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Conflict Under Incapacitation and Revenge: A Game-Theoretic Exploration

  • Puja Mukherjee ORCID logo EMAIL logo
Published/Copyright: February 27, 2024

Abstract

In real life, winning a conflict sometimes does not end the conflict. Revenge motivations can stay and provide momentum to the conflict, thus leading to further escalation of the conflict. This is known as the value effect or vengeance effect of revenge. However, the presence of revenge can lead to de-escalation of the conflict out of self-deterrence and sometimes retaliation out of revenge is not possible if the combatant is incapacitated. Hence, the impact of revenge on the level of violence is a priori not clear. This paper is an attempt to answer that question. Using a two-period game of conflict this paper tries to show how desire and capabilities of the combatants to exact revenge can influence the intensity of the conflict. This paper shows the following: how the strategies of the combatants are influenced by the value effect of revenge, self-deterrence and incapacitation of the combatants; how the stronger combatant is in a favourable position in the conflict and can prevent its opponent from going into second period conflict out of revenge; when the combatants are equally strong the intensity of the conflict starts falling with time. It also lays out some real-life conflicts and existing empirical work to support the results.

JEL Classification: D74; C70

Corresponding author: Puja Mukherjee, Economic Research Unit, Indian Statistical Institute, Kolkata, India, E-mail:

Appendix A: When Desire for Revenge is Sensitive to the Destruction Suffered in the Previous Period

In this framework the revenge-capability function becomes, R i = x j 1 ϕ * ( α i γ i x j 1 ) .

The revenge-capability function can be written in the following way (considering the revenge-capability function of combatant A);

(26) R A = x B 1 ϕ ( α A γ A x B 1 ) > 0 when 0 < x B 1 < ϕ α A ( 1 + ϕ ) γ A .
(27) R A = 0 when x B 1 = 0 .
(28) R A = 0 when x B 1 > = α A γ A .

Like before when x B 1 > = α A γ A combatant B has totally incapacitated its opponent A in the first period conflict and there won’t be any second period conflict out of revenge.

The R P A point is ϕ α A ( 1 + ϕ ) γ A .

If the desire for revenge is more sensitive to the destruction suffered in the previous period that is ϕ > 1 then the R P A point of A will be high. Thus the level of incapacitation where it would have less incentive to retaliate is more for the combatant. A is very revengeful and its desire for revenge will fall only when the incapacitation will be at a high level.

The less the sensitivity (ϕ < 1), the lower the R P point and thus the incapacitation to level where it would have less incentive to retaliate is less.

If the sensitivity to previous destruction is low such that both the combatants can reach its opponent’s revenge peak point then from Proposition 5 we know that the intensity of the conflict falls with time. Thus, when the elasticity is low, the urge to go into second period conflict out of revenge depends mostly on the capability factor and otherwise when the sensitivity is high.

Appendix B: Proof of Proposition 1

(29) X R X ̃ = 2 x B 1 x A 1 V ( x A 1 + x B 1 ) 2 2 R B 3 x B 1 ( α A 2 γ A x B 1 ) ( R A + R B ) 3 + 2 R A 3 x A 1 ( α B 2 γ B x A 1 ) ( R A + R B ) 3 + R A R B ( R A + R B ) V 2 .
(30) X R X ̃ = 2 x B 1 x A 1 V ( x A 1 + x B 1 ) 2 V 2 + R A R B ( R A + R B ) 2 R B 3 x B 1 ( α A 2 γ A x B 1 ) ( R A + R B ) 3 + 2 R A 3 x A 1 ( α B 2 γ B x A 1 ) ( R A + R B ) 3 .
(31) X R X ̃ = V ( x A 1 x B 1 ) 2 2 ( x A 1 + x B 1 ) 2 + R A R B ( R A + R B ) 2 R B 3 x B 1 ( α A 2 γ A x B 1 ) ( R A + R B ) 3 + 2 R A 3 x A 1 ( α B 2 γ B x A 1 ) ( R A + R B ) 3 .

We have to check for the second term in equation (31), since the first term is negative, as shown.

The second part of the equation can be written as:

(32) R A R B ( R A + R B ) 2 2 ( R A 3 x A 1 ( α B 2 γ B x A 1 ) + R B 3 x B 1 ( α A 2 γ A x B 1 ) ) ( R A + R B ) 3 .
(33) R A R B ( R A + R B ) 2 2 ( R A 3 R B γ B x A 1 2 + R B 3 R A γ A x B 1 2 ) ( R A + R B ) 3 .

Further simplifying the equations;

(34) 2 ( R A R B ) 2 R A 3 x A 1 ( α B 3 γ B x A 1 ) R B 3 x B 1 ( α A 3 γ A x B 1 ) ( R A + R B ) 3 .
(35) R A 2 ( R B 2 R A x A 1 ( α B 3 γ B x A 1 ) ) + ( R B 2 ( R A 2 R B x B 1 ( α A 3 γ A x B 1 ) ) ) ( R A + R B ) 3 .

Let x A 1 < α B 3 γ B , x B 1 < α A 3 γ B , and R A  > R B such that R B 2 < R A x A 1 ( α B 3 γ B x A 1 ) , which makes the first term negative. The second term of equation (35) is positive thus if R A is sufficiently large that it dominates the positive value of the second term, making the whole equation (35) negative. Thus, the paradox of revenge can be observed. Similarly, when R B  > R A .

Appendix C: Proof of Proposition 2

First period FOCs of the combatants can be written in the following way:

(36) x i 1 = x B 1 x A 1 V ( x A 1 + x B 1 ) 2 2 R i 3 x i 1 ( α j 2 γ j x i 1 ) ( R A + R B ) 3 , where i j and i = A , B .

We can write (36) in the following way:

(37) x A 1 = x B 1 + 2 R B 3 x B 1 ( α 2 γ x B 1 ) 2 R A 3 x A 1 ( α 2 γ x A 1 ) ( R A + R B ) 3 .

When xA1 = xB1 LHS = RHS of equation (37).

(38) x ̂ A 1 = x ̂ B 1 = x ̂ 1 = V 4 x ̂ 1 ( α 2 γ x ̂ 1 ) 4 .

Total conflict X R is;

(39) X R = 2 x + R A R B ( R A + R B ) = V 2 x ̂ 1 ( α 2 γ x ̂ 1 ) 2 + x ̂ 1 ( α γ x ̂ 1 ) 2 .
(40) X R = V 2 + x ̂ 1 2 γ 2 .

Appendix D: Proof of Proposition 3

In the benchmark case of no revenge, the payoffs are as follows:

(41) π ̃ A 1 = π ̃ B 1 = V 4 .

The payoffs of A and B in the presence of revenge are as follows:

(42) π ̂ A 1 = x A 1 ( x A 1 + x B 1 ) V x A 1 + π ̂ A 2 ,
(43) π ̂ B 1 = x B 1 ( x A 1 + x B 1 ) V x B 1 + π ̂ B 2 .

Comparing these with the payoffs in the benchmark case of no revenge, we get:

(44) π ̂ A 1 π ̃ A 1 = V ( x ̂ B 1 3 x ̂ A 1 ) 4 ( x ̂ A 1 + x ̂ B 1 ) x ̂ A 1 + R A 3 ( R A + R B ) 2 .

From equation (44) it is evident that if the first period equilibrium conflict investment of A is relatively very high from the first period conflict investment of B then the payoff of A is more in the presence of revenge. However, if the equilibrium conflict investment of B is relatively very high than that of A then payoff of A will be lower in the presence of revenge than without revenge. This shows if the B has incapacitated A to a high level in the first period conflict then A will not want to go into a second period conflict out of revenge. The comparison is similar for that of B.

When the capabilities of the combatants is similar we already know from Proposition 2 that the first period conflict investments of the combatants are equal and thus equation (44) becomes;

(45) π ̂ A 1 π ̃ A 1 = V 4 x ̂ 1 + R A 3 ( R A + R B ) 2 .

From Proposition 2 we know the value of x ̂ 1 and from there we can write;

(46) V 4 x ̂ 1 = x ̂ 1 ( α 2 γ x ̂ 1 ) 4 .

Now, putting this value of ( V 4 x ̂ 1 ) in (45) we get:

(47) π ̂ A 1 π ̃ A 1 = x ̂ 1 ( α 2 γ x ̂ 1 ) 4 + R A 3 ( R A + R B ) 2 .

When x ̂ 1 < α 2 γ then equation (46) becomes positive which makes equation (45) positive. Which shows that both the combatants are better off in the presence of revenge when both the combatants are not able to incapacitate each other significantly in the first period conflict. Thus, when the revenge function of both the combatants are rising both of the combatants are willing to go into a second period conflict out of revenge.

If x ̂ 1 > α 2 γ then equation (46) becomes negative and if this negative value is high enough such that it negates the positive desire of exacting revenge, then equation (45) becomes negative. Thus, both the combatants are worse off in the presence of revenge when both the combatants are capable enough to incapacitate each other to a high level in the first period conflict.

Now, let us calculate the value of the first period conflict investment in the symmetric case when the combatants would be (better-off) worse-off in the presence of revenge.

(48) π ̂ A 2 = R A 3 ( R A + R B ) 2 = R 4 = x ̂ 1 ( α γ x ̂ 1 ) 4 .

Since this is a symmetric case and the equilibrium payoff of A equals that of B so re-writing (47) in terms of i where i = A, B and putting the value of π ̂ A 2 in the equation, we get the following:

(49) π ̂ i 1 π ̃ i = x ̂ 1 ( α 2 γ x ̂ 1 ) 4 + x ̂ 1 ( α γ x ̂ 1 ) 4 ,
(50) π ̂ i 1 π ̃ i = x ̂ 1 ( 2 α 3 γ x ̂ 1 ) 4 , where i = A , B .

From equation (50), it is evident that when x ̂ 1 > 2 α 3 γ , equation (50) is negative that is π ̂ i < π ̃ i for both A, B. The SPNE in this case will be to not take revenge. When x ̂ 1 < 2 α 3 γ , equation (50) is positive that is π ̂ i > π ̃ i for both A, B. The SPNE in this case will be to take revenge.

Appendix E: Simplification of equation (25)

We can write equation (20) in the following way;

(51) x ̂ A 1 + 2 R A 3 x A 1 ( α B 2 γ B x A 1 ) ( R A + R B ) 3 = x ̂ B 1 + 2 R B 3 x B 1 ( α A 2 γ A x B 1 ) ( R A + R B ) 3 .
(52) x ̂ A 1 1 + 2 R A 3 ( α B 2 γ B x A 1 ) ( R A + R B ) 3 = x ̂ B 1 1 + 2 R B 3 ( α A 2 γ A x B 1 ) ( R A + R B ) 3 .

We know from the formulation of the revenge function that R i = ( α i 2 γ i x j 1 ) where, i ≠ j, thus we can write the above equation in the following way;

(53) x ̂ A 1 1 + 2 R A 3 R B ( R A + R B ) 3 = x ̂ B 1 1 + 2 R B 3 R A ( R A + R B ) 3 .

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Received: 2023-10-02
Accepted: 2024-02-11
Published Online: 2024-02-27

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