Consider the Following Game Played Between a Boss (B) and an Employee (E)

Home , Ultimatum game

Microeconomics of Banking Solutions to Homework #3 Q1: Consider the following game played between a boss (B) and an employee (E). The boss offers a wage, w ≥ 0. After observing the wage offer, the employee decides how much effort, e ≥ 0, to expend. The payoff functions for the boss and the employee are (α is a constant, α ≥ 0): √ uB(w, e) = 2 e − w e2 u (w, e) = w − + αwe E 2 (a) What is the optimal effort for the employee, as a function of w? The employee chooses e ≥ 0 to maximize

e2 u (w, e) = w − + αwe E 2 This is a downwards facing quadratic. The ﬁrst-order condition is: ∂u E = −e + αw = 0 ∂e Therefore, the maximizing value of e as a function of w is e∗(w) = αw. (b) What are the subgame perfect equilibrium choices of w and e, as a function of α? Given e∗(w) = αw, the boss chooses w to maximize √ 2 αw

The ﬁrst-order condition is: ∂u r α B = − 1 = 0 ∂w w We verify this is a maximum by checking the second-order condition:

2 1 −3 ∂ u α 2 w 2 B = − < 0 ∂2w 2 The optimal choice of w is w∗ = α. Therefore, the unique SPNE is:

e∗(w) = α2, w∗ = α

Q2: Recall the standard ultimatum game, in which two players attempt to split c = 1 between themselves: 1. Player 1 offers an amount x ∈ [0, c]. 2. Player 2 can choose Accept (in which case the pair of payoffs is c − x, x) or Reject (in which case the pair of payoffs is 0, 0).

1 Now, suppose that Player 2 also cares about the equity of the allocation. Let x1, x2 denote the amount of money received by Player 1 and Player 2, respectively. Suppose Player 2’s payoff is given by: u2(x1, x2) = x2 − β|x1 − x2| where β is a positive constant. (Note that if β = 0, then this becomes the original ultimatum game). Now, Player 2 prefers a smaller absolute value of the difference between x1 and x2. (a) Find the set of subgame perfect equilibria, as a function of β > 0. Let x denote the amount offered by Player 1. Then Player 2’s utility if he accepts becomes: u2(x) = x − β|1 − x − x| = x − β|1 − 2x|. Consider the graph of u2(x). It has a kink at x = 1/2, where |1 − 2x| = 0. It will be a piecewise linear function:

( 1 −β + (1 + 2β)x if x < 2 u2(x) = 1 β + (1 − 2β)x if x ≥ 2 The left half of this function is an increasing function, with slope 1 + 2β. Whether the right half is increasing, decreasing, or ﬂat depends on β: 1 β • If β < 2 , it is decreasing, and there are two points where u2(x) = 0, at x = 2β+1 , β and x = 2β−1 . 1 • If β = 2 , it is ﬂat. 1 • If β > 2 , it is increasing. In this and the previous case, there is only one point where β u2(x) = 0, at x = 2β+1 .

Player 2’s optimal strategy is: accept if u2(x) ≥ 0. Player 1 wants to oﬀer the lowest β possible x that still induces Player 2 to accept, which is when x = 2β+1 , which results in β+1 payoﬀs of ( 2β+1 , 0). Now, suppose that there are two types of Player 2: with probability p, the value of β will be 0, and with probability 1 − p, the value of β will be 1. After Player 1 chooses x, Nature will randomly choose the type of Player 2. (b) Draw the graph of this extensive form game (note: this is a game with perfect information).

2 (c) Find the set of subgame perfect equilibria. With probability p, Player 2’s strategy will be to accept any x ≥ 0, and Player 1’s payoff for offering x ≥ 0 will be 1 − x. With probability 1 − p, Player 2’s strategy will be to β 1 accept any x ≥ 2β+1 = 3 . For all x ∈ [0, 1], we can calculate the expected payoff to Player 1. 1 • If 0 ≤ x < 3 , Player 1’s expected payoff is p(1 − x) + (1 − p)0 = p(1 − x), which is maximized at x = 0, giving an expected payoff of p. 1 • If 3 ≤ x ≤ 1, Player 1’s expected payoff is p(1 − x) + (1 − p)(1 − x) = 1 − x, which 1 2 is maximized at x = 3 , giving an expected payoff of 3 . 2 The payoff-maximizing choice of x will depend on p. If p > 3 , Player 1’s optimal offer is 2 1 0. If p ≤ 3 , Player 1’s optimal offer is 3 . Q3: (Monty Hall problem) A contestant is on a TV game show, and is presented with three doors. Behind one of the doors is a car; behind the other two doors, a goat. The contestant values a car more than a goat. The sequence of actions is as follows: 1. The contestant picks a door, but the contents are not revealed. 2. The host, who knows where the car is located, opens one of the other two doors, revealing there is a goat inside. 3. The contestant now has the option of switching the door he has chosen. Should he switch? We will state this problem using conditional probability. Let A be the event that the contestant’s chosen door holds a car. Let B be the event that the host opens one of the doors,

3 revealing a goat. Then, we want to find the conditional probability P (A|B). If this probabil- 1 ity is ≥ 2 , then the contestant should switch. Assume P (A), the unconditional probability of 1 the contestant choosing the door with the car, is 3 . • Use Bayes’ Rule to formulate P (A|B). For the next four questions, calculate P (A|B) using the given assumptions. • Suppose the host always reveals a goat, no matter which door the contestant chooses. • Suppose the host always chooses from the remaining two doors with equal probability, and this is independent of the contestant’s choice (so if the contestant picked the car; he reveals a goat; if the contestant picked a goat, he reveals the car half of the time). • Suppose the host only reveals a goat and offers to switch if the contestant picks a goat. • Suppose the host only reveals a goat and offers to switch if the contestant picks a car. Let A be the event that the contestant chooses the door holding the car, and B be the event that the host opens a door holding a goat. ¬A is the event that the contestant does not choose 1 1 the door holding the car. If P (A|B) < 2 , then the contestant should switch; if P (A|B) = 2 , 1 the contestant is indifferent, and if P (A|B) > 2 , the contestant should not switch. According to Bayes’ Rule, P (B|A)P (A) P (A|B) = P (B|A)P (A) + P (B|¬A)P (¬A) 1 2 In this problem, P (A) = 3 ,P (¬A) = 3 . • Suppose the host always reveals a goat, regardless of whether the contestant chose the 1 door with the car or not. Then P (B|A) = 1,P (B|¬A) = 1, and P (A|B) = 3 , so the contestant should switch. 1 • Suppose the host opens a door randomly. Then P (B|A) = 1,P (B|¬A) = 2 , and 1 P (A|B) = 2 , so the contestant is indifferent. • Suppose the host only reveals a goat when the contestant picks a goat. Then P (B|A) = 0,P (B|¬A) = 1, and P (A|B) = 0, so the contestant should switch (it is certain that the car is behind the other door). • Suppose the host only reveals a goat when the contestant picks a car. Then P (B|A) = 1,P (B|¬A) = 0, and P (A|B) = 1, so the contestant should not switch (it is certain that the car is behind the contestant’s door). Q4: Consider the following version of the entry game:

4 The sequence of actions is as follows: 1. Nature chooses the challenger (Player 1)’s type, which can be weak (W ) or competitive (C), with probability p and 1 − p respectively. A weak-type challenger will cost less for the incumbent to fight. The incumbent (Player 2) does not know what Nature chooses. 2. The challenger knows his type, and chooses Enter (E) or Out (O). 3. The incumbent observes the challenger’s choice, but not his type, and chooses F ight (F ) or Acquiesce (A). There are 4 (pure) strategies for Player 1 and 2 (pure) strategies for Player 2. (a) Suppose p = 1 (Nature always chooses C). Calculate the 4 × 2 matrix of payoffs of the strategic form. (b) Suppose p = 0 (Nature always chooses W ). Calculate the 4 × 2 matrix of payoffs of the strategic form. 1 (c) Now, suppose p = 2 . Calculate the 4 × 2 matrix of expected payoffs of the strategic form. There are two information sets for Player 1, and one information set for Player 2, each with 2 actions. Therefore, Player 1 has 4 pure strategies, and Player 2 has 2 pure strategies. 1. If p = 0:

5 AF OO 0,2 0,2 OE -2,0 -1,1 EO 0,2 0,2 EE -2,0 -1,1

2. If p = 1:

AF OO 0,2 0,2 OE 0,2 0,2 EO -1,-1 1,1 EE -1,-1 1,1

3. If p = 0.5, we can take the weighted combination of the ﬁrst two tables.

AF OO 0,2 0,2 OE -1,1 -0.5,1.5 EO -0.5,0.5 -0.5, 1.5 EE -1.5,-0.5 0,1