Question

Instead of assuming that a player randomizes over his or her pure strategies, an alternative way to interpret mixed strategies is to assume that there are many identical players, each playing a pure strategy. Under this interpretation, one can compute the share (i.e., a number between 0 and 1) of players that play a given pure strategy and interpret that share as the frequency by which (or the probability that) that pure strategy is played. This is just a matter of interpretation, so that this problem will be framed using both interpretations. Assume that a worker’s performance is decreased by the use of social media in the workplace and that this has a negative impact on the employer’s profits. A worker who does not use social media (NS) while working produces output of value π > 0 for the employer. The value generated by a worker who uses social media (S) is 0. A worker is paid w > 0 if he is not discovered using social media, but otherwise is paid 0. The use of social media gives a utility of s > 0. The employer can decide to monitor (M) the workers but this is costly. Monitoring

2

a worker costs c > 0. Let α ∈ [0, 1] be the share of employees who use social media and let β ∈ [0, 1] be the probability that a worker is monitored.

1. Write down the matrix corresponding to the normal-form of this game.

2. Assume that w > c and w > s. How large should the share of workers who use social media in the work force be for random monitoring to be optimal?

3. With what probability should a worker be monitored to be indifferent between using social media or not? Find the Nash equilibrium in mixed strategies.

   Imagine now that the Nash equilibrium in mixed strategies you just found is at play but that the government is concerned about the low productivity of the work force. In particular, the government wants to reduce the use of social media on the job and is choosing between relaxing the privacy laws so that monitoring is made easier (i.e., reduce c), or forcing the social media firms to make their sites more difficult to access during work time (i.e., reduce s).

4. What should the government do? Explain.

Answer 1

Main Lesson If a mixed strategy is a best response then each of the pure strategies involved in the mix must itself be a best response. In particular, each must yield the same expected payo. Before explaining why this must be true, let's just try to rewrite this lesson formally, using our new notation: More Formal statement of the Same Lesson. If player i's mixed strategy pi is a best response to the (mixed) strategies of the other players, pi, then, for each pure strategy si such that pi(si) > 0, it must be the case that si is itself a best response to pi : In particular, Eui(si ; pi) must be the same for all such strategies. Why is this true? Suppose it were not true. Then there must be at least one pure strategy si that is assigned positive probability by my best-response mix and that yields a lower expected payo against pi . If there is more than one, focus on the one that yields the lowest expected payo. Suppose I drop that (low-yield) pure strategy from my mix, assigning the weight I used to give it to one of the other (higheryield) strategies in the mix. This must raise my expected payo (just as dropping the player with the lowest batting average on a team must raise the team average). But then the original mixed strategy cannot have been a best response: it does not do as well as the new mixed strategy. This is a contradiction. So what? An immediate implication of this lesson is that if a mixed strategy forms part of a Nash Equilibrium then each pure strategy in the mix must itself be a best response. Hence all the strategies in the mix must yield the same expected payo. We will use this fact to nd mixed-strategy Nash Equilibria. Finding Mixed-Strategy Nash Equilibria. Let's look at some examples and use our lesson to nd the mixed-strategy NE. Example 1 Battle of the Sexes a b A 2; 1 0; 0 B 0; 0 1; 2 In this game, we know that there are two pure-strategy NE at (A; a) and (B; b). Let's see if there are any other mixed-strategy NE. Suppose that there was another equilibrium in which the row mixed on both A and B. By our lesson of the day, we know that in this case both A and B must be best responses to whatever the column player is doing. But for them both to be best responses, they must both yield the same expected payo for the row player. We will use this fact about row's expected payos to nd what column must be playing! Suppose that column's mixed strategy assigns probability weight q to a and probability weight (1 q) to b. Then, row's expected payo from A against (q; 1 q) = q [2] + (1 q) [0] = 2q row's expected payo from B against (q; 1 q) = q [0] + (1 q) [1] = 1 q But if these expected payo are to be equal, we must have 2q = 1 q or q Handout on Mixed Strategies 2 To summarize so far, if row is mixing on both her strategies in a NE then both must yield the same expected payo, in which case column must be mixing with weights 1 3 ; 2 3 . Notice the trick here: we used the fact that, in equilibrium, row must be indierent between the strategies involved in her mix to solve for column's equilibrium mixed strategy. Now let's reverse the trick for nd row's equilibrium mix. If there is an equilibrium in which the column mixes on both a and b, then (by our lesson of the day) we know that both a and b must be best responses to whatever row is doing. But for them both to be best responses, they must both yield column the same expected payo. We will use this fact about column's expected payos to nd what row must be playing. Suppose that row's mixed strategy assigns probability weight p to A and probability weight (1 p) to B. Then, column's expected payo from a against (p; 1 p) = p [1] + (1 p) [0] = p Row's expected payo from b against (p; 1 p) = p [0] + (1 p) [2] = 2 (1 p) But if these are to be equal, we have 2 (1 p) = p or p = 2 3 . To summarize, if column is mixing on both her strategies in a NE then both must yield the same expected payo, in which case row must be mixing with weights 2 3 ; 1 3 . We used the fact that, in equilibrium, column must be indierent between the strategies involved in her mix to solve for row's equilibrium mixed strategy. I claim that the mixed-strategy prole 2 3 ; 1 3 ; 1 3 ; 2 3 is a NE. To show this I still need to check that neither player has a strictly protable deviation but this turns out to be easy. We constructed the equilibrium so that, given column's mix, 1 3 ; 2 3 , each of row's pure strategies, A and B yields the same expected payo. But, in this case, any mix of those pure strategies (including the equilibrium mix itself) will yield the same expected payo. So all potential deviations yield the same expected payo: none are strictly protable. The same argument applies to column. Example 2. Rock, Scissors, Paper. r s p R 0; 0 1; 1 1; 1 S 1; 1 0; 0 1; 1 P 1; 1 1; 1 0; 0 In this game, we `know' that the mixed-strategy NE is 1 3 ; 1 3 ; 1 3 ; 1 3 ; 1 3 ; 1 3 , but let's use the method from the last example to `nd' the equilibrium as if we did not know. Once again, suppose that there is an equilibrium in which row is mixing on all of R, S and P. By our lesson of the day, we know that in this case R, S and P must each be a best response to whatever the column player is doing. But for them each to be best response, each must yield the same expected payo for the row player. We will use this fact about row's expected payos to nd what column must be playing. Suppose that column's mixed strategy assigns probability weight qr to R, qs to S and (1 qs qr) to P. Then, row's expected payo from R against (qr; qs; 1 qr qs) = qr [0] + qs [1] + (1 qr qs) [1] row's expected payo from S against (qr; qs; 1 qr qs) = qr [1] + qs [0] + (1 qr qs) [1] row's expected payo from P against (qr; qs; 1 qr qs) = Handout on Mixed Strategies 3 Setting these three expected payos equal to one another (and using a little basic algebra) solves to qr = qs = (1 qr qs) = 1 3 . To summarize, if row is mixing on all of her strategies in a NE then each must yield the same expected payo, in which case column must be mixing with weights 1 3 ; 1 3 ; 1 3 . Once again, we used the fact that, in equilibrium, row must be indierent between the strategies involved in her mix to solve for column's equilibrium mixed strategy. We could do the same to nd row's equilibrium mix. That is, we could use the fact, in equilibrium, column must be indierent between the strategies involved in her mix to solve for row's equilibrium mixed strategy. However, since the argument is symmetric, let's skip it. As in the previous example, checking that neither player has a strictly protable deviation is easy. We constructed the equilibrium so that, given column's mix, each of row's pure strategies yields the same expected payo. But, in this case, any mix of those pure strategies (including the equilibrium mix itself) will yield the same expected payo. So all potential deviations yield the same expected payo: none are strictly protable. The same argument applies to column. So we have shown that this is an equilibrium. For nerds only. In this example, it is a little more involved to show that there are no other mixed-strategy equilibria. We have shown that in the only equilibrium in which each player mixes on all of her strategies, each player mixes 1 3 ; 1 3 ; 1 3 . But, in principle, there could be other mixed-strategy equilibria in which one player only mixes on two of her three strategies. Let me sketch the argument why no such equilibrium exists. Suppose there was such an equilibrium. Without loss of generality, let column be the player who mixes on two of her strategies, and without loss of generality, assume that column's mixes only on r and s; that is, column's mixed strategy assigns probability zero to r. Given this, row's expected return from R is strictly greater than that from S. Thus, row's best response must assign probability zero to S. But, given this, column's expected return from p is strictly greater than that from r. Thus, column's best response must assign probability zero to r. But if column assigns zero probability to p (by assumption) and zero probability to r (as we have just shown), then he must be playing the pure strategy s. But we already know that neither can be playing a pure strategy in any equilibrium of rock, scissors paper. What did we learn here? We have found a general method to nd mixed-strategy Nash Equilibria. Method to nd mixed-strategies NE Suppose we conjecture that there is an equilibrium in which row mixes between several of her strategies. If there is such an equilibrium then each of these strategies must yield the same expected payo given column's equilibrium strategy. If we write down these payos [just as we did in the examples above] we can solve for column's equilibrium mix. In other words, row's indierence among her strategies implies column's equilibrium mix. Now we reverse. Look at the strategies that column is mixing on, write down column's indierence condition, and solve for row's equilibrium mix. We are almost done but we still need to check a few (easy) things. First, the equilibrium mix we have found for row must indeed involve those strategies from which we started our conjecture! Second, each of the mixing probabilities we have constructed must indeed be probabilities: they must lie between zero and one! Third, as always, we need to check that neither player has a strictly protable deviation. But, [as we saw in the examples above] if the mix involved all strategies then this last check is for .

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