1
$\begingroup$

Players 1 and 2 are involved in a joint project. Each player i independently chooses an effort $c_{i}$ that can be any number in the interval from 0 to 1; that is, $0 \leq c_{1} \leq 1$ and $0 \leq c_{2} \leq 1$.


(a)

Suppose that their output will depend on their efforts by the formula $y(c_{1}, c_{2}) = 3c_{1}c_{2}$, and each player will get half the output, but each player $i$ must also pay an effort cost equal to $c_{i}^2$ So $$u_{1} (c_{1}, c_{2}) = 1.5c_{1}c_{2} - c_{1}^2$$ $$u_{2} (c_{1}, c_{2}) = 1.5c_{1}c_{2} - c_{2}^2$$. Find all Nash equilibria without randomization.

I know how to solve this problem: In a Nash equilibrium, each player is making the choice that maximizes his/her expected utility, given the other player's choice. That means that to solve this problem, the values of $c_{1}, c_{2}$ will be:

$$\frac{\partial u_{1}}{\partial c_{1}} = 1.5c_{2} - 2c_{1} = 0$$ $$\frac{\partial u_{2}}{\partial c_{2}} = 1.5c_{1} - 2c_{2} = 0$$

Solving these two equations gives: $c_{1} = c_{2} = 0$, which is indeed the solution. But the next question, which is a modification of this problem, is the one that causes some confusion.


(b) Now suppose that their output is worth $y(c_{1}, c_{2}) = 4c_{1}c_{2}$, of which each player gets half, but each player $i$ must also pay an effort cost equal to $c_{i}$. So $$u_{1} (c_{1}, c_{2}) = 2c_{1}c_{2} - c_{1}$$ $$u_{2} (c_{1}, c_{2}) = 2c_{1}c_{2} - c_{2}$$ Find all Nash equilibria without randomization.


By approaching this problem the same way,

$$\frac{\partial u_{1}}{\partial c_{1}} = 2c_{2} - 1 = 0$$ $$\frac{\partial u_{2}}{\partial c_{2}} = 2c_{1} - 1 = 0$$

Solving these two equations gives: $$c_{1} = c_{2} = \frac{1}{2}$$

But apparently, there are two more sets of solutions:

$$c_{1} = c_{2} = 0$$ $$c_{1} = c_{2} = 1$$

My questions are:

  1. How do these solutions satisfy the conditions for the Nash equilibrium?
  2. How would one derive these solutions?
  3. Why do these additional sets of solutions pass as Nash equilibria, whereas in the previous problem, there was only one set of solution?
$\endgroup$

1 Answer 1

0
$\begingroup$

In your title you write mixed equilibria, but the body of the question specifically asks for non-randomized (pure) equilibria. There is a connection with mixed equilibria, which I will point out at the end.


Constrained optimization

This is not specific to game theory, the underlying problem is in constrained optimization. You are trying to solve the problem $$ \max_x 2ax -x $$ subject to the condition $0 \leq x \leq 1$, where $a$ is a parameter.

You can see from the derivative of the goal function the effect that increasing $x$ has on its value: $$ \frac{\text{d}(2ax -x)}{\text{d}x} = 2a - 1 $$ When $a > 1/2$, this is positive, so the goal function it is always strictly increasing in $x$. To maximize the value of the goal function you will want to increase $x$ as much as possible, which in this case is the upper limit permitted by the condition, so 1.

When $a < 1/2$, so the goal function it is always strictly decreasing in $x$. To maximize the value of the goal function you will want to decrease $x$ as much as possible, which in this case is the lower limit permitted by the condition, so 0.

When $a = 1/2$, $x$ has no effect on the goal function's value, it does not matter what value you set it to. This is the only parameter value where a so called interior solution exists, where $x$ can be optimal inside the interval specified by the condition, so where $x \in (0,1)$.


  1. How would one derive these solutions?

One should not forget that since their problem specifies $0 \leq c_i \leq 1$, they are doing constrained optimization, so it is possible that there are extremal (non-interior) solutions, and so one should not assume that straightforward differentiation yields all solutions.

You can do best response mapping, calculate which $c_1$ would maximize 1's payoff to each possible value of $c_2$, and vice-versa. Whereever you find matching pairs (e.g., $c_1 = 1$ is a best response to $c_2 = 1$, and $c_2 = 1$ is a best response to $c_1 = 1$), you have an equilibrium.

A best response mapping


Mixed equilibria

Your professor seems to have thought up a nice exercise, because this fits nicely into mixed strategies.

In game theory, when a player mixes between several strategies, they are indifferent in how they assign the probabilities between their own (mixed) strategies, otherwise they would not mix them but put all probability on their best performing strategy. So mixing is only possible when given the strategies of the other players a player has several strategies which have the same expected payoff. Thus shifting probability from one of these to the other has no effect on the payoff.

In fact, if we look at the game $$\begin{matrix} \begin{array}{c|c|c} &A &B \\ \hline A & 1, 1 & -1, 0 \\ \hline B & 0, -1 & 0, 0 \end{array} \end{matrix} $$ and denote the probability with which player $i$ plays their strategy $A$ by $c_i$, we get the same expected payoffs as in your b) exercise: \begin{align} E(U_1(c_1,c_2)) & = c_1 \cdot c_2 \cdot 1 + c_1 \cdot (1 - c_2) \cdot (-1) + (1 - c_1) \cdot c_2 \cdot 0 + (1 - c_1) \cdot (1 - c_2) \cdot 0 \\ & = c_1 \cdot c_2 \cdot 2 - c_1. \end{align} In this game, when player 2 plays their strategies with a $1/2-1/2$ probability mix, player 1 is indifferent between their own pure strategies, both yield an expected payoff of 0. It does not matter to them how they assign probabilities, when $c_2=1/2$ $$ \frac{\text{d} (c_1 \cdot c_2 \cdot 2 - c_1)}{\text{d}c_1} = 0. $$ The only mixed equilibrium though is when player 1 selects $c_1 = 1/2$. This is because unless they do so, player 2 is not indifferent between their two strategies, and will not mix them. Thus this mixed equilibrium is 'brittle', balancing on the tip of a needle; if one player were to deviate a tiny bit, say, to $c_i = 49/100$, the other would benefit by switching to a pure strategy.

$\endgroup$
5
  • $\begingroup$ Thank you for your comment. I have two qustions about your answer: 1) Is it possible to draw the best response mapping directly from the question, without defining the game matrix? 2) Does the red line in best response map mean player 1's best response to c2, and does the blue line mean player 2's best response to c1? $\endgroup$
    – user9487
    Commented Mar 25 at 7:19
  • $\begingroup$ 1) Yes, you can get it from the solution to the constrained maximization problem. 2) Yes. $\endgroup$
    – Giskard
    Commented Mar 25 at 7:23
  • $\begingroup$ Ok - Is it possible to derive the game matrix for the first question though? The one with the single Nash equilibrium, I mean. $\endgroup$
    – user9487
    Commented Mar 25 at 7:45
  • $\begingroup$ Possibly, but that is a completely different question? $\endgroup$
    – Giskard
    Commented Mar 25 at 7:46
  • $\begingroup$ Having looked at it a bit, that function does not look like it corresponds to a game matrix. Please stop expanding your question. $\endgroup$
    – Giskard
    Commented Mar 25 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.