\documentstyle[12pt,epsf]{msart} \textheight 10in \begin{document} \setcounter{section}{3} \begin{Large} \noindent June 26 1993\\ Janaury 22, 1996 \section*{ 3 Mixed and Continuous Strategies} \newpage \begin{center} {\bf Table 3.1 ``The Welfare Game'' } \begin{tabular}{lllccc} & & &\multicolumn{3}{c}{\bf Pauper}\\ & & & {\it Work} ($\gamma_w$) & & $ Loaf$ ($1-\gamma_w$) \\ & & $ Aid$ ($\theta_a$) & 3,2 & $\rightarrow$ & $-1,3$ \\ & {\bf Government:} &&$\uparrow$& & $\downarrow$ \\ & & {\it No Aid } ($1-\theta_a$) & $-1,1$ & $\leftarrow$ & 0,0 \\ \multicolumn{6}{l}{\it Payoffs to: (Government, Pauper).} \end{tabular} \end{center} In the mixed strategy equilibrium, the pauper selects {\it Work} 20 percent of the time. The way we obtained the number might seem strange: to obtain the pauper's strategy, we differentiated the government's payoff. Understanding why requires several steps. \begin{quotation} \noindent \hspace*{16pt}(1) I assert that an optimal mixed strategy exists for the government. \noindent \hspace*{16pt} (2) If the pauper selects {\it Work} more than 20 percent of the time, then the government always selects {\it Aid}. If the pauper selects {\it Work} less than 20 percent of the time, the government never selects {\it Aid}. \noindent \hspace*{16pt} (3) If a mixed strategy is to be optimal for the government, the pauper must therefore select {\it Work} with probability exactly 20 percent. \end{quotation} \newpage \begin{large} One objection to mixed strategies is that people in the real world do not take random actions. That is not a compelling objection, because all that a model with mixed strategies requires to be a good description of the world is that the actions appear random to observers, even if the player himself has always been sure what action he would take. A more troubling objection is that a player who selects a mixed strategy is always indifferent between two pure strategies. In ``The Welfare Game'', the pauper is indifferent between his two pure strategies and a whole continuum of mixed strategies, given the government's mixed strategy. One way to reinterpret ``The Welfare Game'' is as a many-pauper game with a pure strategy equilibrium: 20 percent of the paupers choose the pure strategy {\it Work} and 80 percent choose the pure strategy {\it Loaf}. The problem persists of how an individual pauper, indifferent between the pure strategies, chooses one or the other, but it is easy to imagine that individual characteristics outside the model could determine which actions are chosen by which paupers. Another interpretation of mixed strategies, which works even in the single-pauper game, is that the pauper is drawn from a population of paupers, and the government does not know his characteristics. The government only knows that there are two types of paupers, in the proportions (0.2, 0.8): the less work-averse who pick {\it Work} if the government picks $\theta_a =0.5$, and the more work-averse who pick {\it Loaf}. It could even be that there is a continuum of types of work-aversion, and that the split in actions is (0.2,0.8) only when $\theta_a =0.5$. See Harsanyi (1973) for a careful exposition of this idea. \newpage \noindent { \bf Existence of Equilibrium} \noindent One of the strong points of Nash equilibria is that they exist, in mixed strategies if not in pure, in practically every game one is likely to encounter. One feature of a game that favors existence is continuity of the payoffs in the strategies. A payoff is continuous in a player's strategy if a small change in his strategy causes a small or zero change in the payoffs. With continuity, the players' strategies can adjust more finely and come closer to being best responses to each other. In `` The Welfare Game'', the payoffs are discontinuous in pure strategies, so no compromise between the players is possible and no pure strategy equilibrium exists. Once mixed strategies are incorporated, the payoffs are continuous in the mixing probability, so an equilibrium does exist. A second feature promoting existence is a closed and bounded strategy set. Suppose in a stock market game that Smith can borrow money and buy as many shares of stock as he likes, so his strategy set, the amount of stock he can buy, is $[0, \infty)$, which is unbounded above (we assume he can buy fractional shares but cannot sell short). If Smith knows that the price today is lower than it will be tomorrow, he wants to buy an infinite number of shares, which is not an equilibrium purchase. If the amount he buys is restricted to be strictly less than 1,000, then the strategy set is bounded (by 1,000) but not closed (because 1,000 is not included), and no equilibrium purchase exists, because he wants to buy 999.999 $\ldots$ shares. An open set like $\{x: 0 \leq x < 1000\}$ has no maximum, because there is no closest real number to 1000. If the strategy set is closed and bounded, on the other hand, its minimum and maximum are well-defined. If Smith can buy up to but no more than 1,000 shares, then 1000 shares is his equilibrium purchase. Sometimes, as in ``The Cournot Game'' later in this chapter, the unboundedness of the strategy sets does not matter because the optimum is an interior solution, but in other games it is good modelling practice to use closed sets instead of open sets. \end{large} \newpage \begin{center} {\bf Table 3.2 ``Chicken'' } \begin{tabular}{lllccc} & & &\multicolumn{3}{c}{\bf Jones}\\ & & & $Continue$ ($\theta$) & & $Swerve$ ($1- \theta$) \\ & & $Continue$ ($\theta$) & $-3,-3$ & $\rightarrow$ & {\bf 2, 0} \\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ & & $Swerve$ ($1-\theta$) & {\bf 0, 2} & $\leftarrow$ & 1, 1 \\ \multicolumn{6}{l}{\it Payoffs to: (Smith, Jones).} \end{tabular} \end{center} ``Chicken'' has two pure strategy Nash equilibria, $(Swerve, Continue)$ and $(Continue, Swerve)$, but they have the defect of asymmetry. As in that game, the best prediction in ``Chicken'' is perhaps the mixed strategy equilibrium, because its symmetry makes it a focal point of sorts, and it does not require any differences between the players. \newpage \begin{normalsize} \noindent {\bf ``The War of Attrition''} Smith and Jones control two firms in an industry which is a natural monopoly, with demand strong enough for one firm to operate profitably, but not two. The possible actions are to $Exit$ or to $Continue$. In each period that both $Continue$, each earns $-1$. If a firm exits, its losses cease and the remaining firm obtains the value of the market's monopoly profit, which we set equal to 3. We will set the discount rate equal to $r > 0$, although that is inessential to the model, even if the possible length of the game is infinite . One simple equilibrium is for Smith to choose ($Continue$ regardless of what Jones does) and for Jones to choose ($Exit$ immediately), which are best responses to each other. But we will solve for a symmetric equilibrium in which each player chooses the same mixed strategy: a constant probability $\theta$ that he picks $Exit$ given that the other player has not yet exited. We can calculate $\theta$ as follows, adopting the perspective of Smith. Denote the expected discounted value of Smith's payoffs by $V_{stay}$ if he stays and $V_{exit}$ if he exits immediately. These two pure strategy payoffs must be equal in a mixed strategy equilibrium (which was the basis for the payoff-equating method). If Smith exits, he obtains $V_{exit} =0$. If Smith stays in, his payoff depends on what Jones does. If Jones stays in too, which has probability $(1-\theta)$, Smith gets $-1$ currently and his expected value for the following period, which is discounted using $r$, is unchanged. If Jones exits immediately, which has probability $\theta$, then Smith receives a payment of 3. In symbols, \begin{equation}\label{e3.6} V_{stay} = \theta \cdot (3) + \left( 1-\theta \right) \left(-1 + \left[ \frac{V_{stay}}{1+r}\right] \right), \end{equation} which after a little manipulation becomes \begin{equation}\label{e3.7} V_{stay} = \left( \frac{1+r}{r+\theta} \right) \left( 4\theta - 1 \right). \end{equation} Once we equate $V_{stay}$ to $V_{exit}$, which equals zero, equation (3.\ref{e3.7}) tells us that $\theta = 0.25$ in equilibrium, and that this is independent of the discount rate $r$. \newpage A{\bf preemption game}: the reward goes to the player who chooses the move which ends the game, and a cost is paid if both players choose that move, but no cost is incurred in a period when neither player chooses it. {\bf ``Grab the Dollar''} is an example. A dollar is placed on the table between Smith and Jones, who each must decide whether to grab for it or not. If both grab, both are fined one dollar. This could be set up as a one-period game, a $T$ period game, or an infinite-period game, but the game definitely ends when someone grabs the dollar. \end{normalsize} \begin{center} {\bf Table 3.3 ``Grab The Dollar'' } \begin{tabular}{lllccc} & & &\multicolumn{3}{c}{\bf Jones}\\ & & & {\it Grab} & & {\it Don't Grab } \\ & & {\it Grab} & $-1, -1$ & $\rightarrow$ & {\bf 1,0} \\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ & & {\it Don't Grab } & {\bf 0,1} & $\leftarrow$ & $0,0$ \\ \multicolumn{6}{l}{\it Payoffs to: (Smith, Jones).} \end{tabular} \end{center} Like ``The War of Attrition,'' ``Grab The Dollar'' has asymmetric equilibria in pure strategies, and a symmetric equilibrium in mixed strategies. In the infinite-period version, the equilibrium probability of grabbing is 0.5 per period in the symmetric equilibrium. Still another class of timing games are DUELS, in which the actions are discrete occurrences which the players locate at particular points in continuous time. Two players with guns approach each other and must decide when to shoot. In the {\bf noisy duel}, if a player shoots and misses, the other player observes the miss and can kill the first player at his leisure. An equilibrium exists in pure strategies for the noisy duel. In a {\bf silent duel}, a player does not know when the other player has fired, and the equilibrium is in mixed strategies. Karlin (1959) has details on duelling games, and Chapter 4 of Fudenberg \& Tirole (1991a) is an excellent discussion of games of timing in general. \newpage \begin{center} {\bf Table 3.6 ``The Civic Duty Game'' } \begin{tabular}{lllccc} & & &\multicolumn{3}{c}{\bf Jones}\\ & & & {\it Ignore} ($\gamma$) & & $Telephone$ ($1-\gamma$) \\ & & {\it Ignore} ($\gamma$) & $0,0$ & $\rightarrow$ & {\bf 10,7} \\ & {\bf Smith:} &&$\downarrow$& & $\uparrow$ \\ & & {\it Telephone } ($1-\gamma$) & {\bf 7,10} & $\leftarrow$ & $7,7$ \\ \multicolumn{6}{l}{\it Payoffs to: (Row, Column).} \end{tabular} \end{center} \begin{normalsize} ``The Civic Duty Game'' has two asymmetric pure-strategy equilibria and a symmetric mixed-strategy equilibrium. In solving for the mixed-strategy equilibrium, let us move from two players to $N$ players. In the N-player version of the game, the payoff to Smith is 0 if nobody calls, 7 if he himself calls, and 10 if one or more of the other $N-1$ players calls. This game also has asymmetric pure-strategy and a symmetric mixed-strategy equilibrium. If all players use the same probability $\gamma$ of $Ignore$, the probability that the other $N-1$ players besides Smith all choose $Ignore$ is $\gamma^{N-1}$, so the probability that one or more of them chooses $Telephone$ is $1-\gamma^{N-1}$. Thus, equating Smith's pure-strategy payoffs using the equating-payoffs method of equilibrium calculation yields \begin{equation}\label{e3.45a} \pi_{Smith} (Telephone) = 7 = \pi_{Smith} (Ignore) = \gamma^{N-1} (0) + (1-\gamma^{N-1}) (10). \end{equation} Equation (3.\ref{e3.45a}) tells us that $ \gamma^{N-1} = 0.3$ and $ \gamma^* = 0.3^{\frac{1}{N-1}}. $ If $N=2$, Smith chooses $Ignore$ with a probability of 0.30, and the probability that neither player phones the police is $ \gamma^{*2}= 0.09$. As $N$ increases, Smith's expected payoff remains constant at 7, since his expected payoff always equals his payoff from the pure strategy of $Telephone$. The probability $\gamma^*$ of $Ignore$, however, increases in $N$. When there are more players, each player relies more on somebody else calling. The probability that nobody calls is $\gamma^{*N} $. Equation (3.\ref{e3.45b}) shows that $\gamma^{*N-1}=0.3$, so $\gamma^{*N} =0.3\gamma^*$, which is increasing in $N$. When there are 38 players, the probability that nobody calls the police is about 0.29, because $\gamma^*$ is about 0.97. The more people that watch a crime, the less likely it is to be reported. The expected payoff per player is 7 whether there is 1 player or 38, whereas if one and only one player called the police it would rise from 7 with 1 player to about 9.9 with 38 (=[1(7) + 37(10]/38). The problem is divided responsibility, lack of a focal point. One person must be made responsible, whether by tradition (e.g., the oldest person on the block always calls the police) or direction (e.g., Smith shouts to Jones: ``Call the police!''). \end{normalsize} %--------------------------------------------------------------- \newpage \begin{center} {\bf Table 3.7 ``Auditing Game I'' } \begin{tabular}{lllccc} & & &\multicolumn{3}{c}{\bf Suspects}\\ & & & {\it Cheat } ($\theta$) & & $ Obey$ ($1-\theta$) \\ & & $Audit$ ($ \gamma$) & $4-C,-F$ & $\rightarrow$ & $4-C, -1 $ \\ & {\bf IRS:} &&$\uparrow$& & $\downarrow$ \\ & & {\it Trust } ($1-\gamma$) & 0,0 & $\leftarrow$ & $4,-1$ \\ \multicolumn{6}{l}{\it Payoffs to: (IRS, Suspects) } \end{tabular} \end{center} ``Auditing Game I'' is a discoordination game, with only a mixed strategy equilibrium. A second way to model the situation is as a sequential game. Let us call this {\bf ``Auditing Game II''}. The equilibrium in ``Auditing Game II'' is in pure strategies, a general feature of sequential games of perfect information. In equilibrium, the IRS chooses $Audit$, anticipating that the suspect will then choose $Obey$. The payoffs are $4-C$ for the IRS and $-1$ for the suspects, the same for both players as in ``Auditing Game I'', although now there is more auditing and less cheating and fine-paying. \newpage \begin{large} \noindent AUDITING GAME III. Suppose the IRS does not have to adopt a policy of auditing or trusting every suspect, but instead can audit a random sample. This is not necessarily a mixed strategy. In ``Auditing Game I'', the equilibrium strategy was to audit all suspects with probability $1/F$ and none of them otherwise. That is different from announcing in advance that the IRS will audit a random sample of $1/F$ of the suspects. For {\bf ``Auditing Game III''}, let the IRS move first, but let its move consist of the choice of the proportion $\alpha$ of tax returns to be audited. We know that the IRS is willing to deter the suspects from cheating, since it would be willing to choose $\alpha = 1$ and replicate the result in `Auditing Game II'' if it had to. It chooses $\alpha$ so that \begin{equation}\label{e3.50} \pi_{suspect} (Obey) \geq \pi_{suspect} (Cheat), \end{equation} i.e., \begin{equation}\label{e3.51} -1 \geq \alpha (-F) + (1-\alpha ) (0). \end{equation} In equilibrium, therefore, the IRS chooses $\alpha = 1/F$ and the suspects respond with $Obey$. The IRS payoff is $4 - \alpha C$, which is better than the $4-C$ in the other two games, and the suspect's payoff is $-1$, exactly the same as before. The equilibrium of ``Auditing Game III'' is in pure strategies, even though the IRS's action is random. It is different from ``Auditing Game I'' because the IRS must go ahead with the costly audit even if the suspect chooses $Obey$. ``Auditing Game III'' is different in another way also: its action set is continuous. In `Auditing Games I'' and ``II'' the action set is {\it \{Audit, Trust\},} although the strategy set becomes $\gamma \in [0,1]$ once mixed strategies are allowed. In ``Auditing Game III'', the action set is $\alpha \in [0,1]$, and the strategy set would allow mixing of any of the elements in the action set, although mixed strategies are pointless for the IRS because the game is sequential. \end{large} %--------------------------------------------------------------- \newpage \begin{center} {\bf ``The Cournot Game''} \end{center} {\bf Players}\\ Firms Apex and Brydox. \noindent {\bf Order of Play}\\ Apex and Brydox simultaneously choose quantities $q_a$ and $q_b$ from the set $[0, \infty)$. \noindent {\bf Payoffs}\\ Production costs are zero. Demand is a function of the total quantity sold, $Q= q_a + q_b$. \begin{equation} \label{e3.8} P(Q) = 120-q_a - q_b . \end{equation} Payoffs are profits, which are given by a firm's price times its quantity, i.e., \begin{equation} \label{e3.9} \begin{array}{l} \pi_{Apex} = 120q_a - q_a ^2 - q_a q_b ;\\ \pi_{Brydox} = 120q_b - q_a q_b - q_b ^2. \end{array} \end{equation} \end{Large} \end{document}