Chapter+2

Chapter 2: Game Theory
1. This is the matrix game: I put stars next to the optimum strategy for row on the left and for column on the right. It can be clearly seen that row would never choose T, so T is a dominated strategy. The pure strategy Nash equilibrium is (B, R) with payoffs (3, 1). It's Pareto efficient since it's impossible to improve either payoff without reducing the other player's payoff. Therefore, by making binding agreements the strategy wouldn't change.
 * || L || R ||
 * T || 2, 5* || 1, 4 ||
 * B || *5, -1 || *3, 1* ||

2. The table below is a zero sum game showing player I's payoff (player II's payoff is the negative) Again, I'm using stars to indicate optimum strategies. The pure strategy Nash equilibrium gives a payoff of 0 to either player, so there is no point playing it.
 * ||||||||= Player II ||
 * || -1 || *2 || -3* || 0 ||
 * Player I || *0* || 1 || 2 || *3 ||
 * || -2 || -3* || *4 || -1 ||

3. For p=50, q=1000; for p=60, q=900. Costs=40. For the same price, they sare the market. For lower price, they get all the market. The payoff matrix is: The Nash equilibrium is both companies charging 50, although they would be both better off coluding and charging 60.
 * || p=50 || p=60 ||
 * p=50 || *(5000, 5000)* || *(10000, 0) ||
 * p=60 || (0, 10000)* || (9000, 9000) ||

If they colude in the first year and offer a most favoured customer clause (i.e., they'll offer the lowest price in the market for their existing customer), the new payoff matrix is:
 * || p=50 || p=60 ||
 * p=50 || (5000, 5000) || (5500, 4500) ||
 * p=60 || (4500, 5500) || (9000, 9000) ||

4. Find pure and mixed strategy for each game:

Pure strategies (T, L) and (B, R) Let r be the probability of row choosing T and c the probability of column choosing c, we have the following table:
 * || L || R ||
 * T || *12, 10* || 4, 4 ||
 * B || 4, 4 || *9,6* ||
 * Combination || Probability || Row Payoff || Column Payoff ||
 * T, L || r.c || 12 || 10 ||
 * T, R || r.(1-c) || 4 || 4 ||
 * B, L || (1-r).c || 4 || 4 ||
 * B, R || (1-r).(1-c) || 9 || 6 ||

Row's Payoff Expected PR = 12rc+4r(1-c)+4(1-r)c+9(1-r)(1-c) PR = 12rc+4r-4rc+4c-4rc+9-9c-9r+9rc PR = 13rc-5r-5c+9 Deriving PR on r : d.PR/d.r = 13c-5 = 0 => c=5/13 The same for PC = 10rc + 4r(1-c)+4(1-r)c+6(1-r)(1-c) PC= 10rc + 4r - 4rc + 4c - 4rc + 6 - 6c - 6r + 6rc PC = 8rc - 2r - 2c + 6 d.PC/d.c = 8r - 2 = 0 => r=1/4 We get the following equilibrium chart:

Changing to -100, the pure strategies won't change, but PC will change:
 * || L || R ||
 * T || *12, 10* || 4, 4 ||
 * B || 4, -100 || *9,6* ||


 * Combination || Probability || Row Payoff || Column Payoff ||
 * T, L || r.c || 12 || 10 ||
 * T, R || r.(1-c) || 4 || 4 ||
 * B, L || (1-r).c || 4 || -100 ||
 * B, R || (1-r).(1-c) || 9 || 6 ||

PC = 10rc + 4r(1-c) - 100c(1-r) + 6(1-r)(1-c) PC = 10rc + 4r - 4rc - 100c + 100rc + 6 - 6c - 6r + 6rc PC = 112rc - 2r -106c + 6 d.PC/d.C = 112r - 106 = 0 => r= 53/56 And the new equilibrium will be:

5. This is the students/teacher homework problem. The payoff table is: Again, I'm using a star to indicate the optimal strategies. a. What is teacher's optimal strategy? Will the students do any work? There is no pure strategy Nash equilibrium. The teacher's best strategy is to check homework only if the students don't do any work and don't bother checking if they work. Students will only work if they think the teacher will check and don't work otherwise.
 * ||  ||||= Teacher ||
 * ||  || check || don't check ||
 * Student || work || *0, -3 || 0, 0* ||
 * || no work || -4, 4* || *1, -2 ||

b. Suppose the teacher tells the students at the beginning of the year that all homewrok will be checked and the students believe her. Will they do the work? Is the teacher likely to stick to this policy? In this case, the students will only consider the "check" column and, therefore, will do their work. The teacher, on the other hand, dismissing the "no work" row won't actually check their homework.

c. Suppose the teacher could commit to checking the homework part of the time but students will not know exactly when. What is the minimal degree of checking o that students are encouraged to do the work? This exercise is in fact asking for the mixed strategy equilibrium. Let s be the probability of students doing the work and t the probability of the teacher checking their work: PS = -4t(1-s) + (1-s)(1-t) PS= -4t + 4ts + 1 - t - s + st PS= 5st - s - 5t + 1 d.PS/d.s = 5t - 1 = 0 => t=1/5 So, the teacher should check at least 20% of the homework to encourage the students to do it.
 * Combination || Probability || Student || Teacher ||
 * Work / Check || s.t || 0 || -3 ||
 * Work / Don't check || s.(1-t) || 0 || 0 ||
 * No work / check || (1-s).t || -4 || 4 ||
 * No work / don't check || (1-s).(1-t) || 1 || -2 ||

6. Given te simultaneous move game below, ind the pure strategy Nash equilibria. How would you play if you were the row player? How would you play if you were the column player? Again, using stars to show optimal strategies we find the Nash equilibria (u, a) and (d, d). Assuming this is a successive game, and disregarding the dominated strategies b and c, we can calculate the mixed strategy Nash equilibrium:
 * || a || b || c || d ||
 * u || *100, 3* || 2, 2 || 2, 1 || 0, -500 ||
 * d || 0, -5000 || *3, -500 || *3, 2 || *1, 10* ||
 * Combination || Probability || Row payoff || Column payoff ||
 * u, a || r.c || 100 || 3 ||
 * u, d || r.(1-c) || 0 || -500 ||
 * d, a || (1-r).c || 0 || -5000 ||
 * d, d || (1-r).(1-c) || 1 || 10 ||

PR = 100rc + (1-r)(1-c) PR = 100rc + 1 - c - r + rc PR = 101rc - r - c + 1 d.PR/d.r = 101c - 1 = 0 => c= 1/101=0.01

PC = 3rc - 500 r(1-c) - 5000c(1-r) + 10 (1-r)(1-c) PC = 3rc - 500r + 500rc - 5000c + 5000rc + 10 - 10c - 10r + 10rc PC = 5513rc - 510r -5010c + 10 d.PC/d.c = 5513r - 5010 = 0 => r=5010/5513=0.91

This suggests that row should play u 91% of the time and column should play d 99% of the time. This would give an expected payoff of 0.99 to row and -453 to column!

7. This exercise asks the expected outcome and the Nash equilibrium for the Insurance problem. The matrix is as below:
 * || low || high ||
 * low || *(l, l)* || (l+2, l-2) ||
 * high || (l-2, l+2) || *(h, h)* ||

There are two Nash equilibrium (low, low) and (high, high). If low is the actual value of the print and high is the highest value they can get away with ($10,000), we can find out what is the best number for them to write down:
 * Combination || Probability || Row payoff || Column payoff ||
 * low, low || r.c || l || l ||
 * low, high || r.(1-c) || l+2 || l-2 ||
 * high, low || (1-r).c || l-2 || l+2 ||
 * high, high || (1-r).(1-c) || 10,000 || 10,000 ||

PR = rcl + (l+2)r(1-c) + (l-2)c(1-r) + 10000(1-r)(1-c) PR = rcl + rl - lrc + 2r - 2rc + lc - lcr - 2c + 2cr + 10000 - 10000c - 10000r + 10000rc PR = rc(l-l-2-l+2+10000) + r(l+2-10000) + c(l-2-10000) + 10000 PR = rc(10000-l) + r(l-9998) + c(l-10002) + 10000 d.PR/d.R = c(10000-l) + (l-9998) = 0 => c=(9998-l)/(10000-l) = (k-2)/k where k is 10000 minus the actual value, suggesting a very high number. In fact, because the $2 penalty is so small, they'll most likely write 10000.

8. The extensive game shows:

Building the payoff matrix, there are 4 Nash Equilibrium: (D,(u,u)), (D,(u,d)), (U,(d,u)), (U,(d,d)). The subgame perfect equilibrium is (U,d).

9. Given the extensive form game below: What are the players' information set? - I couldn't find a decent explanation about information set, based on the best I could find I'd say that: Player 1: its own move (1 information) Player 2: player 1 played T or not (2 informations)

Write this game in normal form and analyse it using the normal form and the game tree
 * ||  ||||||||= Player 2 ||
 * ||  || (t,t) || (t,b) || (b,t) || (b,b) ||
 * || T || *3, 2* || *3, 2* || *2, 1 || *2, 1 ||
 * Player 1 || M || 1, 2 || 1, 4* || 1, 2 || 1, 4* ||
 * || B || 2, 1 || 2, 4* || *2, 1 || *2, 4* ||

There are 3 Nash Equilibrium: (T,(t,t)), (T,(t,b)) and (B, (b,b)). Using the tree, considering the dyamic of this game, because 1 can move first, he will choose T, unless he has reasons to believe player 2's threat of playing b.

10. If the firm polute it gets an extra payoff of g>0. The cost of checking by DE is c>0. If the firm is caught polluting then it pays a penalty p>g and DE gets a payoff of p-c>0. If DE doesn't inspect its payoff is 0. a. Suppose DE can observe whether the firm has poluted before it formaly checks, draw the game tree. What is the equilibrium? This is a sequential game with the firm moving first. The tree is: The equilibrium, shown in bold, is Not Polute and Don't Check.

b. Now DE can't observe before hand. The new tree is: There is no pure strategy equilibrium. To calculate the mixed strategy equilibrium, let f be the probability of the firm poluting and d the probability of DE checking:
 * Combination || Probability || Firm's payoff || DE's payoff ||
 * Polute, Check || f.d || g-p || p-c ||
 * Polute, Don't check || f.(1-d) || g || 0 ||
 * Don't polute, Check || (1-f).d || 0 || -c ||
 * Don't polute, Don't check || (1-f).(1-d) || 0 || 0 ||

PF = fd(g-p) + gf(1-d) d.PF/df = d(g-p)+g(1-d) = g-dp = 0 => d=g/p

PD = fd(p-c)-cd(1-f) d.PD/d.D = f(p-c)-c(1-f)= fp-c = 0 => f=c/p

By increasing the penalty, the firm will decrease its likelihood of poluting, also reducing the necessity of the DE to check.