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Equilibrium in a war of attrition with an option to fight decisively
Operations Research Letters 47 (2019) 326–330
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Operations Research Letters journal homepage: www.elsevier.com/locate/orl
Equilibrium in a war of attrition with an option to fight decisively Geofferey Jiyun Kim a , Bara Kim b , Jeongsim Kim c , a b c
∗
Underwood International College, Yonsei University, 50 Yonsei-ro, Seodaemun-gu, Seoul, 03722, Republic of Korea Department of Mathematics, Korea University, 145 Anam-ro, Seongbuk-gu, Seoul, 02841, Republic of Korea Department of Mathematics Education, Chungbuk National University, 1 Chungdae-ro, Seowon-gu, Cheongju, Chungbuk, 28644, Republic of Korea
article
info
Article history: Received 1 March 2019 Received in revised form 22 May 2019 Accepted 27 May 2019 Available online 4 June 2019 Keywords: War of attrition Incomplete information Bayesian Nash equilibrium
a b s t r a c t We develop a symmetric incomplete-information continuous-time two-player war-of-attrition game with an option to fight decisively. We show that there exists an essentially unique symmetric Bayesian Nash equilibrium. Under equilibrium, the game does not end immediately, and a costly delay persists even with the availability of the fighting option that ends the game if chosen. In addition, there exists a critical time in which a fight occurs unless a player resigns before that time. © 2019 Elsevier B.V. All rights reserved.
1. Introduction The war of attrition is a classic example of a timing game, first analyzed by Maynard Smith [9]. In the standard war of attrition, each of the two players must choose a time at which to resign if the other player has not already resigned. The war of attrition with complete information was analyzed by Bishop and Cannings [2] and Hendricks et al. [7]. Variations of the war of attrition with incomplete information have been investigated by many researchers, including [3,4,6,8,10,11,13,14]. Our war-of-attrition game extends the standard war of attrition by adding the option to fight decisively. If the option to fight decisively is chosen by either player, our game ends immediately. Accordingly, our game ends if a player chooses to resign or fight. Our war-of-attrition model is most closely related to the warof-attrition models developed by Fearon [5] and Özyurt [12]. Both Fearon and Özyurt have extended the standard war-of-attrition model with the attack option, which is similar to our option to fight decisively. The two models and ours are related in that there are two exit options that terminate delays. Unlike our model, in the models by Fearon and Özyurt, if a player chooses to resign, only the resigning player incurs the accumulated cost of delay. Additionally, in the models by Fearon and Özyurt, if the attack option is chosen, no players incur the accumulated cost of delay, and none receives the prize in contention. The essential difference between the models by Fearon and Özyurt is that Özyurt’s model extends Fearon’s model by including a type who never resigns and never attacks. Our model is a straightforward extension of ∗ Corresponding author. E-mail addresses: [email protected] (G.J. Kim), [email protected] (B. Kim), [email protected] (J. Kim). https://doi.org/10.1016/j.orl.2019.05.005 0167-6377/© 2019 Elsevier B.V. All rights reserved.
the standard war-of-attrition model with the option to fight decisively. Of possible interest is Asako’s war-of-attrition game [1]. Asako’s game allows one of the two players to be exogenously defeated. Accordingly, Asako’s game ends if either player resigns or if one of the two players is defeated. Consideration of the fighting option is primarily of theoretical interest and can be of circumstantial substantive interest. We are interested in showing the existence of a costly delay in our warof-attrition game despite the readily-available option to forcefully end the delay. Our analysis can be applied to improve our understanding of various protracted conflicts in the animal kingdom, in firm competitions, and in interstate relations. To account for a specific application, we would need to fine-tune our theoretical model to the specific application of interest. This paper will develop and investigate our incompleteinformation war of attrition with the fighting option. It will be shown that there exists an essentially unique symmetric Bayesian Nash equilibrium (BNE). Under equilibrium, costly delay occurs despite the availability of the fighting option. Moreover, unless a player resigns, a fight takes place at a critical time. 2. A war of attrition with incomplete information In this section, we will briefly review a continuous-time symmetric war of attrition with incomplete information about the valuations of the prize (see Example 6.3 of Fudenberg and Tirole [6]). There are two players, player 1 and player 2. Player i chooses a number ti in [0, ∞], which is referred to as the resignation time. The type of player i, θi , is the private information of player i, and takes values in [0, ∞) with cumulative distribution
G.J. Kim, B. Kim and J. Kim / Operations Research Letters 47 (2019) 326–330
P and density p. The types are independent from each other. The winner receives the prize θi . The payoff of player i is
⎧ ⎪ ⎨ −ti θi ui (θi , ti , tj ) = − ti 2 ⎪ ⎩ θi − tj
if ti < tj ,
si (θi ) = arg maxti
if ti > tj ,
∞
ui (θi , ti , sj (θj ))p(θj )dθj . 0
We now look for a symmetric equilibrium. We say that s(·) is a symmetric BNE if (s(·), s(·)) is a BNE. This game has a unique symmetric BNE and the symmetric BNE is s(θ ) =
θ
∫ 0
xp(x)
dx.
3. A war of attrition with the option to fight decisively In this section, we consider the war of attrition as described in Section 2, but with the addition of the fighting option. When the players fight, each player incurs a cost of ψ > 0, and the probability of winning is 12 . The strategy of player i, (di (·), si (·)) is a function from the set of types to {0, 1} × [0, ∞]. For a type θi of player i, (di (θi ), si (θi )) = (0, ti ) means that player i will resign at ti if the other player does not take any action (resign or fight) until ti . For a type θi of player i, (di (θi ), si (θi )) = (1, ti ) means that player i will choose to fight at ti if the other player does not take any action until ti . We assume that if both players choose to fight or resign at the same time, then the action of each player is chosen with probability 21 . When (dk (θk ), sk (θk )) = (ak , tk ), k = 1, 2, the payoff of player i is given by
⎧ −ti ⎪ ⎪ ⎪ ⎪ ⎪ −ti − ψ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −t + θi ⎪ ⎪ i 2 ⎪ ⎨−t − ψ i ui (θi , (ai , ti ), (aj , tj )) = ⎪ −ti − ψ2 ⎪ ⎪ ⎪ ⎪ ψ ⎪ ⎪−ti − 2 ⎪ ⎪ ⎪−t + θ ⎪ ⎪ j i ⎪ ⎪ ⎩ −tj − ψ
θi 2
+
θi
+
θi
2
d(θ ) =
0 1
(5) i.e., U(θi , (k, t)|(d, s)) is the expected payoff of a player with type θi when the player chooses action k at time t and the other player follows the strategy (d, s). Let t¯ =
2ψ
∫
4 3 4 i
1 − P(x)
+ θ
U(θi , (k, t)|(d, s)) =
2
∫
(θi − s(θj ))p(θj )dθj
− min{t , t¯}(1 − P(g(min{t , t¯}))) ( θi ) + − ψ (1 − P(g(min{t , t¯}))) 2
) 1 × 1 − 1{k=0,t =t¯} − 1{k=0,t
2
d
dt
U(θi , (k, t)|(d, s)) = (θi − t)p(g(t))g ′ (t)
1 s′ (g(t))
=
U(θi , (k, t)|(d, s)) =
1−P(g(t)) , g(t)p(g(t))
(2)
0
There exists a symmetric BNE; if (3)
(7)
0 < t < t¯ , (7) becomes
1 − P(g(t)) ( g(t)
θi − g(t)
+ 1{k=1} (ψ −
if ti > tj , aj = 1.
(6)
where g(t), 0 ≤ t ≤ t¯ , is the inverse function of s(θ ), 0 ≤ θ ≤ 2ψ . Differentiation of (6) with respect to t yields, for 0 < t < t¯ ,
if ti = tj , ai = 1, aj = 0,
ui (θi , (ai , ti ), (dj (θj ), sj (θj )))p(θj )dθj .
g(min{t ,t¯ })
0
d
if ti > tj , aj = 0, θi
dx.
In Eq. (5), we consider three cases of t, t < t¯ , t = t¯ , and t > t¯ , separately, for each value of k, k = 0, 1. For the simplicity of the presentation, we will assume that p(t) is positive and continuous on (0, 2ψ ), even though this technical assumption is not necessary for this proposition. Using (1), we can rewrite (5) as
if ti = tj , ai = 0, aj = 1,
∞
if θ < 2ψ, if θ > 2ψ,
xp(x)
Since g ′ (t) =
if ti = tj , ai = aj = 1,
We now look for a symmetric BNE. As in Section 2, (d(·), s(·)) is a symmetric BNE if ((d(·), s(·)), (d(·), s(·)) is a BNE. The following proposition gives the symmetric BNE of this game. See Fig. 1 for an illustration of the symmetric BNE.
{
0
2
(di (θi ), si (θi ))
Proposition 1.
ui (θi , (k, t), (d(θj ), s(θj )))p(θj )dθj , k = 0, 1, t ∈ [0, ∞],
=
if ti < tj , ai = 1,
Let us look for a BNE ((d1 (·), s1 (·)), (d2 (·), s2 (·)) of this game. For each θi , (di (θi ), si (θi )) must satisfy
∫
(4)
∞
∫
if ti = tj , ai = aj = 0,
+
dx,
− (1 − P(g(t))) + tp(g(t))g ′ (t) ( θi ) − − ψ p(g(t))g ′ (t).
(1)
= arg max(ai ,ti )
1 − P(x)
U(θi , (k, t)|(d, s))
dt
if ti < tj , ai = 0,
+
xp(x)
Proof. Let (d(·), s(·)) be a strategy satisfying (3) and (4). Define
0
1 − P(x)
min{θ,2ψ}
∫
s(θ ) =
then (d(·), s(·)) is a symmetric BNE.
where j is the other player. Here, we assume that if both players resign at the same time, each has an equal chance of winning the prize. The strategy of player i, si (·), is a function from the set of types to the set of resignation times. Let (s1 (·), s2 (·)) be a (pure strategy) BNE of this game. For each θi , si (θi ) must satisfy
∫
and
0
if ti = tj ,
327
θi ) ) , 2
0 < t < t¯ .
(8)
If θi < 2ψ , then U(θi , (0, t)|(d, s)) ≥ U(θi , (1, t)|(d, s)) by (6), and U(θi , (0, t)|(d, s)) takes the maximum at t = s(θi ) by (8), see Fig. 2(a). Therefore (2) holds if θi < 2ψ . On the other hand, if θi ≥ 2ψ , then, by (6) and (8), U(θi , (k, t))|(d, s)) is increasing in t and takes the maximum at (k, t) = (1, t¯ ), see Fig. 2(b) and (c). Therefore (2) holds if θi ≥ 2ψ . □ If both players use the equilibrium strategy, ∫ 2ψ xp(x)then our model, from the beginning of the game to t¯ = 0 1−P(x) dx, is the same as the war of attrition without the fighting option, as described in Section 2. If the game does not end before t¯ , then the game ends by a fight at t¯ . In the proof of Proposition 1, we have shown that the conditions (3) and (4) are sufficient for (d(·), s(·)) to be a symmetric BNE. It can be shown that the conditions (3) and (4) are almost necessary. In the following theorem, we give a necessary and sufficient condition for a strategy (d(·), s(·)) to be a symmetric BNE.
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To prove the necessity, we need a series of lemmas. Recall that U(θ, (k, t)|(d, s)) =
∞
∫
ui (θ, (k, t), (d(θj ), s(θj )))p(θj )dθj , 0
k = 0, 1, t ∈ [0, ∞]. From now on, we will write U(θ, (k, t)) instead of U(θ , (k, t)| (d, s)) for simplicity of notation. Lemma 1. Suppose that 0 ≤ θ ′ < θ ≤ 2ψ , t ∈ [0, t ∗ ) and t ′ ∈ [0, ∞]. If (0, t) is a best response of type θ to the strategy (d, s) and (0, t ′ ) is a best response of type θ ′ to the strategy (d, s), then t ′ ≤ t. Proof. We prove this by contradiction. Suppose that the assumptions are satisfied, but t ′ > t. Then U(θ ′ , (0, t ′ )) − U(θ ′ , (0, t)) Fig. 1. An illustration of the symmetric BNE.
Theorem 1. A strategy (d(·), s(·)) is a symmetric BNE if and only if there exists a null set N ⊂ [2ψ, ∞) such that (3) and (4) for all ∫ 2hold ψ θ ∈ [0, ∞) \ N, and s(θ ) > t¯ for all θ ∈ N, where t¯ = 0 1xp(x) dx. −P(x) Theorem 1 shows that our war of attrition with the fighting option has an essentially unique symmetric BNE. The proof is given in the appendix. In our model, the probability of winning when a fight occurs is assumed to be 1/2, and the fighting cost is assumed to be the same for both players. This symmetry in the payoffs is essential for the BNE to have the structure as shown in Fig. 1. In a model in which the probability of winning when a fight occurs is asymmetric (i.e., the player who chooses to fight wins with a probability greater or less than 1/2) and the fighting cost is also asymmetric (i.e., the player who chooses to fight has the higher or lower fighting cost than the other player), its BNE is expected to have different structure from Fig. 1. An investigation of a model with the previously mentioned asymmetric considerations will be pursued in our future study.
= θ ′ P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) ) (θ′ − ψ P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) + 2 1(θ′
+
2 2
) − ψ P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ })
t′
∫
P(s(Θ ) > τ ) dτ
− t
≥ 0, and U(θ, (0, t)) − U(θ, (0, t ′ ))
= −θ P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) (θ ) − − ψ P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) −
2 1(θ
2 2
) − ψ P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ })
t′
∫
P(s(Θ ) > τ ) dτ
+ t
≥ 0. Hence
Acknowledgments
(θ ′ − θ ) P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) We are grateful to the reviewer for valuable comments and suggestions. B. Kim’s research was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (NRF-2017R1A2B4012676). J. Kim’s research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03029542).
which implies P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) = 0. This contradicts (10). □
Appendix. Proof of Theorem 1
Lemma 2.
The proof of the sufficiency is almost the same as the proof of Proposition 1, so we omit the proof. We only prove the necessity. Suppose that (d(·), s(·)) is a symmetric BNE. Let Θ denote a random variable with the density p(·). Let t ∗ = sup{t ≥ 0 : P(d(Θ ) = 0, s(Θ ) ≥ t) > 0}. Then it is not difficult to show that t ∗ > 0,
P(d(Θ ) = 0, s(Θ ) = t) = 0 for each t ∈ [0, ∞],
(9)
and
P(d(Θ ) = 0, s(Θ ) > t) is continuous and strictly decreasing on [0, t ∗ ).
(10)
+
θ′ − θ
+
θ′ − θ
2 4
P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ }) ≥ 0,
(i) Suppose P(s(Θ ) ≥ t) > 0. If (0, t) is a best response of type θ to the strategy (d, s), then θ ≤ 2ψ . If (1, t) is a best response of type θ to the strategy (d, s), then θ ≥ 2ψ . (ii) Suppose t ∈ [0, t ∗ ) and θ ∈ [0, ∞). If (0, t) is a best response of type θ to the strategy (d, s), then s(θ ′ ) ≤ t for all θ ′ ∈ [0, θ ). Proof. First we prove assertion (i). Suppose P(s(Θ ) ≥ t) > 0. We observe ( ) θ )( 1 U(θ, (0, t)) − U(θ, (1, t)) = ψ − P(s(Θ ) > t) + P(s(Θ ) = t) . 2 2 Since P(s(Θ ) ≥ t) > 0, we have that θ ≤ 2ψ if (0, t) is a best response of type θ to the strategy (d, s), and θ ≥ 2ψ if (1, t) is a best response of type θ to the strategy (d, s).
G.J. Kim, B. Kim and J. Kim / Operations Research Letters 47 (2019) 326–330
329
Fig. 2. Plots of U(θi , (k, t)|(d, s)), k = 0, 1. Note that we use U(θi , (k, t)), k = 0, 1 instead of U(θi , (k, t)|(d, s)), k = 0, 1 in the label of the y-axis.
Next, we prove assertion (ii). Suppose that t ∈ [0, t ∗ ) and 0 ≤ θ ′ < θ < ∞, and that (0, t) is a best response of type θ to the strategy (d, s). If d(θ ′ ) = 1, then U(θ ′ , (1, s(θ ′ ))) − U(θ ′ , (0, s(θ ′ )))
=
(θ′ 2
)( ) 1 − ψ P(s(Θ ) > s(θ ′ )) + P(s(Θ ) = s(θ ′ )) ≥ 0. 2
Since t < t ∗ and (0, t) is a best response of type θ to the strategy (d, s), we have θ ≤ 2ψ by assertion (i), thereby, θ ′ < 2ψ , ′ i.e., θ2 − ψ < 0. Thus, if d(θ ′ ) = 1, then P(s(Θ ) ≥ s(θ ′ )) = 0 and U(θ ′ , (1, s(θ ′ ))) = U(θ ′ , (0, s(θ ′ ))). Hence, for either d(θ ′ ) = 0 or 1, (0, s(θ ′ )) is a best response of type θ ′ to the strategy (d, s). By Lemma 1, s(θ ′ ) ≤ t for all θ ′ ∈ [0, θ ). □ Lemma 3.
We have the following:
(i) P(d(Θ ) = 0, Θ > 2ψ ) = 0, (ii) s(2ψ ) ≥ t ∗ ,
(iii) d(θ ) = 0 and s(θ ) < t ∗ for all θ ∈ [0, 2ψ ). Proof. By Lemma 2(i), if θ > 2ψ and d(θ ) = 0, then s(θ ) ≥ t¯ . Hence P(d(Θ ) = 0, Θ > 2ψ, s(Θ ) < t ∗ ) = 0. By (9) and the definition of t ∗ , P(d(Θ ) = 0, s(Θ ) ≥ t ∗ ) = 0. Therefore, P(d(Θ ) = 0, Θ > 2ψ ) = 0, which is assertion (i). Next, we prove assertion (ii) by contradiction. Suppose s(2ψ ) < t ∗ . Observe that U(2ψ, (0, t)) = U(2ψ, (1, t)). By Lemma 2 (ii), s(θ ) ≤ s(2ψ ) for all θ ∈ [0, 2ψ ). From this and assertion (i), we have t ∗ = sup{t ≥ 0 : P(d(Θ ) = 0, s(Θ ) > t) > 0} ≤ s(2ψ ) < t ∗ , which is a contradiction. Finally, we prove assertion (iii). We have that sup{θ < 2ψ : P(θ < Θ < 2ψ, d(Θ ) = 0) > 0} = 2ψ.
(11)
This is proved by contradiction, which is as follows. Suppose θ˜ ≡ sup{θ < 2ψ : P(θ < Θ < 2ψ, d(Θ ) = 0) > 0} < 2ψ . ˜ 2ψ ) such that d(θ ′ ) = 1 and P(s(Θ ) Then there exists θ ′ ∈ (θ,
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G.J. Kim, B. Kim and J. Kim / Operations Research Letters 47 (2019) 326–330
≥ s(θ ′ )) > 0. By Lemma 2(i), d(θ ′ ) = 0, which is a contradiction. Therefore, we have (11). From this and (9), there exists a strictly increasing sequence {θn } such that limn→∞ θn = 2ψ , d(θn ) = 0 and s(θn ) < t ∗ . By Lemma 2 (ii), s(θ ) < t ∗ for all θ < 2ψ , and by Lemma 2(i), d(θ ) = 0 for all θ < 2ψ . □
Therefore, if θ > 2ψ , then (0, t ∗ ) is not a best response of type θ to the strategy (d, s) and so (d(θ ), s(θ )) = (1, t ∗ ). □ If we show that s(θ ) =
θ
∫
1 − P(x)
0
Let
∫ 2ψ
N = {θ ∈ [0, ∞) : θ ≥ 2ψ, s(θ ) > t ∗ }. From P(d(Θ ) = 0, s(Θ ) > t ∗ ) = 0, it follows that P(d(Θ ) = 1, s(Θ ) > t ∗ ) = 0. Hence P(s(Θ ) > t ∗ ) = 0 and N should be a null set. Lemma 4.
If θ ∈ (2ψ, ∞) \ N, then d(θ ) = 1 and s(θ ) = t ∗ .
Proof. Suppose θ > 2ψ and t ∈ [0, t ∗ ). Then U(θ, (1, t ∗ )) − U(θ , (1, t)) =
(θ
) + ψ P(Θ < 2ψ, s(Θ ) > t) 2 ∫ t∗ − P(Θ > τ )dτ . (12) t
By (9), (10), Lemma 1 and Lemma 3 (iii), s(·) is a strictly increasing and continuous function from [0, 2ψ ) onto [0, t ∗ ). Let g : [0, t ∗ ) → [0, 2ψ ) be the inverse function. Then, for θ ′ ∈ (g(t), 2ψ ), U(θ ′ , (0, s(θ ′ ))) − lim U(θ ′ , (0, t ′ )) t ′ →t +
= θ P(Θ < 2ψ, s(Θ ) ∈ (t , s(θ ′ ))) ){ (θ′ − ψ P(Θ > 2ψ, s(Θ ) ∈ (t , s(θ ′ ))) + 2 ∫ s(θ ′ ) } 1 + P(Θ > 2ψ, s(Θ ) = s(θ ′ )) − P(Θ > τ )dτ ′
2
t
≥ 0. Letting θ ′ → 2ψ− gives 2ψ P(Θ < 2ψ, s(Θ ) > t) ≥
t∗
∫
P(Θ > τ )dτ . t
From this and (12), we obtain U(θ, (1, t ∗ )) − U(θ , (1, t)) ≥
(θ
) − ψ P(Θ < 2ψ, s(Θ ) > t) > 0.
2 Hence, (1, t) cannot be a best response of type θ to the strategy (d, s). By Lemma 2(i), (d(θ ), s(θ )) ̸ = (0, t). Therefore s(θ ) ̸ = t for all t < t ∗ and so s(θ ) ≥ t ∗ for all θ > 2ψ . Hence, s(θ ) = t ∗ since θ∈ / N and s(θ ) ≥ t ∗ . Next, we prove that if θ > 2ψ and s(θ ) = t ∗ , then d(θ ) = 1. Suppose θ > 2ψ . Then U(θ, (1, t ∗ )) − U(θ , (0, t ∗ )) =
1(θ 2 2
xp(x)
) − ψ P(Θ > 2ψ ) > 0.
dx for θ ∈ [0, 2ψ ),
xp(x)
then t ∗ = 0 1−P(x) dx = t¯ , and so the necessity of Theorem 1 is complete. For θ ∈ [0, 2ψ ) and t ∈ [0, t ∗ ), U(θ, (0, t)) =
g(t)
∫
(θ − s(θj ))p(θj )dθj − t(1 − P(g(t))). 0
For each θ ∈ [0, 2ψ ), U(θ, (0, t)) takes a maximum at t = s(θ ). Hence (θ − s(g(s(θ ))))p(g(s(θ )))g ′ (s(θ ))
− (1 − P(g(s(θ )))) + s(θ )p(g(s(θ )))g ′ (s(θ )) = 0. This gives s′ (θ ) = s(θ ) =
θ
∫ 0
θ p(θ ) . 1−P(θ )
xp(x) 1 − P(x)
Since s(0) = 0,
dx for θ ∈ [0, 2ψ ).
Therefore, we complete the proof of Theorem 1. References [1] Y. Asako, One-sided games in a war of attrition, B. E. J. Theor. Econ. 15 (2015) 313–331. [2] D.T. Bishop, C. Cannings, A generalized war of attrition, J. Theoret. Biol. 70 (1978) 85–124. [3] D.T. Bishop, C. Cannings, J. Maynard Smith, The war of attrition with random rewards, J. Theoret. Biol. 74 (1978) 377–388. [4] J. Bulow, P. Klemperer, The generalized war of attrition, Amer. Econ. Rev. 89 (1999) 175–189. [5] J.D. Fearon, Domestic political audiences and the escalation of international disputes, Amer. Political Sci. Rev. 88 (1994) 577–592. [6] D. Fudenberg, J. Tirole, Game Theory, The MIT Press, 1991. [7] K. Hendricks, A. Weiss, C. Wilson, The war of attrition in continuous time with complete information, Internat. Econom. Rev. 29 (1988) 663–680. [8] J. Hörner, N. Sahuguet, A war of attrition with endogenous effort levels, Econom. Theory 47 (2011) 1–27. [9] J. Maynard Smith, Theory of games and the evolution of animal conflicts, J. Theoret. Biol. 47 (1974) 209–221. [10] P.R. Milgrom, R.J. Weber, Distributional strategies for games with incomplete information, Math. Oper. Res. 10 (1985) 619–632. [11] B. Nalebuff, J.G. Riley, Asymmetric equilibria in the war of attrition, J. Theoret. Biol. 113 (1985) 517–527. [12] S. Özyurt, Audience costs and reputation in crisis bargaining, Games Econom. Behav. 88 (2014) 250–259. [13] C. Ponsati, J. Sákovics, The war of attrition with incomplete information, Math. Social Sci. 29 (1995) 239–254. [14] J.G. Riley, Strong evolutionary equilibrium and the war of attrition, J. Theoret. Biol. 82 (1980) 383–400.
Contents lists available at ScienceDirect
Operations Research Letters journal homepage: www.elsevier.com/locate/orl
Equilibrium in a war of attrition with an option to fight decisively Geofferey Jiyun Kim a , Bara Kim b , Jeongsim Kim c , a b c
∗
Underwood International College, Yonsei University, 50 Yonsei-ro, Seodaemun-gu, Seoul, 03722, Republic of Korea Department of Mathematics, Korea University, 145 Anam-ro, Seongbuk-gu, Seoul, 02841, Republic of Korea Department of Mathematics Education, Chungbuk National University, 1 Chungdae-ro, Seowon-gu, Cheongju, Chungbuk, 28644, Republic of Korea
article
info
Article history: Received 1 March 2019 Received in revised form 22 May 2019 Accepted 27 May 2019 Available online 4 June 2019 Keywords: War of attrition Incomplete information Bayesian Nash equilibrium
a b s t r a c t We develop a symmetric incomplete-information continuous-time two-player war-of-attrition game with an option to fight decisively. We show that there exists an essentially unique symmetric Bayesian Nash equilibrium. Under equilibrium, the game does not end immediately, and a costly delay persists even with the availability of the fighting option that ends the game if chosen. In addition, there exists a critical time in which a fight occurs unless a player resigns before that time. © 2019 Elsevier B.V. All rights reserved.
1. Introduction The war of attrition is a classic example of a timing game, first analyzed by Maynard Smith [9]. In the standard war of attrition, each of the two players must choose a time at which to resign if the other player has not already resigned. The war of attrition with complete information was analyzed by Bishop and Cannings [2] and Hendricks et al. [7]. Variations of the war of attrition with incomplete information have been investigated by many researchers, including [3,4,6,8,10,11,13,14]. Our war-of-attrition game extends the standard war of attrition by adding the option to fight decisively. If the option to fight decisively is chosen by either player, our game ends immediately. Accordingly, our game ends if a player chooses to resign or fight. Our war-of-attrition model is most closely related to the warof-attrition models developed by Fearon [5] and Özyurt [12]. Both Fearon and Özyurt have extended the standard war-of-attrition model with the attack option, which is similar to our option to fight decisively. The two models and ours are related in that there are two exit options that terminate delays. Unlike our model, in the models by Fearon and Özyurt, if a player chooses to resign, only the resigning player incurs the accumulated cost of delay. Additionally, in the models by Fearon and Özyurt, if the attack option is chosen, no players incur the accumulated cost of delay, and none receives the prize in contention. The essential difference between the models by Fearon and Özyurt is that Özyurt’s model extends Fearon’s model by including a type who never resigns and never attacks. Our model is a straightforward extension of ∗ Corresponding author. E-mail addresses: [email protected] (G.J. Kim), [email protected] (B. Kim), [email protected] (J. Kim). https://doi.org/10.1016/j.orl.2019.05.005 0167-6377/© 2019 Elsevier B.V. All rights reserved.
the standard war-of-attrition model with the option to fight decisively. Of possible interest is Asako’s war-of-attrition game [1]. Asako’s game allows one of the two players to be exogenously defeated. Accordingly, Asako’s game ends if either player resigns or if one of the two players is defeated. Consideration of the fighting option is primarily of theoretical interest and can be of circumstantial substantive interest. We are interested in showing the existence of a costly delay in our warof-attrition game despite the readily-available option to forcefully end the delay. Our analysis can be applied to improve our understanding of various protracted conflicts in the animal kingdom, in firm competitions, and in interstate relations. To account for a specific application, we would need to fine-tune our theoretical model to the specific application of interest. This paper will develop and investigate our incompleteinformation war of attrition with the fighting option. It will be shown that there exists an essentially unique symmetric Bayesian Nash equilibrium (BNE). Under equilibrium, costly delay occurs despite the availability of the fighting option. Moreover, unless a player resigns, a fight takes place at a critical time. 2. A war of attrition with incomplete information In this section, we will briefly review a continuous-time symmetric war of attrition with incomplete information about the valuations of the prize (see Example 6.3 of Fudenberg and Tirole [6]). There are two players, player 1 and player 2. Player i chooses a number ti in [0, ∞], which is referred to as the resignation time. The type of player i, θi , is the private information of player i, and takes values in [0, ∞) with cumulative distribution
G.J. Kim, B. Kim and J. Kim / Operations Research Letters 47 (2019) 326–330
P and density p. The types are independent from each other. The winner receives the prize θi . The payoff of player i is
⎧ ⎪ ⎨ −ti θi ui (θi , ti , tj ) = − ti 2 ⎪ ⎩ θi − tj
if ti < tj ,
si (θi ) = arg maxti
if ti > tj ,
∞
ui (θi , ti , sj (θj ))p(θj )dθj . 0
We now look for a symmetric equilibrium. We say that s(·) is a symmetric BNE if (s(·), s(·)) is a BNE. This game has a unique symmetric BNE and the symmetric BNE is s(θ ) =
θ
∫ 0
xp(x)
dx.
3. A war of attrition with the option to fight decisively In this section, we consider the war of attrition as described in Section 2, but with the addition of the fighting option. When the players fight, each player incurs a cost of ψ > 0, and the probability of winning is 12 . The strategy of player i, (di (·), si (·)) is a function from the set of types to {0, 1} × [0, ∞]. For a type θi of player i, (di (θi ), si (θi )) = (0, ti ) means that player i will resign at ti if the other player does not take any action (resign or fight) until ti . For a type θi of player i, (di (θi ), si (θi )) = (1, ti ) means that player i will choose to fight at ti if the other player does not take any action until ti . We assume that if both players choose to fight or resign at the same time, then the action of each player is chosen with probability 21 . When (dk (θk ), sk (θk )) = (ak , tk ), k = 1, 2, the payoff of player i is given by
⎧ −ti ⎪ ⎪ ⎪ ⎪ ⎪ −ti − ψ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −t + θi ⎪ ⎪ i 2 ⎪ ⎨−t − ψ i ui (θi , (ai , ti ), (aj , tj )) = ⎪ −ti − ψ2 ⎪ ⎪ ⎪ ⎪ ψ ⎪ ⎪−ti − 2 ⎪ ⎪ ⎪−t + θ ⎪ ⎪ j i ⎪ ⎪ ⎩ −tj − ψ
θi 2
+
θi
+
θi
2
d(θ ) =
0 1
(5) i.e., U(θi , (k, t)|(d, s)) is the expected payoff of a player with type θi when the player chooses action k at time t and the other player follows the strategy (d, s). Let t¯ =
2ψ
∫
4 3 4 i
1 − P(x)
+ θ
U(θi , (k, t)|(d, s)) =
2
∫
(θi − s(θj ))p(θj )dθj
− min{t , t¯}(1 − P(g(min{t , t¯}))) ( θi ) + − ψ (1 − P(g(min{t , t¯}))) 2
) 1 × 1 − 1{k=0,t =t¯} − 1{k=0,t
2
d
dt
U(θi , (k, t)|(d, s)) = (θi − t)p(g(t))g ′ (t)
1 s′ (g(t))
=
U(θi , (k, t)|(d, s)) =
1−P(g(t)) , g(t)p(g(t))
(2)
0
There exists a symmetric BNE; if (3)
(7)
0 < t < t¯ , (7) becomes
1 − P(g(t)) ( g(t)
θi − g(t)
+ 1{k=1} (ψ −
if ti > tj , aj = 1.
(6)
where g(t), 0 ≤ t ≤ t¯ , is the inverse function of s(θ ), 0 ≤ θ ≤ 2ψ . Differentiation of (6) with respect to t yields, for 0 < t < t¯ ,
if ti = tj , ai = 1, aj = 0,
ui (θi , (ai , ti ), (dj (θj ), sj (θj )))p(θj )dθj .
g(min{t ,t¯ })
0
d
if ti > tj , aj = 0, θi
dx.
In Eq. (5), we consider three cases of t, t < t¯ , t = t¯ , and t > t¯ , separately, for each value of k, k = 0, 1. For the simplicity of the presentation, we will assume that p(t) is positive and continuous on (0, 2ψ ), even though this technical assumption is not necessary for this proposition. Using (1), we can rewrite (5) as
if ti = tj , ai = 0, aj = 1,
∞
if θ < 2ψ, if θ > 2ψ,
xp(x)
Since g ′ (t) =
if ti = tj , ai = aj = 1,
We now look for a symmetric BNE. As in Section 2, (d(·), s(·)) is a symmetric BNE if ((d(·), s(·)), (d(·), s(·)) is a BNE. The following proposition gives the symmetric BNE of this game. See Fig. 1 for an illustration of the symmetric BNE.
{
0
2
(di (θi ), si (θi ))
Proposition 1.
ui (θi , (k, t), (d(θj ), s(θj )))p(θj )dθj , k = 0, 1, t ∈ [0, ∞],
=
if ti < tj , ai = 1,
Let us look for a BNE ((d1 (·), s1 (·)), (d2 (·), s2 (·)) of this game. For each θi , (di (θi ), si (θi )) must satisfy
∫
(4)
∞
∫
if ti = tj , ai = aj = 0,
+
dx,
− (1 − P(g(t))) + tp(g(t))g ′ (t) ( θi ) − − ψ p(g(t))g ′ (t).
(1)
= arg max(ai ,ti )
1 − P(x)
U(θi , (k, t)|(d, s))
dt
if ti < tj , ai = 0,
+
xp(x)
Proof. Let (d(·), s(·)) be a strategy satisfying (3) and (4). Define
0
1 − P(x)
min{θ,2ψ}
∫
s(θ ) =
then (d(·), s(·)) is a symmetric BNE.
where j is the other player. Here, we assume that if both players resign at the same time, each has an equal chance of winning the prize. The strategy of player i, si (·), is a function from the set of types to the set of resignation times. Let (s1 (·), s2 (·)) be a (pure strategy) BNE of this game. For each θi , si (θi ) must satisfy
∫
and
0
if ti = tj ,
327
θi ) ) , 2
0 < t < t¯ .
(8)
If θi < 2ψ , then U(θi , (0, t)|(d, s)) ≥ U(θi , (1, t)|(d, s)) by (6), and U(θi , (0, t)|(d, s)) takes the maximum at t = s(θi ) by (8), see Fig. 2(a). Therefore (2) holds if θi < 2ψ . On the other hand, if θi ≥ 2ψ , then, by (6) and (8), U(θi , (k, t))|(d, s)) is increasing in t and takes the maximum at (k, t) = (1, t¯ ), see Fig. 2(b) and (c). Therefore (2) holds if θi ≥ 2ψ . □ If both players use the equilibrium strategy, ∫ 2ψ xp(x)then our model, from the beginning of the game to t¯ = 0 1−P(x) dx, is the same as the war of attrition without the fighting option, as described in Section 2. If the game does not end before t¯ , then the game ends by a fight at t¯ . In the proof of Proposition 1, we have shown that the conditions (3) and (4) are sufficient for (d(·), s(·)) to be a symmetric BNE. It can be shown that the conditions (3) and (4) are almost necessary. In the following theorem, we give a necessary and sufficient condition for a strategy (d(·), s(·)) to be a symmetric BNE.
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To prove the necessity, we need a series of lemmas. Recall that U(θ, (k, t)|(d, s)) =
∞
∫
ui (θ, (k, t), (d(θj ), s(θj )))p(θj )dθj , 0
k = 0, 1, t ∈ [0, ∞]. From now on, we will write U(θ, (k, t)) instead of U(θ , (k, t)| (d, s)) for simplicity of notation. Lemma 1. Suppose that 0 ≤ θ ′ < θ ≤ 2ψ , t ∈ [0, t ∗ ) and t ′ ∈ [0, ∞]. If (0, t) is a best response of type θ to the strategy (d, s) and (0, t ′ ) is a best response of type θ ′ to the strategy (d, s), then t ′ ≤ t. Proof. We prove this by contradiction. Suppose that the assumptions are satisfied, but t ′ > t. Then U(θ ′ , (0, t ′ )) − U(θ ′ , (0, t)) Fig. 1. An illustration of the symmetric BNE.
Theorem 1. A strategy (d(·), s(·)) is a symmetric BNE if and only if there exists a null set N ⊂ [2ψ, ∞) such that (3) and (4) for all ∫ 2hold ψ θ ∈ [0, ∞) \ N, and s(θ ) > t¯ for all θ ∈ N, where t¯ = 0 1xp(x) dx. −P(x) Theorem 1 shows that our war of attrition with the fighting option has an essentially unique symmetric BNE. The proof is given in the appendix. In our model, the probability of winning when a fight occurs is assumed to be 1/2, and the fighting cost is assumed to be the same for both players. This symmetry in the payoffs is essential for the BNE to have the structure as shown in Fig. 1. In a model in which the probability of winning when a fight occurs is asymmetric (i.e., the player who chooses to fight wins with a probability greater or less than 1/2) and the fighting cost is also asymmetric (i.e., the player who chooses to fight has the higher or lower fighting cost than the other player), its BNE is expected to have different structure from Fig. 1. An investigation of a model with the previously mentioned asymmetric considerations will be pursued in our future study.
= θ ′ P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) ) (θ′ − ψ P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) + 2 1(θ′
+
2 2
) − ψ P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ })
t′
∫
P(s(Θ ) > τ ) dτ
− t
≥ 0, and U(θ, (0, t)) − U(θ, (0, t ′ ))
= −θ P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) (θ ) − − ψ P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) −
2 1(θ
2 2
) − ψ P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ })
t′
∫
P(s(Θ ) > τ ) dτ
+ t
≥ 0. Hence
Acknowledgments
(θ ′ − θ ) P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) We are grateful to the reviewer for valuable comments and suggestions. B. Kim’s research was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (NRF-2017R1A2B4012676). J. Kim’s research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03029542).
which implies P(d(Θ ) = 0, s(Θ ) ∈ (t , t ′ )) = 0. This contradicts (10). □
Appendix. Proof of Theorem 1
Lemma 2.
The proof of the sufficiency is almost the same as the proof of Proposition 1, so we omit the proof. We only prove the necessity. Suppose that (d(·), s(·)) is a symmetric BNE. Let Θ denote a random variable with the density p(·). Let t ∗ = sup{t ≥ 0 : P(d(Θ ) = 0, s(Θ ) ≥ t) > 0}. Then it is not difficult to show that t ∗ > 0,
P(d(Θ ) = 0, s(Θ ) = t) = 0 for each t ∈ [0, ∞],
(9)
and
P(d(Θ ) = 0, s(Θ ) > t) is continuous and strictly decreasing on [0, t ∗ ).
(10)
+
θ′ − θ
+
θ′ − θ
2 4
P(d(Θ ) = 1, s(Θ ) ∈ (t , t ′ )) P(d(Θ ) = 1, s(Θ ) ∈ {t , t ′ }) ≥ 0,
(i) Suppose P(s(Θ ) ≥ t) > 0. If (0, t) is a best response of type θ to the strategy (d, s), then θ ≤ 2ψ . If (1, t) is a best response of type θ to the strategy (d, s), then θ ≥ 2ψ . (ii) Suppose t ∈ [0, t ∗ ) and θ ∈ [0, ∞). If (0, t) is a best response of type θ to the strategy (d, s), then s(θ ′ ) ≤ t for all θ ′ ∈ [0, θ ). Proof. First we prove assertion (i). Suppose P(s(Θ ) ≥ t) > 0. We observe ( ) θ )( 1 U(θ, (0, t)) − U(θ, (1, t)) = ψ − P(s(Θ ) > t) + P(s(Θ ) = t) . 2 2 Since P(s(Θ ) ≥ t) > 0, we have that θ ≤ 2ψ if (0, t) is a best response of type θ to the strategy (d, s), and θ ≥ 2ψ if (1, t) is a best response of type θ to the strategy (d, s).
G.J. Kim, B. Kim and J. Kim / Operations Research Letters 47 (2019) 326–330
329
Fig. 2. Plots of U(θi , (k, t)|(d, s)), k = 0, 1. Note that we use U(θi , (k, t)), k = 0, 1 instead of U(θi , (k, t)|(d, s)), k = 0, 1 in the label of the y-axis.
Next, we prove assertion (ii). Suppose that t ∈ [0, t ∗ ) and 0 ≤ θ ′ < θ < ∞, and that (0, t) is a best response of type θ to the strategy (d, s). If d(θ ′ ) = 1, then U(θ ′ , (1, s(θ ′ ))) − U(θ ′ , (0, s(θ ′ )))
=
(θ′ 2
)( ) 1 − ψ P(s(Θ ) > s(θ ′ )) + P(s(Θ ) = s(θ ′ )) ≥ 0. 2
Since t < t ∗ and (0, t) is a best response of type θ to the strategy (d, s), we have θ ≤ 2ψ by assertion (i), thereby, θ ′ < 2ψ , ′ i.e., θ2 − ψ < 0. Thus, if d(θ ′ ) = 1, then P(s(Θ ) ≥ s(θ ′ )) = 0 and U(θ ′ , (1, s(θ ′ ))) = U(θ ′ , (0, s(θ ′ ))). Hence, for either d(θ ′ ) = 0 or 1, (0, s(θ ′ )) is a best response of type θ ′ to the strategy (d, s). By Lemma 1, s(θ ′ ) ≤ t for all θ ′ ∈ [0, θ ). □ Lemma 3.
We have the following:
(i) P(d(Θ ) = 0, Θ > 2ψ ) = 0, (ii) s(2ψ ) ≥ t ∗ ,
(iii) d(θ ) = 0 and s(θ ) < t ∗ for all θ ∈ [0, 2ψ ). Proof. By Lemma 2(i), if θ > 2ψ and d(θ ) = 0, then s(θ ) ≥ t¯ . Hence P(d(Θ ) = 0, Θ > 2ψ, s(Θ ) < t ∗ ) = 0. By (9) and the definition of t ∗ , P(d(Θ ) = 0, s(Θ ) ≥ t ∗ ) = 0. Therefore, P(d(Θ ) = 0, Θ > 2ψ ) = 0, which is assertion (i). Next, we prove assertion (ii) by contradiction. Suppose s(2ψ ) < t ∗ . Observe that U(2ψ, (0, t)) = U(2ψ, (1, t)). By Lemma 2 (ii), s(θ ) ≤ s(2ψ ) for all θ ∈ [0, 2ψ ). From this and assertion (i), we have t ∗ = sup{t ≥ 0 : P(d(Θ ) = 0, s(Θ ) > t) > 0} ≤ s(2ψ ) < t ∗ , which is a contradiction. Finally, we prove assertion (iii). We have that sup{θ < 2ψ : P(θ < Θ < 2ψ, d(Θ ) = 0) > 0} = 2ψ.
(11)
This is proved by contradiction, which is as follows. Suppose θ˜ ≡ sup{θ < 2ψ : P(θ < Θ < 2ψ, d(Θ ) = 0) > 0} < 2ψ . ˜ 2ψ ) such that d(θ ′ ) = 1 and P(s(Θ ) Then there exists θ ′ ∈ (θ,
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≥ s(θ ′ )) > 0. By Lemma 2(i), d(θ ′ ) = 0, which is a contradiction. Therefore, we have (11). From this and (9), there exists a strictly increasing sequence {θn } such that limn→∞ θn = 2ψ , d(θn ) = 0 and s(θn ) < t ∗ . By Lemma 2 (ii), s(θ ) < t ∗ for all θ < 2ψ , and by Lemma 2(i), d(θ ) = 0 for all θ < 2ψ . □
Therefore, if θ > 2ψ , then (0, t ∗ ) is not a best response of type θ to the strategy (d, s) and so (d(θ ), s(θ )) = (1, t ∗ ). □ If we show that s(θ ) =
θ
∫
1 − P(x)
0
Let
∫ 2ψ
N = {θ ∈ [0, ∞) : θ ≥ 2ψ, s(θ ) > t ∗ }. From P(d(Θ ) = 0, s(Θ ) > t ∗ ) = 0, it follows that P(d(Θ ) = 1, s(Θ ) > t ∗ ) = 0. Hence P(s(Θ ) > t ∗ ) = 0 and N should be a null set. Lemma 4.
If θ ∈ (2ψ, ∞) \ N, then d(θ ) = 1 and s(θ ) = t ∗ .
Proof. Suppose θ > 2ψ and t ∈ [0, t ∗ ). Then U(θ, (1, t ∗ )) − U(θ , (1, t)) =
(θ
) + ψ P(Θ < 2ψ, s(Θ ) > t) 2 ∫ t∗ − P(Θ > τ )dτ . (12) t
By (9), (10), Lemma 1 and Lemma 3 (iii), s(·) is a strictly increasing and continuous function from [0, 2ψ ) onto [0, t ∗ ). Let g : [0, t ∗ ) → [0, 2ψ ) be the inverse function. Then, for θ ′ ∈ (g(t), 2ψ ), U(θ ′ , (0, s(θ ′ ))) − lim U(θ ′ , (0, t ′ )) t ′ →t +
= θ P(Θ < 2ψ, s(Θ ) ∈ (t , s(θ ′ ))) ){ (θ′ − ψ P(Θ > 2ψ, s(Θ ) ∈ (t , s(θ ′ ))) + 2 ∫ s(θ ′ ) } 1 + P(Θ > 2ψ, s(Θ ) = s(θ ′ )) − P(Θ > τ )dτ ′
2
t
≥ 0. Letting θ ′ → 2ψ− gives 2ψ P(Θ < 2ψ, s(Θ ) > t) ≥
t∗
∫
P(Θ > τ )dτ . t
From this and (12), we obtain U(θ, (1, t ∗ )) − U(θ , (1, t)) ≥
(θ
) − ψ P(Θ < 2ψ, s(Θ ) > t) > 0.
2 Hence, (1, t) cannot be a best response of type θ to the strategy (d, s). By Lemma 2(i), (d(θ ), s(θ )) ̸ = (0, t). Therefore s(θ ) ̸ = t for all t < t ∗ and so s(θ ) ≥ t ∗ for all θ > 2ψ . Hence, s(θ ) = t ∗ since θ∈ / N and s(θ ) ≥ t ∗ . Next, we prove that if θ > 2ψ and s(θ ) = t ∗ , then d(θ ) = 1. Suppose θ > 2ψ . Then U(θ, (1, t ∗ )) − U(θ , (0, t ∗ )) =
1(θ 2 2
xp(x)
) − ψ P(Θ > 2ψ ) > 0.
dx for θ ∈ [0, 2ψ ),
xp(x)
then t ∗ = 0 1−P(x) dx = t¯ , and so the necessity of Theorem 1 is complete. For θ ∈ [0, 2ψ ) and t ∈ [0, t ∗ ), U(θ, (0, t)) =
g(t)
∫
(θ − s(θj ))p(θj )dθj − t(1 − P(g(t))). 0
For each θ ∈ [0, 2ψ ), U(θ, (0, t)) takes a maximum at t = s(θ ). Hence (θ − s(g(s(θ ))))p(g(s(θ )))g ′ (s(θ ))
− (1 − P(g(s(θ )))) + s(θ )p(g(s(θ )))g ′ (s(θ )) = 0. This gives s′ (θ ) = s(θ ) =
θ
∫ 0
θ p(θ ) . 1−P(θ )
xp(x) 1 − P(x)
Since s(0) = 0,
dx for θ ∈ [0, 2ψ ).
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