Equilibrium

Consider two players choosing actions $x\in\mathbb{R}$ and $y\in\mathbb{R}$ with payoffs $u_1(x,y)$ and $u_2(x,y)$. Assume payoffs exhibit diminishing marginal returns in one's own action: $\tfrac{\partial^2}{\partial x^2}u_1\lt 0$ and $\tfrac{\partial^2}{\partial y^2}u_2\lt 0$.

Step 1: Derive Best Response Functions. Each player maximizes their payoff taking the other's action as given. Since there are no constraints and payoffs have diminishing marginal returns, the first-order conditions are necessary and sufficient: $$\diff{}{x}u_1=0 \quad\text{and}\quad \diff{}{y}u_2=0$$ Solving these yields the best response functions: $x^*=BR_1(y)$ and $y^*=BR_2(x)$.

Step 2: Find the Equilibrium. An equilibrium $(x^e, y^e)$ requires all players to be best responding: $$x^e = BR_1(y^e) \quad\text{and}\quad y^e = BR_2(x^e)$$ This is a system of two equations in two unknowns. The equilibrium is found by solving this system simultaneously. Geometrically, this corresponds to the intersection of the two best response curves.

Please refer to the article on public goods for the model description and set-up. In our example, we shall assume two individuals $n=2$ whose utility for the public good is given by: $$u_i(x_i,G)=x_i\cdot G$$ As we did in class, we use the budget $x_i+g_i=w_i$ to write the utility as a function of the contribution $g_i$. Thus, individual 1 utility is $u_1=(w_1-g_1)\cdot (g_1+g_2)$ since $G=g_1+g_2$. Ignoring non-negativity constraints, $g_1\geq 0$, the individual 1 maximizes utility when $$\diff{}{g_1}u_1=-1\cdot (g_1+g_2)+(w_1-g_1)\cdot 1=w_1-g_2-2g_1=0\Leftrightarrow g_1=\dfrac{w_1-g_2}2$$ That is, $g_1=\dfrac{w_1-g_2}2$ is individual 1's best-response.

The best responses of the individuals are consistent only if there is a solution to the system of equations given by the best responses: $g_1=\dfrac{w_1-g_2}2$ and $g_2=\dfrac{w_2-g_1}2$.

Please refer to the peer-effects model of the previous section for the model description and set-up. In our example, we shall assume two individuals $n=2$ who choose actions $x$ and $y$. Each individual would like to choose her own favorite action, $a_1$ and $a_2$ respectively, but also dislikes choosing an action far from her peer's, with $\lambda\gt 0$ measuring how heavily that peer pressure weighs: $$\begin{aligned} u_1(x,y) &= 1-(a_1-x)^2-\lambda\,(y-x)^2\\[2pt] u_2(x,y) &= 1-(a_2-y)^2-\lambda\,(x-y)^2 \end{aligned}$$

There we asked what maximizes joint utility, assuming the individuals could contract on $x$ and $y$. Here they cannot—each takes the peer's action as given and chooses her own. Since $\tfrac{\partial^2}{\partial x^2}u_1=-2-2\lambda\lt 0$, utility is concave in one's own action and the first-order condition is necessary and sufficient. Individual 1 maximizes utility when $$\diff{}{x}u_1=2\,(a_1-x)+2\lambda\,(y-x)=0\Leftrightarrow x=\dfrac{a_1+\lambda\,y}{1+\lambda}$$ That is, $x=\dfrac{a_1+\lambda\,y}{1+\lambda}$ is individual 1's best-response—a weighted average of her own favorite action and her peer's choice, with the weight on the peer growing in $\lambda$.

The best responses of the individuals are consistent only if there is a solution to the system of equations given by the best responses: $x=\dfrac{a_1+\lambda\,y}{1+\lambda}$ and $y=\dfrac{a_2+\lambda\,x}{1+\lambda}$. Substituting the second into the first, $$x^e=\dfrac{(1+\lambda)\,a_1+\lambda\,a_2}{1+2\lambda} \quad\text{and}\quad y^e=\dfrac{(1+\lambda)\,a_2+\lambda\,a_1}{1+2\lambda}$$ Note that $\lambda=0$ returns $x^e=a_1$ and $y^e=a_2$, each individual ignoring her peer, while as $\lambda$ grows both actions approach the midpoint $\tfrac{a_1+a_2}2$.

In a Tullock contest two players expend effort to win a prize worth $\pi\gt 0$. In our example, we shall assume player 1 exerts effort $x$ and player 2 exerts effort $y$, that each player wins with a probability equal to her share of the total effort, and that the cost of effort is quadratic: $$u_1(x,y)=\dfrac{x}{x+y}\cdot\pi-x^2 \quad\text{and}\quad u_2(x,y)=\dfrac{y}{x+y}\cdot\pi-y^2$$ Effort is spent whether or not one wins—which is what makes a contest wasteful.

Since $\tfrac{\partial^2}{\partial x^2}u_1=-\tfrac{2\,y\,\pi}{(x+y)^3}-2\lt 0$ whenever $y\gt 0$, the payoff is concave in one's own effort and the first-order condition is necessary and sufficient. Ignoring the non-negativity constraint, $x\geq 0$, player 1 maximizes her payoff when $$\diff{}{x}u_1=\dfrac{y}{(x+y)^2}\cdot\pi-2x=0\Leftrightarrow 2x\,(x+y)^2=y\,\pi$$ That is, $2x\,(x+y)^2=y\,\pi$ defines player 1's best-response implicitly.

The best responses of the players are consistent only if there is a solution to the system of equations given by the best responses: $2x\,(x+y)^2=y\,\pi$ and $2y\,(x+y)^2=x\,\pi$. Since the two players face the same problem, look for a symmetric equilibrium $x^e=y^e$; the two equations then coincide and give $8\,(x^e)^3=x^e\,\pi$, so that $$x^e=y^e=\sqrt{\dfrac{\pi}8}$$ Together the players spend $2\,(x^e)^2=\tfrac{\pi}4$ chasing the prize—a quarter of it is burned in the contest.