03 Oct 2015
So far, we’ve seen one-shot Bertrand games with up to \(N\) firms. What happens if the firms are playing a Bertrand game repeatedly for \(T\) periods?
Consider a repeated Bertrand oligopoly with \(N \ge 2\) firms, played for \(T \lt \infty\) periods. We know that the one-shot Nash equilibrium is \(p_{i}=c \ \ \ \forall i\). For \(T\) periods, we start at the last period and work backwards (backwards induction).
In the last period, the game is a one-shot Bertrand game. Thus the equilibrium price will be \(p_{i,T}=c \ \ \ \forall i\). Each firm knows that in period \(T-1\), and since there’s no possibility of colluding tomorrow they each play \(p_{i,T-1}=c\). We can extend the argument to see that \(p_{i,t}=c \ \ \ \forall t,i\).
What if \(T = \infty\)? If the game is infinitely repeated, we can’t apply backwards induction since there is no last period. Instead, we define trigger strategies and look for parameter restrictions that would need to hold to make the strategy a subgame-perfect Nash equilibrium.
Let’s take the same example, and assume that play is history-dependent, i.e. players consider what happened before in deciding what to do today. We’ll consider 3 cases: first when players compete in an infinitely repeated market like above, second when the market as an exogenous probability of ending after each period, and third when firms compete in both markets simultaneously.
Case 1:
Suppose players are competing in a market with inverse demand \(D(p)\), call it market A, and have a discount factor \(\delta \lt 1\). The history at any period \(T\) is given by:
\[H_{T-1} = \{ (p_{i,t}); i=1,2,...,N; t=1,2,...,T-1 \}\]
The trigger strategy I define for each player \(i\) to collude is:
\[p_{i,t} =
\begin{cases}
p_m \ \ \text{if all elements of} \ \ H_{t-1} = p_m, \ \ \text{or if} \ \ t=1 \cr
c \ \ \text{otherwise}
\end{cases}\]
where \(p_m\) is the monopoly price, and \(\pi_m\) is the monopoly profit. We assume that if the firms collude, they do so optimally - they set the price at the monopoly price and divide the profits equally. This is better than pricing at marginal cost and earning 0 profits. But in every period, there is an incentive for firms to deviate and slightly undercut everyone else, capturing the full monopoly profits for one period if they do. However, when they do this everyone goes into the punishment phase and plays \(p=c\), so nobody makes any profits after that. The decision to cooperate or deviate comes down to weighing a stream of discounted payoffs every period against one big payoff today. The discount factor, \(\delta \in (0,1)\) determines how heavily firms weigh future payoffs against present payoffs.
The firms’ payoffs from cooperating and deviating optimally are:
\[\begin{align}
\text{Cooperate:} & \ \ \frac{\pi_m}{N} + \delta\frac{\pi_m}{N} + \delta^2\frac{\pi_m}{N} + ... \cr
& = \frac{\pi_m}{N}\frac{1}{1-\delta} \cr \cr
\text{Deviate:} & \ \ \pi_m + 0 + 0 + ... \cr
& = \pi_m \cr
\end{align}\]
To find the minimum \(\delta\) that can support collusion, we set the cooperation payoff to be greater than or equal to the deviation payoff and solve for \(\delta\). This gives us that the critical discount factor is
\[\delta_c \ge \frac{N-1}{N}\]
So as \(N \to \infty\), \(\delta_c \to 1\). When \(N=2, \delta_c \ge \frac{1}{2}\). Collusion becomes harder to sustain when there are more firms, as it requires each firm to weight future payoffs increasingly heavily against current payoffs.
Case 2:
In this case, there is an exogenous probability \(\alpha \in (0,1)\) that the market will end after each period. Let’s call this market B. The strategy we defined earlier can be applied here and the deviation payoff is the same, but the cooperation payoff is a little different:
\[\begin{align}
\text{Cooperate:} & \ \ \frac{\pi_m}{N} + \delta(1-\alpha)\frac{\pi_m}{N} + \delta^2(1-\alpha)^2\frac{\pi_m}{N} + ... \cr
& = \frac{\pi_m}{N}\frac{1}{1-\delta(1-\alpha)} \cr
\end{align}\]
Solving for the critical discount factor, we get
\[\delta_c \ge \frac{N-1}{N}\frac{1}{(1-\alpha)}\]
When \(N=2\):
-
\[\alpha\to0 \implies \delta_c \ge \frac{1}{2}\]
-
\[\alpha=\frac{1}{2} \implies \delta_c \ge 1\]
-
\[\alpha\to1 \implies \delta_c \ge \infty\]
If the market will survive with probability 1, they are as likely to cooperate as in case 1. If the market is 50% or more likely to end tomorrow, they will not cooperate.
Case 3:
Finally, what if the firms compete in markets A and B simultaneously? Is collusion more or less likely then? The strategies and payoffs are a bit different here.
The history at time T is:
\[H_{T-1} = \{ (p_{i,t,A},p_{i,t,B}); i=1,2,...,N; t=1,2,...,T-1\}\]
The trigger strategy I define is:
\[(p_{i,t,A},p_{i,t,B}) =
\begin{cases}
(p_m,p_m) \ \ \text{if all elements of} \ \ H_{t-1} = (p_m,p_m), \text{or if} \ \ t=1 \cr
c \ \ \text{otherwise}
\end{cases}\]
The payoffs from cooperating and deviating are:
\[\begin{align}
\text{Cooperate:} & \ \ (\frac{\pi_m}{N} + \frac{\pi_m}{N}) + (\delta\frac{\pi_m}{N} + \delta(1-\alpha)\frac{\pi_m}{N}) + (\delta^2\frac{\pi_m}{N} + \delta^2(1-\alpha)^2\frac{\pi_m}{N}) + ... \cr
& = 2\frac{\pi_m}{N} + \frac{\pi_m}{N}\frac{\delta(1+(1-\alpha))}{1-\delta(1+(1-\alpha))} \cr \cr
\text{Deviate:} & \ \ 2\pi_m + 0 + 0 + ... \cr
& = 2\pi_m \cr
\end{align}\]
I’m defining this as simply as I can: firms collude if other firms cooperate in both markets. Any firm who deviates is going to deviate in both markets at the same time. If firms see another firm deviate in either market, all firms go into the punishment phase and play \(p=c\) forever after. This way of defining it makes sense to me because deviating in both markets is better than deviating in one market, given the trigger strategy. I think this trigger strategy also prevents a firm from repeatedly defecting in only one market at a time, which could happen if a double-defection was needed to initiate the punishment phase.
Solving like before, we get
\[\delta_c \ge \frac{2(N-1)}{1+2(N-1)}\frac{1}{(2-\alpha)}\]
As \(N \to \infty\), \(\delta_c \to 1 ~~~ \forall \alpha\). When \(N=2\):
-
\[\alpha\to0 \implies \delta_c \ge \frac{1}{3}\]
-
\[\alpha=\frac{1}{2} \implies \delta_c \ge 1\]
-
\[\alpha\to1 \implies \delta_c \ge \frac{2}{3}\]
As I understand it, \(\alpha=\frac{1}{2}\) represents a state of total uncertainty - market B is equally likely to exist or not exist tomorrow. Any non-uniform distribution over the state of market B tomorrow gives the players more information about what is likely to happen than when \(\alpha=\frac{1}{2}\). When players are completely uncertain about whether or not market B will exist tomorrow, they need to weigh tomorrow at least as heavily as today in order to cooperate (given our assumptions on \(\delta\), they can’t do that). When market B survives with probability 1, they are most likely to cooperate (\(\delta \ge \frac{1}{3}\)). When market B survives with probability 0, they need a discount factor of at least \(\frac{2}{3}\) to cooperate.
Collusion can be more or less likely in case 3 than case 2 or case 1. Depending on how likely it is that market B will exist tomorrow the firms’ payoffs from cooperating can be more than doubled, but their payoffs from deviating are doubled no matter what happens to market B tomorrow.
Summary
The one-shot and finitely-repeated Bertrand games have very vicious competition, but as we see here the infinitely-repeated Bertrand game can support collusion if firms value the future highly enough. As the number of firms increases, collusion gets harder to support. When the market might end at any time, collusion gets harder to support. One way to make collusion more feasible when a market could end at any time is to have the firms compete simultaneously in another market that won’t end. Depending on the probability of the second market ending, the two-market case may be more collusive than a single unending market. Repeated-play makes things interesting.
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02 Oct 2015
In the previous post, I mentioned the concept of pass-through. Pass-through is the partial derivative of price with respect to marginal cost; it tells us how the equilibrium price will change in response to a change in marginal cost. As we will see here, it is closely related to the curvature of the inverse demand function.
Consider a homogeneous product market with \(N \ge 2\) firms which are Cournot competitors with the same constant marginal cost \(c\). The inverse demand function is \(P(Q)\), and it satisfies \(2P'(Q) + QP''(Q) \lt 0\). The first order condition for firm \(i\) when every firm produces positive output is given by \(P(Q) + q_{i}P'(Q) - c = 0\). Because the firms are symmetric, \(q_i = \frac{Q}{N}\), making firm \(i\)’s first order condition \(P(Q) + \frac{Q}{N}P'(Q) - c = 0\). Call this FOC as \(F\).
Define \(\sigma(Q) = -\frac{QP''(Q)}{P'(Q)}\), the curvature of the inverse demand function.
By the implicit function theorem, we can find \(\frac{\partial Q}{\partial c}\), the quantity pass-through, as
\[\begin{align}
\frac{\partial Q}{\partial c} & = - \frac{\frac{\partial F}{\partial c}}{\frac{\partial F}{\partial Q}} \cr
& = \frac{N}{(N+1)P'(Q) + QP''(Q)} \cr
\end{align}\]
Then, by the chain rule,
\[\begin{align}
\frac{\partial P(Q)}{\partial c} & = \frac{\partial P}{\partial Q}\frac{\partial Q}{\partial c} \cr
\implies \frac{\partial P(Q)}{\partial c} & = \frac{\partial P}{\partial Q} \frac{N}{(N+1)P'(Q) + QP''(Q)} \cr
\end{align}\]
Dividing by \(\frac{\partial P}{\partial Q}\) on top and bottom, we get
\[\begin{align}
\frac{\partial P}{\partial c} & = \frac{N}{(N+1) + Q\frac{P''(Q)}{P'(Q)}} \cr
& = \frac{N}{(N+1) - \sigma(Q)} \cr
\end{align}\]
Thus, we see that the pass-through rate is indeed a function of the curvature of the inverse demand function. If we assume inverse demand is linear, we get
\[P'(c) = \frac{N}{N+1}\]
As \(N \to \infty\), \(P'(c) \to 1\). This makes sense, given that we know as \(N \to \infty\), \(P \to c\).
Pass-through is a pretty useful concept, and it seems to be a popular tool these days in analyses of oligopoly and market power.
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01 Oct 2015
Consider a Hotelling duopoly on opposite ends of a linear city of length 1. We know that if the firms and products are symmetric, the indifferent consumer is at the midpoint and the price is greater than marginal cost by the travel cost (\(\tilde{x}=\frac{1}{2}, p=t+c\)). Suppose the firms are symmetric with constant marginal cost \(c\), but consumers value their products at \(v_1\) and \(v_2\), respectively. Assume that \(v_1\) and \(v_2\) are high enough that the market will be covered, and that the firms choose prices simultaneously. As before, we first find the indifferent consumer’s demand and then plug it into the firms’ profit-maximization problems.
The indifferent consumer’s demand:
\[\begin{align}
& v_1 - p_1 - t\tilde{x} = v_2 - p_2 - t(1-\tilde{x}) \cr
\cr
\implies & \tilde{x} = \frac{(v_1-v_2)-(p_1-p_2)}{2t} + \frac{1}{2} \cr
\end{align}\]
The firms’ first-order conditions and optimal prices:
\[\begin{align}
\text{Firm 1:} & \ \ \frac{1}{2}[1 + \frac{(v_1-v_2)-(2p_1-p_2)-c}{t}] = 0 \cr
\text{Firm 2:} & \ \ \frac{1}{2}[1 + \frac{(v_1-v_2)-(p_1-2p_2)-c}{t}] = 0 \cr \cr
p_1 & = \frac{1}{2}[v_1-v_2+p_2^* + \frac{t}{2} + c] \cr
p_2 & = \frac{1}{2}[v_2-v_1+p_1^* + \frac{t}{2} + c] \cr
\end{align}\]
As we’d expect, each firm’s optimal price is increasing in consumers’ valuations of their own product, decreasing in consumers’ valuations of their competitor’s product, and increasing in their marginal cost. Interestingly, their own optimal prices are increasing in their competitor’s price. I think the correct terms are “strategic substitutes” and “strategic complements”, but I could be mistaken.
In Cournot models, if the other firm increases their quantity, you want to decrease yours - we see this from the Cournot best-response, \(q_i^* = \frac{-(p(Q) - c)}{\frac{\partial p}{\partial Q}}\) (which is \(q_i^* = \frac{a - bq_j^* - c}{2b}\) when inverse demand is linear), where \(\frac{\partial q_i}{\partial q_j} \lt 0\). The quantities are “strategic substitutes”. This linkage is through the inverse demand function, where one firm increasing its quantity results in a lower price for all firms.
In Hotelling models the opposite is true regarding prices - they are “strategic complements”, or \(\frac{\partial p_i}{\partial p_j} \gt 0\). My intuition for this is that the other firm raising its price gives you room to raise yours; as long as we assume the market is covered, there is a single indifferent consumer choosing between the two firms who links their prices in this way. If the market isn’t covered, I think you would have two indifferent consumers, one for each firm, each consumer indifferent between buying from the nearer firm or not buying at all. Then the firms’ problems would be decoupled, and \(\frac{\partial p_i}{\partial p_j}\) would be 0.
How much would a change in cost affect the price? We can see this by looking at \(\frac{\partial p_i}{\partial c}\) for any \(i\). This is called the “pass-through rate”, or just the “pass-through”. In a future post, I’ll derive the pass-through rate for a Cournot oligopolist and relate it to the curvature of the inverse demand function. Here, the pass-through is easy: \(\frac{\partial p_i}{\partial c} = \frac{1}{2}\).
A Merger
What if the firms merged and became a monopolist selling two differentiated products? In this case, the firm can capture all of the consumer surplus. The indifferent consumer’s demand becomes:
\[\begin{align}
& v_1 - p_1 - t\tilde{x} = v_2 - p_2 - t(1-\tilde{x}) = 0 \cr \cr
\implies & \tilde{x} = \frac{v_1-p_1}{t}, \cr
& \tilde{x} = \frac{-v_2+p_2}{t} + 1 \cr
\end{align}\]
We can use the two expressions we have for \(\tilde{x}\) to get a function that relates \(p_1\) to \(p_2\). I’m going to go with \(p_2 = v_2 + v_1 - p_1 - t\). Now we solve the firm’s problem and get the prices:
\[\begin{align}
& \max_{p_1} \ [p_1 - c] \ \tilde{x}(p_1) + [p_2(p_1) - c] \ \tilde{x}(p_1) \cr
\implies & p_1 = \frac{1}{2}[v_1 + v_2 - t - 2c] \cr
\implies & p_2 = \frac{1}{2}[v_1 + v_2 - t + 2c] \cr
\end{align}\]
\(p_1\) and \(p_2\) look a little weird to me; I don’t understand why the cost enters them like that. The effect of a marginal cost increase will be exactly offset by the prices, so maybe it doesn’t matter?
Whereas before the firms’ prices were increasing in the travel cost, the new monopoly’s prices are decreasing in travel cost. I think the monopoly is fully internalizing the effect of the travel cost, in that an increase in \(t\) reduces the amount of consumer surplus that the firm can take when it s the marginal consumer’s utility to 0.
As either valuation increases, both prices increase.
I like the addition of the firms merging. I’d like to look at some more models with that.
Variations to explore:
-
A merger between two firms in Triangle City
-
A model of Circle City with \(n\) firms
-
A merger between two firms in an \(n\)-firm Circle City
-
A Hotelling model on a not-fully-connected graph? Consumers and firms would be at the nodes, with the cost of reaching a firm increasing in the number of nodes between a consumer and the firm. I think it would be a lot harder to solve than a Hotelling model on a continuous space, but the linkages in price through demand would be interesting. Maybe a continuous city on a polygon with lines between vertices is an easier way to look at those linkages.
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30 Sep 2015
In most consumer-firm problems, we assume the consumer owns the firm, but doesn’t manage it. Who does? It’s not quite clear. I like to imagine that the firm is a robot that the consumer programmed to produce stuff and whirr MAXIMIZE PROFITS whirr. This model came from thinking about scenarios where a profit-maximizing owner who is perfectly capable of maximizing profits might want to hire a manager in exchange for a share of profits. Why would they do this? When would there be agency problems?
The setting
The underlying problem is a profit-maximizing monopolist facing linear inverse demand \(p=a-bx\) and some positive fixed cost \(F\). That is,
\[\max_{x} (p - c)x - F\]
I’m framing it as a Cournot problem, but we could do it as a Bertrand problem and we’d get the same result because there’s only one firm. The fixed cost goes away and we get the familiar \(x^*=\frac{a-c}{2b}\) and \(p^*=\frac{a+c}{2}\).
The twist is that the fixed cost is \(F(k) = d - k\), where \(d\) is some positive number and \(k\in[0,1]\) is the share of profits the owner will give to a manager who can reduce the fixed costs. (The point wasn’t to be super realistic, I just wanted something simple and decreasing in \(k\).)
The new problems are
\[\begin{align}
\text{Manager:}& \ \max_{x} k[(p(x) - c)x - F] \cr
\text{Owner:} & \ \max_{k} k[(p^* - c)x^* - F(k)] \cr
& s.t. \ k \ge 0 , 1-k \ge 0
\end{align}\]
The owner first decides how much to pay the manager. The more the owner pays the manager, the lower the fixed costs (I was going for “more pay => better manager”). Once this is decided, the manager steps in to maximize profits. I assume that reducing the fixed costs comes naturally to the manager - it is costless to them - so \(F\) will drop out of the manager’s first-order conditions just like it does for a regular problem with fixed costs.
The solution
The manager’s problem is a standard Cournot monopolist’s problem. The profit-share, \(k\), is irrelevant to the optimal quantity choice because it enters the manager’s problem as a constant scaling the profits.
\[\begin{align}
\text{Manager's FOC:} \ & k[p(x) + x\frac{\partial p}{\partial x} - c] = 0 \cr
\implies & x^* = \frac{a-c}{2b}, \cr
& p^* = \frac{a+c}{2}
\end{align}\]
The owner’s problem is a little trickier because of the restrictions on \(k\). We need to set up the following Lagrangian:
\[L = (1-k)[(p^* - c)x^* - d + k] + \lambda_1(k) + \lambda_2(1-k)\]
Which yields the following saddle-point conditions:
\[\begin{align}
\mathsf{k:} & \ -[(p^* - c)x^* - d + 2k] + \lambda_1 - \lambda_2 = 0 \cr
\mathsf{\lambda_1 :} & \ \lambda_1(k) = 0 \cr
\mathsf{\lambda_2 :} & \ \lambda_2(1-k) = 0 \cr
\end{align}\]
This gives us 3 cases which fully characterize the parameter space.
Case 1: k = 0
If we assume \(k=0\), we get the following saddle point:
\[(k,\lambda_1,\lambda_2) = (0,\frac{(a-c)^2}{4b} - d,0)\]
The owner gives the manager nothing. The lagrange multipliers have to be non-negative, which gives us the parameter restriction: \(d \le \frac{(a-c)^2}{4b}\).
Case 2: k \(\in\) (0,1)
If we assume \(k \in (0,1)\), we get the following saddle point:
\[(k,\lambda_1,\lambda_2) = (\frac{1}{2}(d - \frac{(a-c)^2}{4b}),0,0)\]
The owner gives the manager some positive share and the lagrange multipliers are 0. \(k \gt 0\) gives us the following parameter restriction: \(d \gt \frac{(a-c)^2}{4b}\).
Case 3: k = 1
If we assume \(k=1\), we get the following saddle point:
\[(k,\lambda_1,\lambda_2) = (0,0,d - \frac{(a-c)^2 + 8b}{4b})\]
The owner gives the manager full control of the company. \(\lambda_2 \ge 0\) gives us the following parameter restriction: \(d \ge \frac{(a-c)^2 + 8b}{4b}\).
Summary
The owner and manager both will implement monopoly pricing - there’s only one firm, so this is optimal. If the fixed cost without a manager is greater than the monopoly profit, then the owner will hire the manager to reduce the fixed cost. The owner will pay the manager half the difference between the original fixed cost and the monopoly profit - this is the owner’s profit-maximizing wage for the manager. If the fixed cost is high enough, the owner will give the manager the company because the business isn’t worth it.
Given the relationship between the wage paid to the manager and the fixed cost in this model, we can think of \(k\) not only as the profit-maximizing wage paid to the manager, but also as the optimal level of managerial ability that the owner will choose when it costs profit share.
What could this be a model of? I don’t know. There’s a manager who can make it worthwhile to get into a market where you’ll be a monopolist if you can cross the high fixed cost to entry. Maybe a manager who has connections or knowledge that would reduce the fixed costs?
I think the agency problem involved in non-owners running a firm (so any professional management) is interesting. Eventually, I’d like to construct a model where rational actors cause an agency problem. In this model, the manager does their job (picking the profit-maximizing business decision) perfectly. There is no agency problem because the manager is paid a fraction of profits, so the manager’s wage-maximizing outcome is aligned with the owner’s profit-maximizing desires. The manager is a price-taker in the sense that they simply take the share they are offered as given. This is reasonable if the market for managers is perfectly competitive.
####Variations to explore:
-
What if the manager was paid a share of sales or revenue instead of profits? I’d expect agency problems to be possible here, since the objective functions don’t necessarily align.
-
What if the manager could reduce marginal costs instead of fixed costs? I think of this as a manager who brings persistent efficiencies, whereas this one brings one-time efficiencies. I’d expect qualitatively similar results to this model.
-
What if the firm was taxed (increasing in output) and the manager could reduce the tax? Again I’d expect qualitatively similar results to this model, but it seems like a more plausible model for hiring managers with government connections.
-
O.E. Williamson, in his paper “A Model of Rational Managerial Behavior”, describes a pretty cool model with staffing and reported profits. I may blog about it at some point.
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29 Sep 2015
Models of perfect competition are kinda dull. Firms price at marginal cost, no firm makes any profits, and there’s just not a whole lot to be said there. Imperfect competition is much more interesting. There can be markups, firms can do cool things. In this post, we’re going to look at some simple models of oligopoly (imperfect competition where there are a few firms). I’m going to start with 2 firms, then n firms, and then a couple variations.
There are two popular models of oligopoly: Cournot (firms choose quantities) and Bertrand (firms choose prices). Although the firms’ objective functions are the same (to maximize profits), the two models produce very different outcomes and require different solution techniques.
Cournot models give us a degree of competition, becoming perfectly competitive as the number of firms goes to \(\infty\), and can usually be solved with standard optimization techniques. Bertrand models become perfectly competitive when there is more than one firm involved (!! 2 firms is perfectly competitive??!), and because of discontinuities induced by assumptions on demand, cannot usually be solved by standard optimization techniques. We’ll use game theory to solve both.
Although both of these models pre-date Nash’s solution concept, their equilibria are still Nash equilibria (NE or NEs - in equilibrium, no one can increase their payoff by unilaterally deviating from the equilibrium strategy profile). This is interesting to me because I learned to solve Bertrand models by looking for NEs. I think it would be harder to see this solution without that equilibrium concept specifically in mind, but maybe not.
Cournot competition
Cournot models are nice because we can use a pretty simple optimization program - set up profit, take first order conditions, solve for best-response (BR) functions, and get quantity. The firm’s profit maximization problem is usually unconstrained, which makes it even nicer.
In a Cournot model, each firm chooses a quantity to produce and then they play that quantity at the same time. Because they move simultaneously, they have to guess what the other will do. This gives us their best-response functions. Best-response functions are what they sound like: functions which tell you your best response to your opponent’s move (“If he does this, I should do that”). They are sometimes correspondences instead of functions. We find the optimal quantity by plugging one firm’s best-response into the other’s. The resulting intersection gives us the quantities that the firms will play in equilibrium.
2 symmetric firms
Let the inverse demand function be linear as \(P = a - bQ\), where \(Q = q_i + q_j\), and the common constant marginal cost is \(c\). \(a\), \(b\), and \(c\) are all strictly greater than 0 and \(a>c\). Firm \(i\) solves:
\[\max_{q_i} (p - c)q_i\]
The FOC is:
\[a - 2bq_i - bq_j^* - c = 0\]
Firm \(i\)’s best-response function is:
\[q_i^* = \frac{a - bq_j^* - c}{2b}\]
Since the firms are symmetric, \(q_i = q_j\), giving us firm i’s equilibrium quantity:
\[q_i = \frac{a - c}{3b}\]
Again by symmetry of the firms, we don’t need to solve firm \(j\)’s problem: it’s identical to firm \(i\)’s.
The equilibrium is: \((P, q_i, q_j, \pi_i, \pi_j) = (a-\frac{2}{3}(a-c), \frac{a - c}{3b}, \frac{a - c}{3b}, \frac{(a-c)^2}{9} , \frac{(a-c)^2}{9})\)
\(n\) symmetric firms
The conditions and results are similar for \(n\) symmetric firms. Inverse demand is now \(P = a - b\sum_{i=1}^n q_i\). Firm \(i\) solves:
\[\begin{align}
& \max_{q_i} (p - c)q_i \cr
\text{FOC:} & a - 2bq_i - b\sum_{j \neq i}^n q_j^* - c = 0 \cr
\text{BR:} & q_i^* = \frac{a - (n-1)bq_j - c}{2b} \cr
\implies & q_i = \frac{a-c}{(n+1)b}, \cr
& P = \frac{a}{n+1} + \frac{n}{n+1}c, \cr
& \pi = \frac{(a-c)^2}{(n+1)^2} \cr
\end{align}\]
We can see that the 2 firm example was a special case of the \(n\) firm example. As \(n \to \infty\), \(P \to c\) and \(\pi \to 0\). When \(n = 1\), we get the monopoly outcome, \(q = \frac{a-c}{2b}\), \(P = \frac{a + c}{2}\), and \(\pi = \frac{(a-c)^2}{4}\).
Bertrand competition
In Bertrand models, we assume that the firm with the lower price captures all of the market and that firms split the market evenly when they charge the same price. This creates a discontinuity in profits as a function of price, which means we can’t apply a regular optimization program.
Instead, we have to consider the optimal price in cases, looking for NEs. This means asking the question: assuming the strategy profile we’re considering is an equilibrium, can any player improve their payoff by unilaterally deviating? If yes, then that strategy profile isn’t a Nash equilibrium.
2 symmetric firms
Suppose prices are \(p_i = p_j > c\). Firm \(j\) can deviate to a \(p_j \in (c,p_i)\) \(\forall p_i > c\) and capture all of the demand and increase profits. So \(p_i > c\) can’t be an equilibrium.
We can rule out any case where \(p_i < c\) \(\forall i\), since the firm would then make negative profits and could improve its payoff by deviating to \(p_i = c\) to make 0 profits.
So we’re left with \(p_i = c\). If both firms charge \(c\), neither can improve their payoff by deviating unilaterally. So the equilibrium is that they charge marginal cost and make 0 profits. Deviating can only lead to losses or vicious price competition.
\(n\) symmetric firms
The same argument from the 2 firm case holds for 2 of the \(n\) firms. As a result, all of the other \(n-2\) firms are indifferent between charging \(p_i = c\) and \(p_i > c\). They either split the demand evenly at 0 margin (0 profits), or have a margin and no demand (0 profits).
A couple variations
Now that we know the standard results, let’s look at some variations on the 2-firm models: Cournot competition with asymmetric firms, and Bertrand competition with integer pricing.
Cournot competition: 2 firms with different marginal costs
Same as the earlier 2-firm setup, except now \(c_i > c_j\).
\[\begin{align}
BR_i : & q_i^* = \frac{a - bq_{j}^* - c_i}{2b} \cr
BR_j : & q_j^* = \frac{a - bq_{i}^* - c_j}{2b} \cr
\implies & q_i = \frac{a- 2c_i + c_j}{3b} \cr
& q_j = \frac{a- 2c_j + c_i}{3b} \cr
\implies & q_i < q_j \cr
\end{align}\]
If \(c_i = c_j\), we would be back in the symmetric 2 firm case. Without calculating the profits, we can argue by revealed preference that \(\pi_i < \pi_j\). The argument is straightforward: firm \(j\) could have chosen \(q_i\), but didn’t. Since firms are profit-maximizers, this means \(j\) must make higher profits selling the higher quantity.
Another neat thing we can do here is look at the derivatives of the equilibrium quantities with respect to costs and quantities. This tells us how each firm’s optimal quantity will change with respect to their own and their competitor’s marginal costs. We get:
\[\begin{align}
\frac{\partial q_i}{\partial c_i} & = -\frac{2}{3b} < 0 \cr
\frac{\partial q_i}{\partial c_j} & = \frac{1}{3b} > 0
\end{align}\]
Firms will produce less as their own costs increase, and more as their competitor’s costs increase. Seems intuitive.
Bertrand competition: 2 symmetric firms with integer pricing
The setup is the same as the earlier 2 firm case, except now firms can only charge prices that are integers.
As before, we can rule out any prices below marginal cost, since 0 profits are better than negative profits.
If both firms price at marginal cost, neither firm can deviate an increase their payoff. So $$p_i = p_j = c)\ is one Nash equilibrium.
Is there an equilibrium with positive profits? Let \(d>c\) be the smallest integer greater than \(c\). Suppose both firms price at \(d\). Then deviating downward to \(c\) would yield 0 profits (all demand, no margin), as would deviating upward (some margin, no demand). So there is another Nash equilibrium at \(p_i=p_j=d\).
We could simplify that argument by saying \(d=c+1\), which it is in this case. Doing it this way, we can see that this equilibrium exists whenever prices are discrete. Say prices are discretized into intervals of \(\delta\), replace “smallest integer greater than \(c\)” with “smallest discrete price greater than \(c\)”, and the result follows.
In conclusion
Cournot models may seem less realistic in most settings at face value because it’s more natural (for me at least) to assume firms choose prices rather than quantities. But the results are more plausible: with only a few firms, each firm has some market power and uses it to get profits. As the number of firms increases, each firm’s market power decreases and in the limit we get perfect competition.
Bertrand models, though they accord with my initial ideas of how firms compete, produce less realistic results: only 2 firms is enough to create a perfectly competitive outcome (prices at marginal cost, zero profits). Competition is brutal in Bertrand models: firms force each other to the lowest price possible, only making profits (in the examples we considered) when the space of prices has discontinuities that prevent continuous undercutting.
I find Cournot models easier to think about and solve, since it’s pretty much algorithmic. Bertrand models require a little more thought, but can often be condensed into pretty compact and elegant arguments.
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