ps7solution

d) What wage demands and wage offers will be presented to the arbitrator after the railroad is built? Given this, if you were Amtrak, for what values of k would you build the new line? Why is your answer different from that in parts b and c ? Solution: The shut-down price is w = 120 , 000 as in c . The current going wage is w = 50 , 000. Therefore the government makes w = ( w + w ) / 2 = 85 , 000. The new line will be built if and only if k + 10 , 000 , 000 + 85 , 000 × 1 , 000 ≤ 130 , 000 , 000 That is, k ≤ 35 , 000 , 000. Now half of the surplus(sunk cost not considered) goes to Amtrak instead of all of the surplus as in b or none of the surplus as in c , and hence the threshold of k should be different. 2. An inspection game. This example combines mixed strategies in two-player, zero-sum games with the use of Zermelo’s algorithm. An environmental protection agency knows that a firm is determined to discharge a pollutant into a river on one of 4 days. The agency will learn immediately when the river is polluted because of the telephone complaints it will receive from local residents. However, to obtain a conviction, the agency has to catch the firm red-handed. This means that it must try and guess in advance the day on which the firm will pollute the river so as to have an inspector on the spot. Unfortunately, the agency’s resources are so overstretched that it can only afford to dispatch an inspector to the side on one of the 4 days, and the firm knows this. This situation can be modeled as a two-player game in which player I is the agency and player II is the firm. It will be assumed that the agency wins if it chooses to inspect the firm on the day the firm pollutes the river. Otherwise the firm wins. Both players assign a payoff of +1 if winning and -1 to losing. Compute the subgame perfect Nash equilibrium of this game in mixed strategies. [Hint: start with de- termining what happens if day 4 is reached and neither player made a move yet; this determines a payoff for both players in case this subgame is reached; then work backwards, using the fact that mixing requires that the player is indifferent between the actions that she mixes over.] What are the expected payoffs of the players in this mixed equilibrium? Solution: If day 4 is reached and neither player made a move, then player 1 wins for sure. Payoff is (1 , − 1). If day 3 is reached and neither player made a move, assume player 1 mixes and executes inspection with probability α ; player 2 mixes and pollutes with probability β . To make player 1 indifferent, we have β − (1 − β ) = − β + (1 − β ), i.e, β = 1 2 . Similarly, α = 1 2 . In this case, both players get zero expected payoff. If day 2 is reached and no one moves in the previous day, assume they mix with probability ( p, q ). Similarly, to make player 1 indifferent, we have q − (1 − q ) = − q + 0 i.e, q = 1 3 . Same logic, p = 1 3 . Their expected payoff would be ( − 1 3 , 1 3 ). Now back to day 1, assume they mix with ( m, n ), to make player 1 indifferent, we will have n − (1 − n ) = − n − 1 3 (1 − n ), namely, n = 1 4 . To make player 2 indifferent, we will have − m + (1 − m ) = m + 1 3 (1 − m ), thus, m = 1 4 . In conclusion, in day 1, each players are executing inspection/poluting with probability 1 4 . If game is still on in day 2, both players will continue to mix with probability 1 3 . If the game goes into day 3, they mix with probability 1 2 . The expected payoffs of both players in this mixed equilibrium are ( − 1 2 , 1 2 ). 2