v(2,1, j 1 is the value of the gae v(1,0, j 1 f ( j 1, that is, of v(1,1, j 1 v(1,1, j 2 r j 1 p r j 1 f ( j 1. 1 p 0 r j 1 (39 12

13 The two gaes in (38 and (39 correspond to the two cells in the botto row of (37 and both ust have the sae probability p of inspection if this is to be applied to the gae without full inforation. That is, according to (18, 1 p p = 1 p 1 = f ( j + r j = f ( j + 1 = f ( j 1 + r j 1 = f ( j (40 r j r j r j 1 r j 1 For j = 2, this shows f (2/r 2 = f (1/r 1 =: b, where b > 1 because 1/p 1 > 0. For j = 3 it shows f (3/r 3 = f (2/r 2, and so on, so that f ( j/r j = b for all 1 j k, as claied. Another question is if there is an intuitive reason that the inspector s optial strategy in Γ(n,,k does not depend on k (for any, not just for = 1 as in the proof of Theore 2. For exaple, Ferguson and Melolidakis [14] have applied a gae with finite resources due to Gale [15] to a different inspection gae where the solution also applies when one of the players lacks inforation. However, we have not been able to apply the highly syetrical strategy in this gae to our setting. At present, the very canonical proof (see equation (28 and onwards of the explicit representation (7 and (6 sees to be the best explanation. To conclude this section, we discuss the solution of the gae Γ(n,,k for soe siple special cases. If = 1, then it is easy to see that the inspector uses his single inspection in the first n 1 tie periods with equal probability, which for the last tie period is ultiplied with 1 + b, so if b > 0 then higher probability is given to the last period. The case b = 0, where a caught violation terinates the gae but no further penalty applies, has also soe easily described properties. Then s(n, = ( n and thus p = /n in (7, which eans that all -sets of the n tie periods are equally likely to be inspected. Moreover, if r i = 1 for k i 1, then t(n,,k = k ( n k i=1 by (5, and v(n,,k = t(n,, k/s(n, can be interpreted as the expected nuber of successful violations, as follows: The inspectee is indifferent between all possible tie periods for choosing his k violations, and thus gets payoff v(n,,k if he violates in the first k tie periods. Then, if the -set of inspections does not include period 1 (with ( n 1 ( out of n choices, the first violation succeeds. If this set also does not include period 2 (with further ( n 2 choices, then the first and second violation succeed, and so on. The probability q of violation in the first period depends on k, and for b = 0 and r i = 1 for k i 1 has the following for. If k = 1, then q = 1/n, independently of. If k = n (where the inspectee violates as often as possible, then q = 1/( + 1, independently of n. Unfortunately, there is no straightforward siple extension of these values for interediate values of k. In general, we have only found coplicated expressions for the inspectee s strategy, which is why we have left it in the for (8 derived fro the well-known representation (19 in ters of the gae payoffs. 4 Non-zero-su payoffs In this section, we extend the zero-su-gae Γ(n,,k to a non-zero-su gae ˆΓ(n,,k. The reason to consider non-zero-payoffs is that a caught violation as the outcoe of the 13

14 gae is typically less preferred by both inspector and inspectee than legal action, because for the inspector it eans the failure of the inspection regie. This is a standard assuption in inspection gaes, first proposed by Maschler [19]. We denote the equilibriu payoffs in ˆΓ(n,,k by v(n,,k for the inspector and by w(n,,k for the inspectee (which are unique as shown in Theore 4 below. The reference payoff for legal action throughout is zero for both players. As before, we assue that the inspectee acts legally if the inspector can inspect in every reaining period, that is, v(n,,k = w(n,,k = 0 if n. (41 Also as before, if the inspector has run out of inspections, then the inspectee collects a nonnegative reward r k,r k 1,... which is a cost to the inspector for each reaining period up to the axiu nuber k of intended violations, that is, v(n,0,k = w(n,0,k = in{k,n} r k+1 i. (42 i=1 For the case that a violation is caught we introduce two paraeters a and b as costs to inspector and inspectee (scaled by the reward r k for a successful violation, where 0 < a < 1, b 0, (43 so that for n > > 0 and k > 0 the gae ˆΓ(n,,k (with full inforation has the following recursive description: inspectee inspector inspection no inspection v(n 1, 1, k v(n 1,,k legal action w(n 1, 1, k w(n 1,,k a r k violation w(n 1,,k 1 + r k v(n 1,,k 1 r k b r k In (44, the arrows represent the circular preferences of the players, which have been proved for Γ(n,,k as (17. In particular, if k = 1, then the botto-right cell in (44 for an uncaught violation has payoff r k to the inspector, whereas the top-right cell has payoff a r k. Because a < 1, the inspector therefore prefers a caught violation to an uncaught one, as it should be the case. Due to (43, the gae ˆΓ(n,,k does not include the zero-su gae Γ(n,,k as a special case. However, the ore general conditions a < 1 and b > 1 do include it when a = b. The following theore is essentially a corollary to Theore 1. (44 14

15 Theore 3 Let n,,k be nonnegative integers, let the reals a and b be as in (43, and let r k,r k 1,,r 1 0. Define s(n,,k as in (4, t(n,,k as in (5, and ŝ(n, by ŝ(n, = i=0 ( n ( a i. (45 i Then the non-zero-su gae ˆΓ(n,,k defined by (44 for n > > 0 and k > 0 and by (41 and (42 otherwise has equilibriu payoffs to inspector and inspectee v(n,,k = t(n,,k ŝ(n,, w(n,,k = t(n,,k s(n,. (46 For n > > 0 and k > 0, the gae (44 has a copletely ixed equilibriu where the inspector inspects with probability p according to (7, and the inspectee violates with probability q according to (8. This is the unique subgae perfect Nash equilibriu (SPNE of the gae, unless r k+1 i = 0 for 1 i in{k,n }, in which case all entries in (44 are zero and the players can play arbitrarily. Each player s strategy is the in-ax strategy for the payoffs of his opponent. Proof. Proof. If we odify the gae ˆΓ(n,,k to a zero-su gae with the payoffs v(n,,k to the inspector (and thus v(n,,k to the inspectee, then it fulfills the assuptions of Theore 1 with b = a > 1. In this gae, the inspector prefers not to inspect when the inspectee acts legally and to inspect when the inspectee violates, as shown with the vertical arrows in (44. The inspectee s strategy is as in (8 and is a inax strategy for the inspector s payoff. It akes the inspector indifferent between his two actions, with the inspector s payoff v(n,,k as in (46. Note that in (45, ŝ(n, > 0 for 0 n due to the alternative representation (13 where 1 + b = 1 a > 0. Siilarly, if we odify the gae ˆΓ(n,,k to a zero-su gae based on the payoffs w(n,,k to the inspectee (and thus w(n,,k to the inspector, then it fulfills also the assuptions of Theore 1 with b 0 > 1. Then the inspectee prefer to act legally when he is inspected and to violate otherwise, as shown with the horizontal arrows in (44. In this gae, the inspector s strategy is given by (7, and is a in-ax strategy for the inspectee s payoff. It akes the inspectee indifferent between his two actions in (44, with the inspectee s payoff w(u,,k as in (46. So the gae in (44 has a circular preference structure and a unique ixed equilibriu as described (except when r k+1 i = 0 for 1 i in{k,n }, which by induction defines the unique SPNE of ˆΓ(n,,k. In Theore 3, the inspector s strategy in ˆΓ(n,,k does not depend on k. As argued in 3, this strategy can therefore also be applied to the gae ˆΓ (n,,k without full inforation. That is, we obtain the analogous stateent to Corollary 1. Corollary 2 The equilibriu payoff and the equilibriu strategies for the non-zero-su inspection gae described in Theore 3 with full inforation also apply in the gae ˆΓ (n,,k without full inforation. 15

16 As entioned after Corollary 1, the gae Γ (n,,k without full inforation ay have additional equilibriu strategies for the inspectee, which applies in the sae way to the gae ˆΓ (n,,k. Because the gaes ˆΓ(n,,k and ˆΓ (n,,k are not zero-su, the question arises if they have other Nash equilibriu payoffs. The following theore asserts that this is not the case. Theore 4 All Nash equilibria of the non-zero-su inspection gae ˆΓ(n,, k and the gae ˆΓ (n,,k without full inforation have the payoffs described in Theore 3. Proof. Proof. Consider first the gae ˆΓ(n,, k with full inforation. Let the gae be represented as an extensive gae. Call a stage of the gae a particular tie period together with the history of past actions. At each stage, we let, as in [26], the inspector ove first and the inspectee second, where the decision nodes of the inspectee belong to a two-node inforation set so that the inspectee is not infored about the action of the inspector at the current stage, but knows everything else. The inforation set of the inspector is a singleton (this is different in the gae ˆΓ (n,,k that we consider later. Consider a Nash equilibriu of this gae. Suppose that there is a stage of the gae that is reached with positive probability where the players do not behave according to the SPNE described in Theore 3, and let there be no later such stage. That is, each of four cells in (44 at this stage either has the SPNE payoffs as entries or is reached with probability zero. We clai that the equilibriu property is violated at this stage. If all cells have positive probability, then at least one player gains because they do not play the unique equilibriu at this stage. If soe cells have probability zero, then one player plays deterinistically. For exaple, suppose the inspectee acts legally. Then if the inspector inspects with positive probability, he gets the SPNE payoff v(n 1, 1,k. However, this is not his best response, because when he does not inspect at this stage, he gets at least v(n 1,,k because that is also his in-ax payoff which he can guarantee by playing a ax-in strategy after no inspection at the current stage. Because we are in equilibriu, the inspector therefore responds with no inspection to the inspectee s certain legal action at this stage. However, then the inspectee can iprove on his SPNE payoff w(n 1,,k by violating and subsequently playing a ax-in strategy, which contradicts the assued equilibriu. This reasoning follows fro the strictly circular payoff structure in (44 and holds for any assued unplayed strategy. Hence, players always ix and the SPNE of the gae ˆΓ(n,,k is its unique Nash equilibriu (in behavior strategies, as always. The crucial condition used is that SPNE payoffs are in-ax payoffs, and the arguent is siilar to an analogous known result on finitely repeated gaes where all stage equilibriu payoffs are in-ax payoffs (see Osborne and Rubinstein, [20, Proposition 155.1]. Before we consider the gae ˆΓ (n,,k, we discuss a potential threat of the inspectee to use a violation even in the case n when the inspector can inspect in every reaining period. If we assue that in this case the inspector has the choice not to inspect, this defines a gae where legal action gives payoff zero to both players but violation gives a negative payoff to both players. The in-ax payoff to the inspector is then a r k rather than zero, when the inspectee irrationally violates later (which is his threat and 16

17 the inspector inspects. If we use this payoff a r k in the botto-left cell of (44 rather than the assued zero SPNE payoff v(n 1,n 1,k, we show that nevertheless a r k > v(n 1,n 2,k so that the preceding reasoning still applies, that is, the inspector s best response to legal action at the current stage (and violation later is still no inspection. Naely, by (46, and (9 and (10 with b = a and = n 1, v(n 1,n 2,k = t(n 1,n 2,k ŝ(n 1,n 2 = ( r n 2 k n 2 ((1 a n 1 ( n 1 n 1 /( a = a r k 1 (1 a n 1 and thus a r k > v(n 1,n 2,k as claied because 0 < a < 1 by (43. That is, the inspector still prefers not to inspect in response to legal action and a later threatened violation that will be caught. Hence there is no threat of the inspectee that could induce a Nash equilibriu other than the SPNE. Consider now a Nash equilibriu in behavior strategies of the gae ˆΓ (n,,k without full inforation. The inspector s lack of inforation is represented by inforation sets of the inspector that coprise ultiple decision nodes with the sae history of the inspector s own past actions, but different past actions of the inspectee at the stages where the inspector did not inspect. Consider such an inforation set h of the inspector that is reached with positive probability where the inspector does not use the in-ax strategy against the inspectee in Theore 3, and assue that there is no later inforation set of this kind. Then at this stage, that is, for all inforation sets of the inspectee that iediately follow this ove of the inspector at h, the inspectee will have the sae action (legal action or violation as a best response, which he therefore chooses with certainty because we are in equilibriu. However, in response the inspector would have to ake a ove at h against which the oves of the inspectee are not optial. This contradicts the equilibriu property. Hence, the inspector has to choose the in-ax strategy throughout, so that the inspectee s payoff is as in Theore 3. Now suppose that the inspector s payoff is different fro his in-ax payoff. This has to be a larger payoff because the inspector can guarantee his in-ax payoff by playing a ax-in strategy. Then at soe inforation set h of the inspector that is reached with positive probability, again looking at the latest such set, the inspectee does at this stage not play a in-ax strategy against the inspector, that is, a strategy that does not equalize the inspector s payoffs. To this the inspector plays a unique pure best response at h. This response is different fro the inspector s strategy in Theore 3, but we have just shown that this cannot be the case. Hence, the players payoffs are uniquely given according to Theore 3, as claied. 5 Inspector leadership The gae by Dresher [10] with a single intended violation has been studied by Maschler [19] in a leadership variant where the inspector can announce and coit to his ixed strategy. We extend these considerations to our gae with k intended violations, and siplify soe of Maschler s arguents. 17

18 A two-player gae in strategic for is changed to a leadership gae by declaring one player as leader and the other as follower. The leader chooses and coits to a strategy about which the follower is fully infored and chooses, as in a subgae perfect equilibriu, a best response to every coitent of the leader. Both players receive the payoffs of the original gae. A leadership gae is often called a Stackelberg gae, following von Stackelberg [25] who odified in this anner the siultaneous odel of quantity copetition by Cournot [7] to a sequential gae. We consider the leadership gae for the ixed extension of a finite two-player where the leader can coit to a ixed strategy, as analyzed in full generality by von Stengel and Zair [28]. Inspection gaes odel situations where an inspectee is obliged to act legally and hence cannot openly declare that he will violate. However, the inspector can becoe a leader and coit to a ixed strategy, using a roulette wheel or other randoization device that decides with a verifiable probability in each tie period (siultaneously to the choice of the inspectee whether to inspect or not. Maschler [19] observed that in a non-zerosu recursive gae, siilar to (44 for k = 1, the inspector can coit to essentially the sae ixed strategy as before, but that the inspectee acts legally with certainty as long as inspections reain. We first consider a general 2 2 gae as it arises in our context. d a p 1 p 1 q q a A B c C D b d L N c C p * p b B p D 0 A 1 Figure 1 Left: 2 2 gae with probability p for playing the top row and q for playing the right colun. Right: Payoffs to colun and row player if (47 and (49 hold in the leadership gae where the row player coits to p. Proposition 1 Consider the 2 2 gae on the left in Fig. 1 where the payoffs A,B,C,D to the row player and a,b,c,d to the colun player fulfill A < C, B > D, a > b, c < d. (47 Let p = d c a b + d c (48 18

19 and assue that p A + (1 p C > p B + (1 p D. (49 Then the gae has a unique ixed equilibriu where p is the equilibriu probability that the row player plays the top row, with Nash payoff N = B D+C A BC AD to the row player. In the leadership gae where the row player is the leader and can coit to a ixed strategy p, the unique subgae perfect equilibriu is that the row player coits to p and the colun player responds with q = 1 if p < p and q = 0 if p p, in particular with q = 0 on the equilibriu path where p = p. In the leadership gae, the payoff to the leader is L = p A + (1 p C, and L > N. The payoff to the follower is p a + (1 p c, the sae as in the siultaneous gae. Proof. Proof. By (47, the gae has a unique ixed equilibriu where the row player plays p and the colun player plays q = B D+C A C A, and the row player gets payoff N and the colun player gets p a + (1 p c. The clais about the leadership gae can be seen fro the right picture in Figure 1, which shows the players payoffs as a function of p. For illustration, the colun player s payoffs are assued to be positive (as it typically holds in our inspection gaes, with the exception of the payoff b and those of the colun player as negative, here for the case that B > A (so that (49 can only hold if C > D. For p < p the follower s best response is the right colun (q = 1, with expected payoff pb + (1 pd to the follower and pb + (1 pd to the leader. For p > p the follower s best response is the left colun (q = 0, with expected payoff pa+(1 pc to the follower and pa+(1 pc to the leader. For the coitent to p = p the follower is indifferent and in principle could reply with any q in [0,1]. The payoff to the leader as a function of p is shown as the bold line in the figure, including the vertical part for p = p for q [0,1]. By (47, this leader payoff is increasing in p for p < p and decreasing in p for p > p, so it has its axiu L if p = p and, by (49, if the follower s response is q = 0, shown by a full dot in the picture. This reaction of the follower defines in fact the unique SPNE in the leadership gae, because for p = p the follower, even though indifferent, has to choose the response, here q = 0, that axiizes the leader s payoff, because otherwise the leader could induce this behavior by changing his coitent to p + ε for soe arbitrarily sall positive ε, which contradicts the SPNE condition. This has essentially been observed by Maschler [19], who postulated a Pareto-optial response of the follower if he is indifferent, and noted that otherwise the leader can get a payoff arbitrarily close to L with a coitent to p + ε. The SPNE arguent has been ade by Avenhaus, Okada, and Zair [3], and in generality by von Stengel and Zair [28] who also give further references. In addition to the leader payoff L, Fig. 1 shows the Nash payoff N (with a hollow dot further below on the vertical line, which is less than L because it is given by N = q L + (1 q M where M = p B+(1 p D is the iniu payoff to the leader if the follower responds to p by choosing the right colun q = 1, and L > M by (49. If B > A and thus C > D, then N is also the ax-in payoff to the row player where his expected payoffs are the sae for both coluns, when he plays his ax-in strategy ˆp = B A+C D C D, also shown in the picture. 19

20 We want to apply Proposition 1 to the inspection gae (44. Siilar to Maschler [19, p. 18], the leadership gae for n tie periods, inspections, and k intended violations is described as follows: inspectee inspector legal action violation roulette wheel calls for inspection u(n 1, 1, k w(n 1, 1, k c(k b r k (50 roulette wheel calls for no inspection u(n 1,,k w(n 1,,k w(n 1,,k 1 + r k u(n 1,,k 1 r k If the assuptions of Proposition 1 are et, then in this leadership gae the inspector chooses the sae strategy as in the siultaneous gae so that the inspectee is indifferent between legal action and violation. However, the inspectee acts legally as long as > 0, that is, there will never be a caught violation. For that reason, the result will hold for any negative cost c(k to the inspector in that cell of the table. In the gae (50, the inspectee as follower should, by Proposition 1, get the sae recursively defined payoff w(n,,k as in Theore 3, but the inspector gets a new payoff u(n,,k. The following consideration shows what this payoff should be. First, if = 0, then the inspectee can and will safely use his k intended violations, as far as possible, in each of the reaining n tie periods, so that as in (42, as well as u(n,0,k = w(n,0,k = in{k,n} r k+1 i, (51 i=1 u(n,,k = w(n,,k = 0 if n. (52 For n > > 0 and k > 0, the gae (50 applies, where the inspectee gets the sae payoff w(n,,k for legal action and violation, given by (46. In particular, this is the inspectee s payoff if he always acts legally, as we assue he does in the leadership gae. Once the inspector has run out of inspections, the inspectee gets the sae payoff w(n,0,k as in (42 and (51, which is the negative of the inspector s payoff. By induction, the inspector s payoff should therefore in general siply be u(n,,k = w(n,,k. In the following theore, subgae perfection refers to the leadership gae that assues best responses of the follower even off the equilibriu path, naely for all other inspection probabilities that the inspector could coit to. In ters of inforation about the history of the gae, the probability of the roulette wheel at each stage is a function of n and but not of k, as before. Theore 5 Let n,,k be nonnegative integers, r k,r k 1,,r 1 0, b 0 and c(k > 0. Then in the leadership gae defined by (52, (51, and (50 for n > > 0 and k > 0, the 20