Terry College of Business - ECON 7950 Lecture 5: More on the Hold-Up Problem + Mixed Strategy Equilibria Primary reference: Dixit and Skeath, Games of Strategy, Ch. 5.
The Hold Up Problem Let there be two players, a seller (S) of a single intermediate input and a buyer (B) who is capable of making a final good. The seller s cost of producing the input is c. The buyer s value of selling the final good is V (I ), where I is an investment that costs I. Let the investment be an increasing function of I, such that V (I ) > 0. Assume also that there are diminishing marginal returns. Suppose the investment is completely specialized...b cannot get value V (I ) unless it uses the intermediate input produced by S.
The Hold Up Problem The buyer and seller may vertically integrate. If they do, they must pay cost K but may jointly decide the level of I. The choice of whether to integrate occurs prior to any decision about I. If the buyer and seller do not integrate, the investment is non-contractible...b chooses it in advance of Nash bargaining, which determines the terms of trade. Let the buyer and seller have equal bargaining power.
The Hold Up Problem Identify an equation which establishes the values of K for which vertical integration is optimal. Use whatever notation you like but explain it clearly as you go.
The Holdup Problem Generally - Avoiding Both Bad Bargaining Positions and Underinvestment Intel, the pioneer of integrated circuit production, sells to manufacturers who must spend resources to tailor their products to Intel s chips. By tailoring, firms risk being locked in...so in early years Intel feared that this might make firms reluctant to do so. Their solution was to second source their technology to firms like Advanced Micro Devices (AMD). In many instances, a good solution to hold-up problems is vertical integration. Power plants located near coal mines are subject to holdup problems. Such plants and mines are frequently jointly owned (see Joskow 1985, 1987).
The Card Game An example of bargaining with more than 2 players. I have 26 black cards. The 26 red cards are distributed among 26 individuals. Each pair of cards turned in is worth $100. Students cannot get together and bargain as a group. They must bargain individually. How will the surplus be split? If I offer you $20, should you take it, or reject and wait for another offer? Who holds all the cards?
The Card Game You should reject my offer of $20 and be patient. I need you to get value from one of my black cards. If each of you are patient, you should expect to roughly split the surplus evenly. But what if I were to destroy 3 cards?
Adding Value If I destroy 3 cards, now there is just $2300 in value, less than $2600. However, I have changed the game dramatically. When there were 26 black cards, each red card added $100 in value (without it, there s no pair). With only 23 cards, each of you adds zero value at the margin. I don t need any individual one of you. I have outside options, while you do not. If I offer you $20, you should take it. By being on the short side of the market, I dramatically increase the amount of value I capture from bargaining.
The Card Game and Business Strategy NFL franchises: fewer teams than interested cities. To attract teams, cities routinely build $300 million stadiums. Rams to St. Louis. Browns to Baltimore. Oilers to Nashville. The NFL can do this (and many other things) because it essentially has a monopoly (a cartel, more specifically) on professional football in the US. The USFL sued for antitrust in the mid-1980s, and won $1 in damages (trebled to $3).
Matching Pennies No dominant strategies. No dominated strategies. No pure strategy equilibrium.
A Mixed Strategy Equilibrium Iterating around the matrix, the players play cat-and-mouse. If player 1 always plays heads, player 2 s best-response is to match. Player 2 does well (+1), player 1 does poorly (-1). Since neither player benefits from committing to a particular strategy, it is better to randomize. By playing heads and tails each 50% of the time, player 1 leaves player 2 without a response that is strictly better. Indeed, if player 1 randomizes this way, player 2 is indifferent between playing heads, tails, or any probabilistic mixture of the two strategies. Thus, if both players randomize 50/50 between heads and tails, then each player cannot improve upon their payoff by deviating...hence, they are both playing best responses and these mixed strategies form a Nash equilibrium.
Mixed Strategies in Coordination Games Consider the game played by Fred and Barney earlier, but now assume rabbits are unavailable. There remain two pure-strategy Nash equilibria. There is also a mixed strategy equilibrium...can you figure it out?
Mixed Strategies in Coordination Games Solving for the mixed strategy requires a bit of backward induction. Take the position of Fred and ask...how can I randomize so that Barney is indifferent between choosing Stag and Bison? Let Fred s probability of choosing Stag be p. Then, if he randomizes this way, Barney earns 3p by playing Stag and 30(1 p) if he plays Bison. These payoffs are the same if 3p = 30(1 p), which holds for p = 10 11. Then, if Barney also plays Stag with probability 10 11, Fred is indifferent between Stag and Bison. Intuitively, the probabilities must be lopsided because the payoffs are lopsided (in favor of coordinating on Bison).
Mixed Strategies and Business Strategy in Practice Although we have illustrated mixed strategies in a static setting, it seems clear than there is likely to remain a lot of guessing in actual play. Consider drug testing of employees. It is very expensive to conduct drug tests, so always testing employees is likely to be inefficiently expensive even if it reduces drug use to zero. On the other hand, if tests are never given, employees may harm the company by harming their own productivity through drug use. Again, we have a cat and mouse game. But since this setting involves a lot of repeated interaction, employees can draw inferences about how often tests are conducted. In this way, managers can reveal their commitment to a mixed strategy.
The War of Attrition - A Simple Model Consider a game with two players competing for a prize. In each period, the players decide whether to stay in or exit. The last firm in (by itself) wins $1 million. If both exit at the same time, they split the prize ($500,000 each). Staying in requires cost c < $500, 000.
The War of Attrition - A Simple Model - Equilibrium If the other firm is committed to staying in forever, then it is in my interest to quit now. If the other firm is likely to quit soon, it is in my interest to stay in forever. Thus, two equilibria involve one firm giving up without a fight. A third equilibrium involves battle...
The War of Attrition - A Simple Model - Equilibrium Consider a strategy of exiting with probability p. Let s see if we can identify an equilibrium where both firms play it. What is my best response if the other firm exits with probability p? If I exit now, my payoff is 500, 000p, because I get half the prize if the other firm also exits. If I stay in, I pay cost c to stay in, win 1, 000, 000 with probability p, and find myself back at the beginning of the game with probability 1 p. If exiting yields the same payoff as staying in, which it will in equilibrium, then if I get back to the beginning of the game, my payoff will be 500, 000p. Thus, the payoff to exiting now is 500, 000p, while the payoff to staying in is 1, 000, 000p c + (1 p)500, 000p, so in equilibrium, p must satisfy 500, 000p = 1, 000, 000p c + (1 p)500, 000p.
The War of Attrition - A Simple Model - Equilibrium Solving this yields p = 1 1 2c 1, 000, 000. Payoffs are increasing in the cost of waiting c. The game could last a long time...if c is 10, 000, the equilibrium p is about.02. The probability that neither firm concedes for 100 periods is about 13.3%...and expected profits from the fight are 10, 050. The game is a useful way of thinking about litigation, competition for small markets, fights among animals, etc.
Repeated Simultaneous Move Games - The Prisoner s Dilemma Return to the prisoner s dilemma game played earlier. Now imagine it is played twice.
Repeated Simultaneous Move Games - The Prisoner s Dilemma Will players cooperate given that the horizon is longer? Think forward and reason backward. In the final period, the players face a one-shot prisoner s dilemma. The unique Nash equilibrium is Confess, Confess. If the players anticipate this type of play in the final period, then there is no reason to play Deny in the current period. This same logic holds for any finite time horizon. In experiments, though, players often do cooperate in early periods. This result may change with a bit of asymmetric information about things like altruism.
Infinitely Repeated Simultaneous Move Games - The Prisoner s Dilemma Consider the following strategy play Deny until my opponent plays Confess once, then play Confess from that point forward. This is called a Nash reversion strategy. It is sufficient to see if this strategy maximizes a player s payoff, given that the other player is playing it, too. The payoff to playing Deny every period is 2 + δ( 2) + δ 2 ( 2) +... = 2 1 δ The payoff to cheating now and playing Confess is 1 + δ( 5) + δ 2 ( 5) +... = 1 + δ 1 δ ( 5)
Infinitely Repeated Simultaneous Move Games - The Prisoner s Dilemma Working a little math, we see playing Deny instead of confess is a better strategy as long as δ 1 6. That is, sufficiently patient players will be willing to sustain cooperation. What about the decision in period 2? It is the same decision as in period 1!
Infinitely Repeated Simultaneous Move Games This is a useful framework for thinking about repeat interactions between firms. Obviously, duopolists do not literally play prisoner s dilemma games, but there are types of rivalry where the payoff structure is similar. Such games help us understand things like collusion and price wars.