Possible output of a firm using game theory

Question

Let there be two firms A and B. Let the price(P) output(X) graph be a linear one $P =a-bX=5-(1)X,c=2$ The parameter

1. a is MAX price.

2. b is the slope of the curve profit vs output(line)

3. c is the price at perfect competitive market ie large number of firms.

Now firm A wants to put a production unit.Both firms A and B are rational and A knows that B is rational. No firm has negative amount of production. Using the knowledge of game theory and Nash equilibrium choose the options which the firm A can produce. $$\text{The option set(in some proper units)}$$

$A\in(0, 0.25, 0.5, 0.75, 1 ,1.25, 1.5, 1.75 ,2)$ $$\text{Attempt}$$ Clearly the firm wont produce 0 units so it isn't one of the choices.Now we know from Nash equilibrium that total output for $N$ firms is $(1-\frac{1}{N+1})(\frac{a-c}{b})$.Here $N=2$ and putting in other values we see the total ouput$=2$. Also at equilibrium the players(firms)mutually take the best option available so each firm produces same quantity. Hence each firm's produce is $=1$.So $1$ is the equilibrium production. Now we know firms are rational but that doesn't mean that A will have the best reply right from the beginning. It may eventually start producing $1$ units. So other options do exist for firm A. We know that the profit vs output curve is quadratic $\text{Profit of A}=k(\frac{a-c}{b}-B-A)A$ where $$\text{A,B are outputs of the respective firms.}$$ I dont know how to proceed to see which other options are available to A. $$\text{Note:}$$ Assume I have no knowledge of economics so if possible limit your answer to basics of maths and game theory. $$\text{Edit}$$ In part b I was asked that if B doesn't know that he is rational then which of the above outputs would the firm A never choose.

Answer 1

1. Write out a payoff matrix, often called the strategic form. Here it is a $9 \times 9$ matrix/tableau with the quantities along the left side and the top, with two numbers in the $(i,j)$ box, one the payoff for row and the other the payoff for column when row plays $q_i \in A$ and column plays $q_j \in A$: $$ \pi_{row}(q_i,q_j) = (A-b(q_i+q_j))q_i - cq_i $$ and $$ \pi_{col}(q_i,q_j) = (A-b(q_i+q_j))q_j - cq_j. $$ Include every strategy, including zero: there are often surprising equilibria where agents adopt strategies you might not expect.

2. Construct the best reply correspondence for each player. Fix a column, and find the highest payoff for the row player, and underline; do this for all columns. Fix a row, and find the highest payoff for the column player, and underline; do this for all rows. In general, this correspondence exists and is upper hemi-continuous (by extending the pure strategies to random, mixed strategies) by Berge's Theorem of the Maximum.

3. Solve for a fixed point of the best reply correspondences, which is a Nash equilibrium. Find all boxes where both payoffs are underlined. In your game there are probably three-ish. In general, a pure-strategy equilibrium might not exist, but if you allow for random strategies, you can use Kakutani's fixed-point theorem along with convexity and upper hemi-continuity of the best-reply correspondences to prove existence of a fixed point (i.e., a Nash eqm).

That is all of game theory in a nutshell: solve for best reply correspondences for the players (use Berge's Theorem), then find a fixed point using Brouwer, Kakutani, Tarski, Glicksberg, Schauder, Banach, Eilenberg-Montgomery, or whatever the appropriate FPT is.

This game is actually solvable by iterated deletion of weakly dominated strategy, which is an epistemically stronger concept than Nash equilibrium: for each player, remove any strategy where there exists an alternative strategy that always gives a weakly higher payoff, and repeat until no further strategies can be deleted for either player. That is one way of picking a unique equilibrium in situations where there are multiple Nash equilibria, and it will be the closest one to the continuous version of the eqm you posted.

I guess the issue is that you solved the continuous version of the model using calculus. The discrete version ($A$ is finite) is going to have multiple eqa, and the question is asking, "What values in $A$ correspond to a Nash eqm for the row player?" A more sophisticated question would be, ``What values in $A$ are rationalizable: i.e., constitute a best reply to some strategy that the column player might use?" If you take the iterated deletion route, the unique answer as a result of common knowledge of rationality will be the Nash strategies.

Answer 2

Consider the game $G = \langle\{1,2\},(A_1,A_2),(u_1,u_2)\rangle$ where $i \in \{1,2\}$ denotes a firm, $A_i = \{0,0.25,\ldots,1.75,2.0\}$ the action space of firm $i$ and $u_i : A_1 \times A_2 \to \mathbb R$ with $u_i(a_1,a_2) = (5 - a_1 - a_2)a_i - 2a_i$ the utility/payoff function of firm $i$.

Approach: Solve the continuous version for an equilibrium $(\bar a_1,\bar a_2) \in \mathbb R^2$ and check whether $(\bar a_1,\bar a_2) \in A_1 \times A_2$.

Claim: The game $G$ exhibits an equilibrium at $(a_1,a_2) = (1,1)$.

Let $\text{BR}_2(a_1) = (3-a_1)/2 = \arg\max_{a_2}u_2(a_1,a_2)$ denote the best response of firm 2. Consider the first order condition for maximizing $u_1$ evalutated at $a_2 = \text{BR}_2(a_1)$ \begin{align} \frac{\partial u_1(a_1,\text{BR}_2(a_1))}{\partial a_1} = 3(1-a_1)/2 = 0. \end{align} We thus found an equilibrium candidate at $(a_1,a_2) = (1,\text{BR}_2(1)) = (1,1)$. Since the following conditions are true, it is an equilibrium: \begin{align} \frac{\partial u_i(1,1)}{\partial a_i} &= 0 \quad \forall i,\\[3mm] \frac{\partial^2 u_i(a_1,a_2)}{\partial a_i^2} &= -2 \quad \forall i. \end{align}

Edit

Consider the payoff matrices below. Now we have to carefully identify the optimal response of each agent given the action of the other agent. Agent 1 is the row player. Best responses: If $a_2 = 0$, then the best response of 1 is BR$_1(0) = 1.5$ with payoff $u_1(1.5,0) = 2.25$. If $a_2 = 0.25$, then the best response of 1 is BR$_1(0.25) = \{1.25,1.5\}$ with payoff $u_1(a_1,0.25)|_{a_1 \in \text{BR}_1(0.25)} = 1.88$. And so forth.

You will see that BR$_1(0.75)=\{1,1.25\}$, BR$_1(1)=\{1\}$, BR$_1(1.25)=\{0.75,1\}$ and BR$_2(0.75)=\{1,1.25\}$, BR$_2(1)=\{1\}$, BR$_2(1.25)=\{0.75,1\}$.

Since $1.25 \in \text{BR}_1(0.75)$ and $0.75 \in \text{BR}_2(1.25)$, there exists a Nash equilibrium at $(a_1,a_2) = (1.25,0.75)$. In total there are three equilibria \begin{align} \text{NE}(G) = \{(0.75,1.25),(1,1),(1.25,0.75)\}. \end{align}

u1 =

[     0,     0,     0,     0,     0,     0,      0,      0,      0]
[ 0.688, 0.625, 0.562,   0.5, 0.438, 0.375,  0.312,   0.25,  0.188]
[  1.25,  1.12,   1.0, 0.875,  0.75, 0.625,    0.5,  0.375,   0.25]
[  1.69,   1.5,  1.31,  1.12, 0.938,  0.75,  0.562,  0.375,  0.188]
[   2.0,  1.75,   1.5,  1.25,   1.0,  0.75,    0.5,   0.25,      0]
[  2.19,  1.88,  1.56,  1.25, 0.938, 0.625,  0.312,      0, -0.312]
[  2.25,  1.88,   1.5,  1.12,  0.75, 0.375,      0, -0.375,  -0.75]
[  2.19,  1.75,  1.31, 0.875, 0.438,     0, -0.438, -0.875,  -1.31]
[   2.0,   1.5,   1.0,   0.5,     0,  -0.5,   -1.0,   -1.5,   -2.0]

u2 =

[ 0, 0.688,  1.25,  1.69,  2.0,   2.19,   2.25,   2.19,  2.0]
[ 0, 0.625,  1.12,   1.5, 1.75,   1.88,   1.88,   1.75,  1.5]
[ 0, 0.562,   1.0,  1.31,  1.5,   1.56,    1.5,   1.31,  1.0]
[ 0,   0.5, 0.875,  1.12, 1.25,   1.25,   1.12,  0.875,  0.5]
[ 0, 0.438,  0.75, 0.938,  1.0,  0.938,   0.75,  0.438,    0]
[ 0, 0.375, 0.625,  0.75, 0.75,  0.625,  0.375,      0, -0.5]
[ 0, 0.312,   0.5, 0.562,  0.5,  0.312,      0, -0.438, -1.0]
[ 0,  0.25, 0.375, 0.375, 0.25,      0, -0.375, -0.875, -1.5]
[ 0, 0.188,  0.25, 0.188,    0, -0.312,  -0.75,  -1.31, -2.0]