Game Theory & Industrial Organization

1. Fix a value for \(p\) and solve for \(q\) to get \[q(p)=-\frac{p_{0}}{p_{1}} + \frac{1}{p_{1}} p.\] Since \(p_{1}<0\), demand is non-negative if and only if \(p \le p_{0}\).

2. We calculate

   \begin{align*} E(p) &= \frac{\mathrm{d} q}{\mathrm{d} p} \frac{p}{q(p)} \\ &= q'(p) \frac{p}{q(p)} \\ &= \frac{1}{p_{1}} \frac{p}{-\frac{p_{0}}{p_{1}} + \frac{1}{p_{1}} p} \\ &= \frac{p}{-p_{0} + p} \end{align*}

   For every \(0 \le p\le p_{0}\) is non-positive.

3. We have

   \begin{align*} E(q) &= \frac{\mathrm{d} q}{\mathrm{d} p} \frac{p(q)}{q} \\ &= \frac{1}{\frac{\mathrm{d} p}{\mathrm{d} q}} \frac{p(q)}{q} \\ &= \frac{1}{p'(q)} \frac{p(q)}{q} \\ &= \frac{1}{p_{1}} \frac{p_{0} + p_{1}q}{q} \end{align*}

4. Substituting gives

   \begin{align*} E(q(p)) &= \frac{p_{0} + p_{1}q(p)}{p_{1} q(p)} \\ &= \frac{p_{0} - p_{0} + p}{- p_{0} + p} \\ &= \frac{p}{-p_{0} + p} \end{align*}
5. Using part (c), we have \(E(q)=-1\) if and only if \[p_{0} + p_{1}q = -p_{1}q,\] or equivalently \[q = -\frac{p_{0}}{2p_{1}}.\]

1. The monopolist solves \[\max_{q} \left\{ p(q)q - c(q) \right\}\]

2. The first order condition is \[p'(q)q + p(q) = c'(q) \iff p_{0} + 2p_{1}q = c_{1},\] which gives

   \begin{align*} q_{m} &= -\frac{p_{0} - c_{1}}{2p_{1}}, \\ p_{m} &= \frac{c_{1} + p_{0}}{2}, \end{align*}

   and \[\pi_{m} = -\frac{(p_{0} - c_{1})^{2}}{4p_{1}}.\]

3. The demand function is \[q(p) = -\frac{1}{p_{1}} (p_{0} - p)\] and the monopolist problem is \[\max_{p} \left\{ q(p)p - c(q(p)) \right\}.\]
4. The first order condition is \[q'(p)p + q(p) = c'(q)q'(p) \iff \frac{1}{p_{1}}(2p - p_{0}) = \frac{c_{1}}{p_{1}},\] which leads to the same price, quantity, and profit as in part (b). The result is the same because inverse demand is affine (more generally because inverse demand is one-to-one).

5. The consumer welfare is

   \begin{align*} C_{m} &= \int_{0}^{q_{m}} (p(q) - p_{m}) \mathrm{d} q \\ &= p_{0}q_{m} + \frac{p_{1}}{2} q_{m}^2 - p_{m} q_{m} \\ &= \frac{1}{2} q_{m} (p_{0} - p_{m})\\ &= -\frac{(p_{0} - c_{1})^{2}}{8p_{1}}, \end{align*}

   so the total welfare is \[W_{m} = C_{m} + \pi_{m} = - 3 \frac{(p_{0} - c_{1})^{2}}{8p_{1}}.\]

6. In perfect competition, \(p_{c} = c_{1}\). Then, \[q_{c}=\frac{c_{1} - p_{0}}{p_{1}}\] and profits are zero. The competition total welfare is

   \begin{align*} W_{c} &= C_{c} \\ &= \int_{0}^{q_{c}} (p(q) - c_{1}) \mathrm{d} q \\ &= p_{0}q_{c} + \frac{p_{1}}{2} q_{c}^2 - c_{1} q_{c} \\ &= \frac{1}{2} q_{c} (p_{0} - p_{c})\\ &= -\frac{(p_{0} - c_{1})^{2}}{2p_{1}}. \end{align*}

   Finally, the deadweight loss is \[D = W_{c} - W_{m} = -\frac{(p_{0} - c_{1})^{2}}{8p_{1}}.\]

1. There are two pure Nash equilibria in this game, namely \((Bach, Bach)\) and \((Stravinsky, Stravinsky)\).

   There is also a mixed strategy equilibrium in which player \(A\) chooses \(Bach\) with \(p=\frac{2}{3}\) and \(Stravinksy\) with \(p=\frac{1}{3}\), while player \(B\) chooses \(Bach\) with \(q=\frac{1}{3}\) and \(Stravinksy\) with \(q=\frac{2}{3}\).

2. The objective of the players is to coordinate. The players would like to choose the same action. Games of this type are called coordination games.

1. The normal for of the game is as follows.

2. We distinguish three cases based on the value of the parameter \(c\). For \(01\), the game has two pure strategy Nash equilibria. These are \(\{Dove, Hawk\}\) and \(\{Hawk, Dove\}\). The last case combines the pure strategy equilibria of the first and second cases. That is, for \(c=1\), the equilibria in pure strategies are \(\{Hawk, Hawk\}\), \(\{Dove, Hawk\}\), and \(\{Hawk, Dove\}\)

   To calculate the mixed equilibria, suppose that Animal \(A\) mixes its actions by choosing \(Dove\) with probability \(p\). In equilibrium, Animal \(B\) should be indifferent between choosing \(Dove\) and \(Hawk\), therefore, \[p\frac{1}{2} = p + (1-p)\frac{1}{2}(1 -c).\] Solving for \(p\) gives \[p = \frac{c-1}{c}.\] Thus, if \(c < 1\), there is no mixed strategy equilibrium. If \(c=1\), then \(p=0\), so we essentially get one of the pure strategy equilibria of this case (namely \(\{Hawk, Hawk\}\)). Lastly, if \(c>1\), we get an additional mixed equilibrium.

3. If \(c<1\), the cost of fighting is small, and the animals attempt to get the value for their own. This leads to a prisoners' dilemma type of game with Pareto inefficient outcome. If \(c>1\), the cost of fighting is great enough to deter fighting. The animals then resort to playing a coordination game (Similar to Bach or Stravinksy).

The best responses become \[q_{i} = \frac{c_{1, i} - p_{0} - p_{1} q_{j}}{2 p_{1}}.\] Equilibrium ceases to be symmetric. The equilibrium quantities are given by \[q_{i} = \frac{2c_{1, i} - c_{1, j} - p_{0}}{3 p_{1}}.\] The symmetric equilibrium quantities are obtained as a special case from the last formula by setting \(c_{1, i} = c_{2, j}\).

Firm \(i\) produces more than \(j\) if and only \[\frac{2c_{1, i} - c_{1, j} - p_{0}}{3 p_{1}} \ge \frac{2c_{1, j} - c_{1, i} - p_{0}}{3 p_{1}},\] which, because \(p_{1} < 0\), is equivalent to \(c_{1, i} \le c_{1, j}\). Thus, the lower cost firm produces more.

The total market quantity is \[q = \frac{2c_{1, i} - c_{1, j} - p_{0}}{3 p_{1}} + \frac{2c_{1, j} - c_{1, i} - p_{0}}{3 p_{1}} = \frac{c_{1, i} + c_{1, j} - 2p_{0}}{3 p_{1}},\] and the market price \[p(q) = \frac{c_{1, i} + c_{1, j} + p_{0}}{3}.\] We can the calculate the profit of firm \(i\) as

\begin{align*} \pi_{i} &= \left(\frac{c_{1, i} + c_{1, j} + p_{0}}{3} - c_{1, i}\right) \frac{2c_{1, i} - c_{1, j} - p_{0}}{3 p_{1}} \\ & = -\frac{(2c_{1, i} - c_{1, j} - p_{0})^{2}}{9 p_{1}}. \end{align*}

The firm that produces more makes the greatest profit. The easiest way to get this result is to rewrite profits as \(\pi_{i} = -q_{i}^{2}p_{1}\). Since \(p_{1}<0\), we have \(\pi_{i}\ge \pi_{j}\) if and only if \(q_{i}\ge q_{j}\).

1. The Nash equilibrium is characterized by \(p_{1}^{\ast} = v_{1}-v_{2}>0\) and \(p_{2}^{\ast}=0\). We work our way to this result by showing that there is no profitable unilateral deviation. In the candidate equilibrium, the profits are \(\pi_{1}^{\ast} = v_{1}-v_{2}\) and \(\pi_{2}^{\ast} = 0\).

   Suppose that firm \(2\) deviates to \(p_{2}>0\). We then have \(p_{1} < p_{2} + v_{1} - v_{2}\). This implies that \(v_{2} - p_{2} < v_{1} - p_{1}\), so the consumers prefer commodity \(1\). As a result, \(\pi_{2} = 0 = \pi_{2}^{\ast}\). Thus, firm \(2\) has no profitable deviation.

   Consider the deviations of firm \(1\). There are two potential deviations. Firstly, suppose that firm deviates to \(p_{1} > v_{1} - v_{2} = p_{2} + v_{1} - v_{2}\). Then, commodity \(2\) is preferred, and firm \(1\) makes zero profit, which is less than \(\pi_{1}^{\ast}\). Secondly, consider a deviation \(p_{1} < v_{1} - v_{2} = p_{2} + v_{1} - v_{2}\). Then, commodity \(1\) is preferred, and firm \(1\) earns \(\pi_{1} = p_{1} < v_{1} - v_{2} = \pi_{1}^{\ast}\).

2. In equilibrium, it holds \[v_{1}-p_{1}^{\ast\ast} = v_{2}-p_{2}^{\ast\ast} = v_{b}-p_{1}^{\ast\ast}-p_{2}^{\ast\ast}\] because at least one of the two firms has a profitable deviation otherwise. For example, suppose that \(v_{1}-p_{1} < v_{2}-p_{2}^{\ast\ast}\). Then, only commodity \(2\) is consumed, and the profit of firm \(1\) is zero. Firm \(1\) can deviate to \(p_{1}^{\ast\ast}\) and earn profit \(p_{1}^{\ast\ast} = v_{b} - v_{2} > 0\). Similarly, other cases can be excluded.
3. The consumer surplus of the market in part one is \[C^{\ast} = v_{1} - p_{1}^{\ast} = v_{1} - v_{1} + v_{2} = v_{2}.\] The consumer surplus of the market in part two is \[C^{\ast\ast} = v_{1} - p_{1}^{\ast\ast} = \underbrace{v_{1} - v_{b}}_{<0} + v_{2} > 0.\] Therefore, \(C^{\ast}>C^{\ast\ast}\) and consumers are better off in the first case.

We find the subgame perfect equilibrium using backward induction. At the last date, player \(B\) solves \[\max_{y_{B}} \left\{-(y_{A}-y_{B})^{2}\right\}\] and gets the best response \(y_{B}(y_{A})=y_{A}\). At the same date, player \(A\) solves \[\max_{y_{A}} \left\{y_{A}y_{B} + x y_{A} - y_{A}^{2} - \frac{x^{3}}{3}\right\}\] to get the best response \(y_{A}(y_{B};x)=x\). Solving the system of best responses, we find that \(y_{A}^{\ast}=y_{B}^{\ast}=x\) in equilibrium.

At the initial date, player \(A\) anticipates the behavior of date \(2\) and maximizes \[\max_{x} \left\{ x^{2} - \frac{x^{3}}{3}\right\}.\] This gives the first order condition \(2x - x^{2}=0\), from which we get the solution \(x = 2\).

Consequently, we have a unique Nash equilibrium given by \(\left\{(x=2,y_{A}=2), y_{B}=2\right\}\).

There are two subgames. The subgame perfect equilibrium is \(\left\{(U,E,H),(B,C)\right\}\).

1. Suppose that we are at time \(t\) and player \(B\) makes an offer. Player \(A\) accepts the offer if and only if \(1-y \ge 0\). Thus, player \(B\), aiming at maximizing her share, solves \[\max_{1-y \ge 0} y\] and offers \(y=1\). At time \(t-1\), player \(B\) knows that if the game proceeds to time \(t\), player \(A\) will accept the offer \(y=1\). Therefore, if she waits for the next time point, player \(B\) can gain \(\delta_{B}\cdot 1\) in present value terms. Hence, she accepts any offer of player \(A\) securing her at least this gain, namely \(1-x \ge \delta_{B}\). Player \(A\) anticipates this behavior, solves \[\max_{1-x \ge \delta_{B}} x,\] and offers \(x=1-\delta_{B}\).

   At time \(t-2\), player \(A\) knows that if she delays, she can gain \(1-\delta_{B}\), which corresponds to gaining \(\delta_{A}(1-\delta_{B})\) discounted in time \(t-2\) terms. She, therefore, accepts an offer if \(1-y \ge \delta_{A} - \delta_{A}\delta_{B}\). Thus, player \(B\) solves \[\max_{1-y \ge \delta_{A} - \delta_{A}\delta_{B}} y\] and offers \(y = 1 - \delta_{A} + \delta_{A}\delta_{B}\). In a similar manner, one argues that, at time \(t-3\), player \(A\) solves \[\max_{1-x \ge \delta_{B} - \delta_{A}\delta_{B} + \delta_{A}\delta_{B}^2} x,\] and offers \(x=1 - \delta_{B} + \delta_{A}\delta_{B} - \delta_{A}\delta_{B}^2\).

   Repeating the aforementioned recursive arguments, we find that player \(A\)'s offer at date \(t-(2n-1)\) (observe that, because we started with player \(B\), player \(A\) moves at odd offer-rounds) is

   \begin{align*} x &= 1 - \delta_{B} + \delta_{A}\delta_{B} - \delta_{A}\delta_{B}^2 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1} - \delta_{A}^{n-1}\delta_{B}^{n} \\ &= 1 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1} - \delta_{B}\left(1 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1}\right) \\ &= \left(1 - \delta_{B}\right) \sum_{j=0}^{n-1} \delta_{A}^{j}\delta_{B}^{j}. \end{align*}

   Letting \(n\) go to infinity, the share of player \(A\) converges to

   \begin{align*} x = \frac{1 - \delta_{B}}{1 - \delta_{A}\delta_{B}}. \end{align*}

2. We can find the effect of \(\delta_{A}\) on player \(A\)'s share by calculating

   \begin{align*} \frac{\partial x}{\partial \delta_{A}} = \delta_{B}\frac{1 - \delta_{B}}{\left(1 - \delta_{A}\delta_{B}\right)^{2}}, \end{align*}

   which is positive.

3. From

   \begin{align*} \frac{\partial x}{\partial \delta_{B}} = - \frac{1 - \delta_{A}}{\left(1 - \delta_{A}\delta_{B}\right)^{2}} < 0, \end{align*}

   we see that player \(A\)'s share is decreasing in \(\delta_{B}\).

1. If the firms compete simultaneously using prices, they set \(p_{1} = p_{2} = 1\) and make zero profits. If they compete sequentially using quantities, the follower makes profit \[\pi_{f} = \frac{(5-1)^{2}}{16} = 1,\] and the leader makes profit \[\pi_{l} = \frac{(5-1)^{2}}{8} = 2.\]
2. The extensive form of the game is
3. There are two pure strategy Nash equilibria, namely \[\left\{\left\{Early, Late\right\}, \left\{Late, Early\right\}\right\},\] and a mixed strategy equilibrium in which each firm chooses \(Early\) with probability \(p=\frac{2}{3}\).

1. By Fermat's theorem, the welfare is maximized when \(W'(n) = 0\). Using the fundamental theorem of calculus, we get \[p(n q(n))\left(q(n) + n q'(n)\right) - c(q(n)) - n c'(q(n))q'(n) = \bar c.\] The optimal quantity produced by each firm when \(n\) firms enter the market is \[q(n) = -\frac{p_{0} - c_{1}}{(n + 1) p_{1}},\] hence, one calculates \[\frac{q'(n)}{q(n)} = \frac{1}{n + 1}.\] We will use the last expression to simplify the first order condition. Rearranging the necessary condition and using the linearity of costs gives

   \begin{align*} \bar c &= p(n q(n))\left(q(n) + n q'(n)\right) - c(q(n)) - n c'(q(n))q'(n) \\ &= p(n q(n))q(n) - c(q(n)) + n \left(p(n q(n)) - c'(q(n))\right)q'(n) \\ &= \pi(n) + n \left(p(n q(n))q(n) - c(q(n))\right) \frac{q'(n)}{q(n)} \\ &= \pi(n) \left(1 + n \frac{q'(n)}{q(n)}\right) \\ &= \pi(n) \left(1 - n \frac{1}{n + 1}\right) \\ &= \pi(n) \left(\frac{1}{n + 1}\right). \end{align*}

   The profit of each firm when \(n\) firms enter the market is \[\pi(n) = -\frac{(p_{0} - c_{1})^{2}}{(n + 1)^{2} p_{1}},\] so, we get

   \begin{align*} \bar c &= -\frac{(p_{0} - c_{1})^{2}}{(n + 1)^{3} p_{1}}. \end{align*}

   Therefore, the socially optimal number of firms is given by

   \begin{align*} n &= \sqrt[3]{-\frac{(p_{0} - c_{1})^{2}}{p_{1} \bar c}} - 1. \end{align*}

1. In the stage game, each firm maximizes \[\max_{q_{i}} (220 - q_{i} - q_{j} - 10 ) q_{i},\] which implies that \(q_{i} = q_{j} = q_{c} = 70\). When firms do not cooperate, they choose to produce these quantities at all dates. In this case, the market price is \[p(2q_{c}) = 220 - 140 = 80,\] and each firm makes profit \[\pi_{c} = (80 - 10 ) 70 = 4900.\]

   The monopolist would instead maximize \[\max_{q} (220 - q - 10 ) q,\] which gives \(q_{m} = 105\). The market price is then \[p(q_{m}) = 220 - 105 = 115,\] and the monopolist makes profit \[\pi_{m} = (115 - 10) 105 = 11025.\]

2. The trigger strategy that can lead to collusion is for each firm \(i\) to produce according to the rule

   \begin{align*} q_{i,t} = \left\{ \begin{aligned} \frac{q_{m}}{2} & \quad\quad q_{j,s}=\frac{q_{m}}{2} \text{ for all } s < t \\ q_{c} & \quad\quad \text{ otherwise } \end{aligned}\right. . \end{align*}
3. Given that firm \(j\) plays the trigger strategy, firm \(i\) maximizes \[\max_{q_{i}} \left(220 - q_{i} - \frac{105}{2} - 10\right) q_{i},\] which implies that the best firm \(i\) can do, if it deviates from the trigger strategy, is to produce \(q_{d} = \frac{315}{4}\). The profit of firm \(i\) is then \[\pi_{d} = \frac{99225}{16},\] which is greater than \(pi_{m}/2\). Thus, the firm can make greater flow profit at the initial date if it deviates from coordination.
4. This practice, however, does not make the firm overall better off. Only short-sighted firms prefer this scenario. Firms that value future profits prefer coordination. Specifically, coordination can be sustained if and only if \[\frac{1}{1-\delta} \frac{\pi_{m}}{2} \ge \pi_{d} + \frac{\delta}{1 -\delta}\pi_{c}\] or equivalently \[\delta \ge \frac{\frac{\pi_{m}}{2} - \pi_{d}}{\pi_{c}- \pi_{d}} = \frac{\frac{11025}{2} - \frac{99225}{16}}{4900 - \frac{99225}{16}} = \frac{11025}{20825} = \frac{9}{17}.\]

1. If \(p=1\), the firms compete only for one date, and the game reduces to Bertrand competition. The firms set \(p_{1}=p_{2}=c\).
2. The most profitable deviation of each firm at the initial date is to slightly undercut the monopoly price by \(\varepsilon\). By doing so, the firm can take all the market demand and achieve almost the monopolistic profit. For simplicity, assume that the deviating firm earns exactly the monopolistic profit. Then, the deviation is not profitable if and only if \[\frac{1}{p}\frac{\pi_{m}}{2} \ge \pi_{m} \iff p \le \frac{1}{2}.\] Collusion is sustainable if the probability that the market closes down is small and the firms have the potential to benefit from future profits.
3. With arguments analogous to the \(\mu=\frac{1}{2}\) case, the condition for firm \(1\) is \[\frac{1}{p}\mu \pi_{m} \ge \pi_{m} \iff p \le \mu,\] and for firm \(2\) is \[\frac{1}{p}(1-\mu) \pi_{m} \ge \pi_{m} \iff p \le 1-\mu.\] Therefore, collusion is sustainable if and only if \[p \le \min\{\mu, 1-\mu\}.\] If \(\mu < \frac{1}{2}\), the equilibrium condition becomes less stringent when \(\mu\) increases. In contrast, the equilibrium condition becomes more stringent when \(\mu\) increases for \(\mu>\frac{1}{2}\). The most lenient equilibrium condition is obtained for \(\mu=\frac{1}{2}\). The further is \(\mu\) from \(\frac{1}{2}\), the less are the flow profits that one of the two firms makes and, as a result, coordination based on future benefits becomes less profitable.

1. As in the grim trigger strategy case, if the firms do not coordinate, they produce \[q_{c} = \frac{c - p_{0}}{3 p_{1}},\] make flow profits \[\pi_{c} = -\frac{(p_{0} - c)^{2}}{9 p_{1}},\] and total discounted profits \[V_{c} = -\frac{1}{1-\delta}\frac{(p_{0} - c)^{2}}{9 p_{1}}.\]
2. In this case, each firm produces \[\frac{q_{m}}{2} = \frac{c - p_{0}}{4 p_{1}},\] makes flow profit \[\frac{\pi_{m}}{2} = -\frac{(p_{0} - c)^{2}}{8 p_{1}},\] and total discounted profit \[V_{m} = -\frac{1}{1-\delta}\frac{(p_{0} - c)^{2}}{8 p_{1}}.\]
3. The most profitable deviation of firm \(i\) is calculated by solving \[\max_{q_{i}} \left(p_{0} + p_{1} \left(q_{i} + \frac{q_{m}}{2}\right) - c\right) q_{i},\] which leads to \[q_{1} = \frac{3}{8}\frac{c-p_{0}}{p_{1}},\] and flow profit \[\pi_{1} = -\frac{9}{64}\frac{(c-p_{0})^{2}}{p_{1}}.\]

4. Firm \(i\) is producing half the monopolistic quantity, and firm \(j\) is producing the Cournot quantity, punishing the initial date deviation. Thus, firm \(i\) makes profit

   \begin{align*} \pi_{2} &= \left(p_{0} + p_{1} \left(\frac{q_{m}}{2} + q_{c}\right) - c\right) \frac{q_{m}}{2} \\ &= -\frac{5}{48}\frac{(c-p_{0})^{2}}{p_{1}}. \end{align*}

5. There are two cases we should consider. The deviating firm may revert back to producing half the monopolistic quantity after the deviation, or it can resort to producing the Cournot quantity. In the first case, the payoff is \[V_{r} = \pi_{1} + \delta \pi_{2} + \delta^{2} V_{m},\] while in the second case, it is \[V_{n} = \pi_{1} + \delta V_{c}.\] The second case is excluded if and only if

   \begin{align*} V_{r} \ge V_{n} &\iff \pi_{2} + \delta V_{m} \ge V_{c} \\ &\iff -\frac{5}{48}\frac{(c-p_{0})^{2}}{p_{1}} - \frac{\delta}{1-\delta} \frac{1}{8}\frac{(c-p_{0})^{2}}{p_{1}} \ge - \frac{1}{1-\delta} \frac{1}{9}\frac{(c-p_{0})^{2}}{p_{1}} \\ &\iff \delta \ge \frac{1}{3}. \end{align*}

   Moreover, for collusion to be sustainable, we need

   \begin{align*} V_{m} \ge V_{r} &\iff V_{m} \ge \pi_{1} + \delta \pi_{2} + \delta^{2} V_{m} \\ &\iff -\frac{1 - \delta^{2}}{1-\delta} \frac{1}{8} \frac{(c-p_{0})^{2}}{p_{1}} \ge -\frac{9}{64}\frac{(c-p_{0})^{2}}{p_{1}} - \delta \frac{5}{48}\frac{(c-p_{0})^{2}}{p_{1}} \\ &\iff (1 + \delta) \frac{1}{8} \ge \frac{9}{64} + \delta \frac{5}{48} \\ &\iff \delta \ge \frac{3}{4}. \end{align*}

1. The Bayesian normal form is
2. The unique Bayesian Nash equilibrium of the game is \(\left\{BottomTop, Right\right\}\).

1. We will show that setting a bid \(b_{i}^{\ast}=v_{i}\) weakly dominates any other bidding strategy. That is, we will show that bidding strategies \(b_{i}\neq v_{i}\) do not give greater payoffs than \(b_{i}^{\ast}=v_{i}\). Suppose that player \(i\) has valuation \(v_{i}\) and considers a bid \(b_{i}\). We examine the following two alternative cases.

   Firstly, consider \(b_{i}>v_{i}\). There are four potential outcomes for this bid. Firstly, if the other player's bid is greater than \(b_{i}\), i.e., \(b_{j}>b_{i}\), then both \(b_{i}\) and \(b_{i}^{\ast}\) result in zero payoffs for player \(i\). Secondly, if the two bids are equal, then player \(i\)'s expected payoff is \(\frac{1}{2}(v_{i}-b_{i})<0\). Thus, player \(i\) is doing better by deviating to \(b_{i}^{\ast}\). Thirdly, if the other player's bid is between \(b_{i}\) and \(v_{i}\), i.e., \(b_{i}>b_{j}>v_{i}\), then player \(i\) wins the object and pays a price \(b_{j}\). Her payoff is then \(v_{i}-b_{j}<0\), and the player can do better (zero payoff) if she deviates to \(b_{i}^{\ast}\). Fourthly, if the other player's bid is less than player \(i\)'s valuation, i.e., \(v_{i} \ge b_{j}\), then player \(i\) wins the object and has a payoff \(v_{i}-b_{j}\ge 0\). Her payoff is the same if she bids \(b_{i}^{\ast}\).

   Secondly, consider \(b_{i} < v_{i}\). Due to symmetry, the previous paragraph implies that we should focus to cases for which \(b_{j}\le v_{i}\). Firstly, if \(b_{j}=v_{i}\), player \(i\) has zero payoff. Deviating to \(b_{i}^{\ast}\) gives payoff \(\frac{1}{2}(v_{i}-b_{i})=0\). Thus, the player is indifferent between \(b_{i}\) and \(b_{i}^{\ast}\). Secondly, if \(b_{i} < b_{j} < v_{i}\), player \(i\) has zero payoff. Deviating to \(b_{i}^{\ast}\) gives payoff \(v_{i}-b_{j} > 0\), which makes player \(i\) better off. Thirdly, if \(b_{i}=b_{j} < v_{i}\), player \(i\) has expected payoff \(\frac{1}{2}(v_{i}-b_{i}) > 0\). Her payoff is doubled if she deviates to \(b_{i}^{\ast}\). Finally, if \(b_{j} < b_{i} < v_{i}\), player \(i\) has expected payoff \(v_{i}-b_{j} > 0\). Player \(i\) achieves the same payoff by deviating to \(b_{i}^{\ast}\).

2. Yes, the auction mechanism allocates the object to the buyer with the highest valuation.

1. We will show that the auction has a symmetric equilibrium of the form \(b_{i}^{\ast}(v)=\alpha v\) using the method of undetermined coefficients. Buyer \(i\) maximizes her expected payoff \[u_{i} = \max_{b_{i}} \mathbb{E} (v_{i} - b_{i}) = \max_{b_{i}} (v_{i} - b_{i}) \mathbb{P}(b_{i} \ge b_{j}).\] The bid equality has zero probability and can be ignored. Thus,

   \begin{align*} u_{i} &= \max_{b_{i}} (v_{i} - b_{i}) \mathbb{P}(b_{i} > \alpha_{j} v_{j}) \\ &= \max_{b_{i}} (v_{i} - b_{i}) \frac{b_{i}}{\alpha}. \end{align*}

   The necessary condition is \[2 b_{i} = v_{i},\] which implies that \(\alpha=1/2\). The objective is strictly concave, so the solution represents a unique maximum.

2. Yes, the auction mechanism allocates the object to the buyer with the highest valuation.

1. Firm \(2\) observes its marginal cost and solves \[\max_{q_{s,2}} \left\{ p(q_{1} + q_{s,2})q_{s,2} - c_{s,2}q_{s,2}\right\} \quad\quad (c_{s,2} = 0,2).\] From this, we can calculate the two first order conditions (one for each state) as \[q_{s,2} = \frac{-c_{s,2} + 10 - q_{1}}{2} \quad\quad (c_{s,2} = 0,2).\] Further, we find that the expected quantity produced by firm \(2\) is \[\mathbb{E}q_{2} = \frac{9 - q_{1}}{2}.\] Firm \(1\) does not observe the marginal cost of firm \(2\), so it maximizes the expected profit \[\max_{q_{1}} \left\{ \mathbb{E} p(q_{1} + q_{2})q_{1} - q_{1}\right\},\] which gives the best response \[q_{1} = \frac{9 - \mathbb{E}q_{2}}{2}.\] Substituting the expected quantity produced by firm \(2\) gives \(q_{1} = 3\). Finally, substituting in the best response of firm \(2\), we get \[q_{0,2} = \frac{21}{6},\] when nature draws the low cost and \[q_{2,2} = \frac{15}{6},\] when nature draws the high cost.
2. When the marginal cost of firm \(2\) is zero, the market quantity is \[q_{0} = q_{1} + q_{0,2} = 3 + \frac{21}{6} = \frac{39}{6},\] and the market price is \[p(q_{0}) = 10 - q_{0} = \frac{21}{6}.\] When the marginal cost of firm \(2\) is equal to \(2\), the market quantity is \[q_{2} = q_{1} + q_{2,2} = 3 + \frac{15}{6} = \frac{33}{6},\] and the market price is \[p(q_{2}) = 10 - q_{2} = \frac{27}{6}.\]
3. The expected profit of firm \(2\) is \[\mathbb{E}\pi_{2} = \frac{1}{2} \frac{21}{6} \frac{21}{6} + \frac{1}{2} \left(\frac{27}{6} - 2\right)\frac{15}{6} = \frac{37}{4}.\] The profit of firm \(1\) is \[\mathbb{E}\pi_{1} = \frac{1}{2} \left(\frac{21}{6} - 1\right)3 + \frac{1}{2} \left(\frac{27}{6} - 1\right) 3 = 9.\] Firm \(2\) makes, on average, more profit than firm \(1\). Firm \(2\) has more information about the state of nature and can adjust its decisions better than firm \(1\).

1. Firm \(2\) observes its marginal cost and solves \[\max_{q_{\theta,2}} \left\{ p(q_{1} + q_{\theta,2})q_{\theta,2} - c_{\theta,2}q_{\theta,2}\right\}.\] From this, we can calculate the state dependent first order conditions (one for each state) as \[q_{\theta,2} = \frac{c_{\theta,2} - p_0 - p_{1}q_{1}}{2 p_{1}}.\] Further, we calculate the expected quantity produced by firm \(2\) as \[\mathbb{E}q_{2} = \frac{\mathbb{E}c_{2} - p_0 - p_{1}q_{1}}{2 p_{1}}.\] Firm \(1\) does not observe the marginal cost of firm \(2\), so it maximizes the expected profit \[\max_{q_{1}} \left\{ \mathbb{E} p(q_{1} + q_{2})q_{1} - c_{1}q_{1}\right\},\] which gives the best response \[q_{1} = \frac{c_{1} - p_0 - p_{1}\mathbb{E}q_{2}}{2 p_{1}}.\] Substituting the expected quantity produced by firm \(2\) gives \[q_{1} = \frac{2c_{1} - p_0 - \mathbb{E}c_{2}}{3 p_{1}}.\] Finally, substituting in the best response of firm \(2\), we get \[q_{\theta,2} = \frac{3 c_{\theta,2} + \mathbb{E} c_{2} - 2c_{1} - 2p_0}{6 p_{1}}.\]

2. The total market quantity is

   \begin{align*} q_{\theta} &= q_{1} + q_{\theta,2} \\ &= \frac{2c_{1} - p_0 - \mathbb{E}c_{2}}{3 p_{1}} + \frac{3 c_{\theta,2} + \mathbb{E} c_{2} - 2c_{1} - 2p_0}{6 p_{1}} \\ &= \frac{2c_{1} + 3 c_{\theta,2} - \mathbb{E}c_{2} - 4p_0}{6 p_{1}}, \end{align*}

   and the market price is

   \begin{align*} p(q_{\theta}) &= p_{0} + p_{1}q_{\theta} = \frac{2c_{1} + 3 c_{\theta,2} - \mathbb{E}c_{2} + 2p_0}{6}. \end{align*}

3. The profit of firm \(2\) is

   \begin{align*} \pi_{\theta,2} &= \left( \frac{2c_{1} + 3 c_{\theta,2} - \mathbb{E}c_{2} + 2p_0}{6} - c_{\theta,2} \right) \frac{3 c_{\theta,2} + \mathbb{E} c_{2} - 2c_{1} - 2p_0}{6 p_{1}} \\ &= - \frac{\left(2c_{1} - 3 c_{\theta,2} - \mathbb{E}c_{2} + 2p_0\right)^{2}}{36 p_{1}}. \end{align*}

   The profit of firm \(1\) is

   \begin{align*} \pi_{\theta,1} &= \left( \frac{2c_{1} + 3 c_{\theta,2} - \mathbb{E}c_{2} + 2p_0}{6} - c_{1} \right) \frac{2c_{1} - p_0 - \mathbb{E}c_{2}}{3 p_{1}} \\ &= \frac{-4c_{1} + 3 c_{\theta,2} - \mathbb{E}c_{2} + 2p_0}{6} \frac{2c_{1} - p_0 - \mathbb{E}c_{2}}{3 p_{1}} . \end{align*}
4. We have \(q_{1} > \mathbb{E} q_{2}\) if and only if \[\frac{2c_{1} - p_0 - \mathbb{E}c_{2}}{3 p_{1}} > \frac{3 \mathbb{E} c_{2} + \mathbb{E} c_{2} - 2c_{1} - 2p_0}{6 p_{1}},\] which simplifies to \[c_{1} < \mathbb{E} c_{2}.\] By the distributional assumptions of \(c_{2}\), the last condition is equivalent to \[0 < \mathbb{E} \theta = 2p - 1.\] If the average shock to the marginal cost of firm \(2\) is positive (or equivalently \(p>1/2\)), then firm \(2\)'s production cost is on average greater than that of firm \(1\), and the expected production of firm \(2\) is less than that of firm \(1\).