The Calculus of Nash Equilibria

post by Heighn · 2022-04-01T14:40:59.937Z · LW · GW · 0 comments

Contents

  Prisoner's dilemma
  Nonlinear payoff functions
    Making things a bit more complicated
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Now that we know a bit about derivatives [? · GW], it's time to use them to find dominant strategies and Nash equilibria. It helps if the reader is familiar with Nash equilibria already.

Prisoner's dilemma

The payoff matrix of the Prisoner's dilemma [? · GW] can be as follows:

We can see that the payoff for Prisoner 1 depends on her own action (Cooperate/Defect) but also on the action of Prisoner 2. Therefore, the payoff function for Prisoner 1 is a multivariable function: $V1(a_1,a_2)$, where $a_n$ is the action of Prisoner $n$ (and $n\in \{1,2\}$). Let's say $a_n=0$ when the action of Prisoner $n$ is Cooperate, and $a_n=1$ for Defect. So $a_n\in \{0,1\}$. Then $V1(a_1,a_2)=20-20a_2+10a_1$, and crucially, $V{1}_{a1}^{\prime }(a_1,a_2)=10$. So for Defect ($a_1=1$), Prisoner 1's payoff will be $1$0 higher than for Cooperate ($a_2=0$), as can be confirmed in the table. Note that $a_2$ doesn't show up in $V{1}_{a_1}^{\prime }(a_1,a_2)$: Defect gives $\$10$ more for Prisoner 1 regardless of what Prisoner 2 does, which makes Defect a dominant strategy. Don't get me wrong: Prisoner 1's payoff certainly does depend on what Prisoner 2 does. The point is that no matter what Prisoner 2 does, Prisoner 1's payoff will be $10 higher when she (Prisoner 1) defects - and that's what's reflected in $V{1}_{a1}^{\prime }(a_1,a_2)=10$.

Since the payoff matrix is symmetrical, $V2(a_1,a_2)=20-20a_1+10a_2$ and $V{2}_{a_2}^{\prime }(a_1,a_2)=10$. Prisoner 2 therefore also has a dominant strategy: Defect. The Prisoner's dilemma, then, has a Nash equilibrium: when both prisoners defect. With the partial derivatives, we demonstrated that when both prisoners defect, no one prisoner can do better by changing her action to Cooperate. If e.g. Prisoner 1 were to do this, then $a_1$ would go from $1$ to $0$, and since $V{1}_{a_1}^{\prime }(a_1,a_2)>0$, that would lower $V1$ (regardless of $a_2$). By symmetry, the same is true for Prisoner 2.

Nonlinear payoff functions

In the Prisoner's dilemma, the payoffs of both players (prisoners) can be modelled by linear payoff functions. What if the payoffs are nonlinear?

Let's say $V1(a_1,a_2)=-a_1^2$ and $V2(a_1,a_2)=-a_2^2+a_2$. Then $V{1}_{a_1}^{\prime }(a_1,a_2)=-2a_1$ and $V{2}_{a_2}^{\prime }(a_1,a_2)=-2a_2+1$. A Nash equilibrium is a point where no player can do better by doing another action given the action of the other player; therefore, $V1(a_1,a_2)$ should be maximized with respect to $a_1$ while keeping $a_2$ constant, whereas $V2(a_1,a_2)$ should be maximized with respect to $a_2$ while keeping $a_1$ constant. If $V1(a_1,a_2)$ has a peak value with respect to $a_1$, $V{1}_{a_1}^{\prime }(a_1,a_2)=-2a_1$ must be $0$ in that point. $V{1}_{a_1}^{\prime }(a_1,a_2)=-2a_1=0$ gives $a_1=\frac{0}{-2}=0$. So $a_1=0$ could represent a peak, but also a valley, since $V{1}_{a_1}^{\prime }(a_1,a_2)$ would be $0$ in both. If $V{1}_{a_1}^{\prime }(a_1,a_2)=-2a_1$, $V{1}_{a_1}^{\prime \prime }(a_1,a_2)=-2<0$. So $a_1=0$ represents a local maximum in $V1(a_1,a_2)$ (when $a_2$ is held constant)! Since $V1(a_1,a_2)$ is quadratic, we can be sure this local maximum is the global maximum too (so there are no values for $a_1$ for which $V1(a_1,a_2)$ is higher when $a_2$ is held constant).

$V{2}_{a_2}^{\prime }(a_1,a_2)=-2a_2+1=0$ gives $2a_2=1$ and $a_2=\frac{1}{2}$. $V{2}_{a_2}^{\prime \prime }(a_1,a_2)=-2<0$, so $a_2=\frac{1}{2}$ again represents a local maximum. $V2(a_1,a_2)$ is quadratic, so this is a global maximum as well.

So $a_1=0$ represents a global maximum for $V1$ (for a constant $a_2$), and $a_2=\frac{1}{2}$represents a global maximum for $V2$ (for a constant $a_1$). That means $a_1=0$ is a dominant strategy for player 1, $a_2=\frac{1}{2}$ is a dominant strategy for player 2 and we have a Nash equilibrium in $(a_1=0,a_2=\frac{1}{2})$.

Making things a bit more complicated

Let's now define $V1(a_1,a_2)=-a_1^2\ast a^2$ and $V2(a_1,a_2)=-(a_2-1{)}^2$. Then for $V{1}_{a_1}^{\prime }(a_1,a_2)=-2a_2\ast a_1=0$, we have $a_2=0\vee a_1=0$. $V{1}_{a_1}^{\prime \prime }(a_1,a_2)=-2a_2$, which is negative when $a_2>0$.

For $V{2}_{a_2}^{\prime }=-2a_2+2=0$ we have $a_2=1$. $V{2}_{a_2}^{\prime \prime }=-2<0$, so this is a local optimum - and also the global one, since $V2(a_1,a_2)$ is quadratic. For $a_2=1$, $V{1}_{a_1}^{\prime }=-2\ast 1\ast a_1=-2a_1$. Solving for $0$ gives $a_1=0$ (which we found earlier as well). And since $a_2=1>0$ and therefore $V{1}_{a_1}^{\prime \prime }(a_1,a_2)<0$, we now have a local maximum for $V1(a_1,a_2)$! For a constant $a_2$, $V1(a_1,a_2)$ is quadratic, so this is the global maximum as well. We found a Nash equilibrium: $(a_1=0,a_2=1)$.

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