A Game-theoretic approach to non-life insurance.

1. A Game-theoretic approach to non-life insurance. Lorna Pamba & Karol Rakowski

2. Introduction: • Goal of research as mathematics majors with minors in economics • Nash Equilibrium: a stable state of a system involving the interaction of different participants, in which no participant can gain by a unilateral change of strategy if the strategies of the others remain unchanged

3. Some background Information; • Hawk and Dove Game • Player choice either Hawk(H) or Dove (D) • Four pure strategy combinations ; HH,HD,DH,HH is the reproductive value of territory is the cost of being injured in a fight

4. In bird's 1 point of view, lets consider the cases: • DH( Bird1 plays dove and Bird 2 plays Hawk) Payoff = 0 • HD( Bird 1 plays Hawk and Bird 2 plays Dove) Payoff = ρ • DD Expected payoff=0*Prob(F=I)+ρ*Prob(F=II)=ρ/2 • HH Exp=ρ*Prob(S=I)+(-C)*Prob(S=II)=(( ρ-C))/2

5. Payoff Matrices; Bird 2 Bird 1

6. Probabilities of mixed strategies; Bird II, play H with probability vand play D with probability If I and II choices don’t affect each other; = = Bird I, play H with probability uand play D with probability

7. Then f1(u,v)= E(F1) = = Similarly, f2 (u,v) = E(F2 ) Then f2 (u,v)= E(F2 ) = = For a fixed v For a fixed u

8. Therefore bird I’s reaction set is; R1 = And bird II’s reaction set is; R2 = { R1= blue line R2= Green line Intersection: Nash equilibrium

9. Insurance Market • 2 companies • Fixed number of customers in market • Possible strategies (increase or decrease in premiums) Represented by i andd. • Disclaimer of quantatative “real world” application of following work

10. Definitions

11. Payoff – Change in Revenue • Change in Revenue = (Change in Number of Customers) * (Change in Premium) - (Previous Revenue). • Change in Revenue =((∆rI,II- ∆rII,I) * (.01 * T) + PI,II) * (rI,II + ∆rI,II) – ((PI,II) * (RI,II))

12. Payoff Matrices

13. Payoff Matrices

14. Payoff Matrices • 4 possible outcomes for both companies • Want to create a payoff function so we can examine which strategy is best choice • dominant strategy (a strategy is dominant if it is always better than any other strategy)

15. Constructing Payoff Functions • we need to establish the probabilities, for both companies, of determining their decision to increase or decrease their premiums. • Let: u = ( Prob{I}=i) and v = ( Prob{II}=i ) • where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1 1 – u = ( Prob{I}=d ) and 1 – v = ( Prob{II}=d )

16. Constructing Payoff Functions

17. Constructing Payoff Functions

18. Evaluation of Payoff Function (1) • ƒI(u,v) = ((((-xI+ xII)*(.01T) + PI)*(rI+ xI)) - RI)(uv) + ((((-xI- yII)*(.01T) + PI)*(rI+ xI)) - RI)(1-v)u + ( (((yI+ xII)*(.01T) + PI)*(rI- yI)) - RI)(u-1)v + ((((yI- yII)*(.01T) + PI)*(rI- yI)) - RI)(1-u)(1-v) • ƒI(u,v) = u(v((xI+rI)(-.01xIT + .01xIIT + PI) – RI) + (1-v)((xI+rI)(-.01xIT - .01yIIT + PI) – RI) + (v((rI-yI)(.01T(xII + yI) + PI) – RI) – ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI))) – v((rI-yI)(.01T(xII + yI) + PI) – RI) + ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI)))

19. Evaluation of Payoff Function (1) cont. • ⍵ = (v((xI+rI)(-.01xIT + .01xIIT + PI) – RI) + (1-v)((xI+rI)(-.01xIT - .01yIIT + PI) – RI) + (v((rI-yI)(.01T(xII + yI) + PI) – RI) – ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI))) • ƒI(u,v) = u⍵– v((rI-yI)(.01T(xII + yI) + PI) – RI) + ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI)))

20. Evaluation of Payoff Function (2) • ƒII(u,v) = ((((-xII+ xI)*(.01T) + PII)*(rII+ xII)) - RII)(uv) + ((((-xII- yI)*(.01T) + PII)*(rII+ xII)) - RII)(1-v)u + ( (((yII+ xI)*(.01T) + PII)*(rII- yII)) - RII)(u-1)v + ((((yII- yI)*(.01T) + PII)*(rII- yII)) - RII)(1-u)(1-v) • ƒII(u,v) = v(u((xII+rII)(-.01xIIT + .01xIT + PII) – RII) - u((xII+rII)(-.01xIIT - .01yIT + PII) – RII) + (u-1)((rII-yII)(.01T(xI + yII) + PII) – RII) – ((1-u)((rII – yII)(.01yIIT - .01yIT + PII) – RII))) + u((rII-yII)(.01T(xI + yII) + PII) – RII) + ((1-u)((rII – yII)(.01yIIT - .01yIT + PII) – RII)))

21. Evaluation of Payoff Function (2) cont. • ⍺ = (u((xII+rII)(-.01xIIT + .01xIT + PII) – RII) - u((xII+rII)(-.01xIIT - .01yIT + PII) – RII) + (u-1)((rII-yII)(.01T(xI + yII) + PII) – RII) – ((1-u)((rII – yII)(.01yIIT - .01yIT + PII) – RII))) • ƒII(u,v) = v⍺+ u((rII-yII)(.01T(xI + yII) + PII) – RII) + ((1-u)((rII – yII)(.01yIIT - .01yIT + PII) – RII)))

22. Evaluation of Payoff Functions • Company 1 has no control over the value of v and hence has no control over the value of the expression v((rI-yI)(.01T(xII + yI) + PI) – RI) + ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI))) • The goal is to maximize u⍵ (when we maximize u⍵, we maximize ƒI). Again, we also know 0 ≤ u ≤ 1. So, when ⍵ is positive, negative, or equal to 0, we will have three different optimal values for u

23. Relation Between ⍵ and u • If, ⍵ > 0  Then, u = 1 • If, ⍵ = 0  Then, All u ∈ [0,1] • If, ⍵ < 0  Then, u = 0

24. Relation Between ⍺ and v • If, ⍺ > 0  Then, v = 1 • If, ⍺ = 0  Then, All v ∈ [0,1] • If, ⍺ < 0  Then, v = 0

25. Further Evaluation • A closer look at what determines the values of ⍵ and ⍺ • The ⍵ and ⍺ equalities are crucial to examine as they have a direct impact on the u, v strategies that will be taken by Companies 1 and 2.

26. Evaluation of ⍵ • ⍵ = (v((xI+rI)(-.01xIT + .01xIIT + PI) – RI) + (1-v)((xI+rI)(-.01xIT - .01yIIT + PI) – RI) + (v((rI-yI)(.01T(xII + yI) + PI) – RI) – ((1-v)((rI – yI)(.01yIT - .01yIIT + PI) – RI))) • ⍵ = v(-2 RI + (xI + rI) (PI - 0.01 xI T + 0.01 xII T) + (-yI + rI) (PI + 0.01 (xII + yI)T) - (xI + rI) (PI - 0.01 xI T - 0.01 yII T) + (-yI + rI) (PI + 0.01 yI T - 0.01 yII T))+ (xI + rI) (PI - 0.01 xI T - 0.01 yII T) - (-yI + rI) (PI + 0.01 yI T - 0.01 yII T)

27. Evaluation of ⍵ cont. • δ = (-2 RI + (xI + rI) (PI - 0.01 xI T + 0.01 xII T) + (-yI + rI) (PI + 0.01 (xII + yI)T) - (xI + rI) (PI - 0.01 xI T - 0.01 yII T) + (-yI + rI) (PI + 0.01 yI T - 0.01 yII T)) • a = (xI + rI) (PI - 0.01 xI T - 0.01 yII T) - (-yI + rI) (PI + 0.01 yI T - 0.01 yII T) • ⍵ = v δ + a

28. Evaluation of ⍺ • ⍺ = (u((xII+rII)(-.01xIIT + .01xIT + PII) – RII) - u((xII+rII)(-.01xIIT - .01yIT + PII) – RII) + (u-1)((rII-yII)(.01T(xI + yII) + PII) – RII) – ((1-u)((rII – yII)(.01yIIT - .01yIT + PII) – RII))) • ⍺ = u(-2 RII + (xII + rII) (PII - 0.01 xII T + 0.01 xI T) + (-yII + rII) (PII + 0.01 (xI + yII)T) - (xII + rII) (PII - 0.01 xII T - 0.01 yI T) + (-yII + rII) (PII + 0.01 yII T - 0.01 yI T))+(xII + rII) (PII - 0.01 xII T - 0.01 yI T) - (-yII + rII) (PII + 0.01 yII T - 0.01 yI T)

29. Evaluation of ⍺ cont. • λ = (-2 RII + (xII + rII) (PII - 0.01 xII T + 0.01 xI T) + (-yII + rII) (PII + 0.01 (xI + yII)T) - (xII + rII) (PII - 0.01 xII T - 0.01 yI T) + (-yII + rII) (PII + 0.01 yII T - 0.01 yI T)) • b = (xII + rII) (PII - 0.01 xII T - 0.01 yI T) - (-yII + rII) (PII + 0.01 yII T - 0.01 yI T) • ⍺ = u λ + b

30. Relation between v and ⍵ • v = 1  ⍵ = δ + a • v = 0  ⍵ = a • v = V0 ⍵ = 0

31. Relation between u and ⍺ • u = 1  ⍺ = λ + b • u = 0  ⍺ = b • u = U0 ⍺ = 0

32. Rational Reaction Set • for every v which Company 1 has no control over, there exists a corresponding u which is a strategy that will make Company 1’s payoff/benefit as large as possible given the situation. So a rational reaction set is a set of all of these possible combinations, given an opposing strategy that a player has no control over

33. Rational Reaction Set • ZI = { (u,v) | 0 ≤ u, v ≤ 1 , ƒI(u,v) = max0 ≤ ū ≤ 1ƒI(ū,v) } • For each (u,v) in ZI, if Company 2 selects v then a best reply for Company 1 is to select u (a best reply rather than the best reply because there may be more than one). Note that ZI is obtained in practice by holding v constant and maximizing ƒIas a function of a single variable (whose maximum will depend on v). • ZII = { (u,v) | 0 ≤ u, v ≤ 1 , ƒII(u,v) = max0 ≤ ῡ ≤ 1ƒI(u,ῡ) } *Similar Explanation*

34. First Example

35. Example 2

36. Third Example

37. Multiple Nash Equilibriums • Rational Introspection: A NE (Nash Equilibrium) seems a reasonable way to play a game because my beliefs of what other players do are consistent with them being rational. This is a good explanation for explaining NE in games with a unique NE. However, it is less compelling for games with multiple NE. • Markus Mobius, Lecture IV: Nash Equilibrium II - Multiple Equilibria. (lecture., Harvard, 2008), http://isites.harvard.edu/fs/docs/icb.topic449892.files/lecture42.pdf. • Risk Aversion method in decision

38. A continuous Non-coperative game. • 2 Insurance Companies; Geico and Progressive • Battleground for the two companies; Based on; Area under the curve should be 1 ; • Since we are trying to estimate some realistic situation , we’ll say; Therefore our probability distribution function;

39. This is our pdf if the mean is chosen to be 13000. X axis represents the number of miles driven in a year Y axis represents the population in (100000000)

40. Some assumptions made • Homogenous product but different level of satisfaction; • Best premium rate for the Geico depends upon Progressive premium rate and vice versa. • Geico and Progressive do not communicate with one another • Utility is measurable.

41. Let player 1 be Geico and player 2 be Progressive. • Let be the premium rate set by Geico and be the premium set by Progressive • Let u be the customers’ utility Accordingly, we can say ans makes progressive the better choice. Similarly, we can say and makes Geico the better choice.

42. Then we have 3 cases,

43. Without loss of generality, let’s assume all through that; , Let denote a cumulative distribution function Let F1 denote Geico’ s payoff function;

44. Similarly progressive’s payoff function; Let be Geico’ s and Nan’s strategy simultaneously, = =

45. And progressive’s reward function is Recall, that if Similarly;

46. The set of all feasible strategy combinations Or similarly; | Restricted price

47. Recall, with constant R2 = with constant This would help us find points in R1 if exact solutions were possible (would give potential p1 values to choose for a fixed p2).

48. This would help us find points in if exact solutions were possible (would give potential values to choose for a fixed). For example

49. Nash Equilibrium This shows R1 in blue and R2in red when k1=15/1000 and k2=17/1000.

50. Conclusion • Weaknesses of research • Justifications