Given $n$ firms ($n\geq 3$). Assume $x_i\in R^{+}$. The market price is $1-\sum_{i=1}^{n} x_i$ if $1-\sum_{i=1}^{n} x_i\geq 0$, and $0$ otherwise. Each firm has a constant marginal cost of $c<1$.
The payoff function is $\pi_{i}(x_1, x_2,\ldots, x_n) = q_i(\text{max} \left\{1-\sum_{j=1}^{n} q_j, 0\right\} - c)$.
(a) Let $q^m$ be symmetric quantity by each firm that maximizes the joint profits. Compute $q^m$ and the corresponding symmetric profit for each firm.
(b) Compute the min-max payoff for firm $i$.
Now, assume that this game is repeated infinitely many times. Each firm discounts each period payoffs by $\delta \in (0,1)$. Assume each firm chooses the same quantity as each competitors after each history. Given a symmetric $q\in R^{+}$, define
$$u(q) = q(\max\left\{1-nq,0\right\} - c).$$
Also, given $u^d(q) = max_{s_i\in R^{+}}\ \pi_i(s_i,q,\ldots, q)$ is the stage-game deviation payoff by a single firm.
(c) Find any q so that $u(q) = u^d(q)$. Graph the two functions $u(q)$ and $u^d(q)$ (label $q^c$ and $q^m$ on the figure).
(d) Assume each firm has a collusive output $q^m$ in each stage of the game, and we consider a SPNE of this infinitely repeated game. Find the minimum $\delta$ in which such outcome is supported as an SPNE in an infinitely repeated game. What is the trigger strategy to support this outcome as a SPNE?
Given an alternative SPNE, where in the first period all players choose some symmetric quantity $q^S\in R^+$. In any period thereafter, each player chooses some $q^C\in R^+$. Any deviation from this prescribed choices in any period of the game causes the players to revert to the first period play in the subsequent period, starting with $q^S$ and revert to $q^C$ again until another deviation occurs (aka. carrot-and-stick strategy).
(e) Given a subgame starting at time $t$, in which the carrot-and-stick strategy prescribes playing $q^C$ in period $t$. Prove that $u^d(q^C)\leq (1+\delta)u(q^C) - \delta u(q^S)$
(f) Same assumption as in part (e), but assume carrot-and-stick strategy prescribes playing $q^S$. Prove that $u^d(q^S)\leq (1-\delta)u(q^S) + \delta u(q^C)$
(g) Can a firm be held to its min-max payoff by the stick punishment? If that is the case, what is the corresponding stick quantity $\underline{q}$?
(h)If $q^m$ is a carror and $\underline{q}$ is a stick, find the lower bound of the discount factor $\delta$ above with the carrot-and-stick strategy.
My attempt
(a) Due to symmetry at equilibrium and firms have constant cost, we need to find $q^m = q_1 = \ldots = q_n$ is the solution to the optimization problem: $\max_{q_1,\ldots, q_n} (1-\sum_{i=1}^{n} q_i)(q_1+\ldots + q_n) - c(q_1 + \ldots + q_n) = \max_{q_1,\ldots, q_n} (1-nq^m)nq^m - cnq^m$.
Taking the first-order derivative w.r.t. $q^m$ and set it equal to $0$ to solve for $q^m$, we obtain $q^m = \fbox{$(1-c)/2n$}$
Then the symmetric profit for each firm $i$ is $\pi_i = [1- (1-c)/2](1-c)/2 -c(1-c)/2 =$ $\fbox{$\frac{(1-c)^2}{4}$}$
(b) We start from the definition of min-max payoff for firm i: $v_i = \min_{\sigma_{-i}} [\max_{q_i} \pi_i(q_i, \sigma_{-i})]$. From the definition of $\pi_i$, we get:
$\max_{q_i} \pi_i(q_i, \sigma_{-i}) = \left\{ \begin{array}{ll} \frac{1-c}{n+1}[\frac{n+c}{n+1} - \sum_{j\neq i}^{n} q_i - c],\ \text{if}\ (n+c)/(n+1) > \sum_{j\neq i}^{n} q_j\\ -c(1-c)/(n+1), \text{otherwise} \end{array} \right.$
Thus, $v_i = \min_{q_1,q_2,\ldots, q_{i-1}, q_{i+1}\ldots, q_n} \left\{ \begin{array}{ll} \frac{1-c}{n+1}[\frac{n+c}{n+1} - \sum_{j\neq i}^{n} q_i - c],\ \text{if}\ (n+c)/(n+1) > \sum_{j\neq i}^{n} q_j\\ -c(1-c)/(n+1), \text{otherwise} \end{array} \right.$
Since the first case is always greater than the second for all $q_j\neq q_i$, and the minimum of the first case, in the sense that $\sum_{j\neq i}^{n} q_j\rightarrow (n+c)/(n+1)$, is the same as that of the second case, we conclude that $\fbox{$v_i = -c(1-c)/(n+1)$}$
(c) First, $\pi_i(s_i, q, \ldots, q) = \left\{ \begin{array}{ll} s_i(1-(n-1)q-s_i-c)\ \text{if}\ s_i\leq 1-(n-1)q\\ -cs_i < 0\ \text{otherwise} \end{array} \right.$
In the first case, by taking the first-order derivative w.r.t $s_i$ and set it equal to $0$ to solve for $s_i$, we obtain $s_i = \frac{1-(n-1)q-c}{2}$. Thus, for $s_i\leq 1-(n-1)q, max_{s_i\in R^{+}}\ \pi_i(s_i,q,\ldots, q) = \frac{(1-(n-1)q\ -\ c)^2}{4}$. For $s_i\geq 1-(n-1)q$, $\pi_i (s_i,q,\ldots, q) = -cs_i$, which is a strictly decreasing function, so $\max \pi_i (s_i,q,\ldots, q) = -c(1-(n-1)q)$ over this interval.
Finally, it is easy to see that $(1-(n-1)q-c)^2/4> -c(1-(n-1)q)$, so $\max_{s_i\in R^+} \pi_i (s_i,q,\ldots, q) = \fbox{$\frac{(1-(n-1)q\ -\ c)^2}{4}$}$. Now, since $u(q) = \left\{ \begin{array}{ll} q(1-nq-c)\ \text{if}\ 1/n > q\\ -qc < 0\text{otherwise} \end{array} \right.$,
we need to find $q$ that satisfies $(1-(n-1)q-c)^2/4 = q(1-nq-c). After some algebraic manipulations, we are left with
$$q^2(n+1)^2 + (1-c)^2 - 2q(1-c)(1+n) = 0$$ $$\iff [q(n+1)-(1-c)]^2 = 0$$ $$\iff q = (1-c)/(n+1) > 0$$
I am not sure what is $q^c$, so I have no idea how to plot this point. But $q^m$ is just the same as in part (a), and the plot of $u(q)$ is comprised of two parts: one is a parabola with maximum at $(1/n, -qc)$ and the second part is a straight line with slope is $-c$, from $q=0$ to $q=1/n$.
My question. Is my solution to parts (a) - (c) above correct?
For (e) and (f), I spent so many hours trying to solve it using the definition of $u^d$ and $u$ derived in part (c), but I am still unsuccessful. I think the reason is because I cannot derive somehow the relationship between $q^C$ and $q^S$, although $(1+\delta)u(q^C) - \delta u(q^S)$ has $4$ parts corresponding to whether $q^S, q^C > 1/n$ or not. Finally, for (d), (g) and (h) I am completely stuck. Can someone please help me with any of these five remaining parts?