<p>The key is the definition of the Nash equilibrium solution concept that you are applying to solve your game. In non-formal terms, a Nash equilibrium is a set of strategies such that no player can increase his or her payoff by deviating to some alternative strategy.</p>

<p>Let's consider $\{30,30,30\}$. In this 'putative' equilibrium, firm A's profit is</p>

<p>$$\frac{D(p)}{3}(p-c)=\frac{20}{3}(30-20)=\frac{200}{3}&gt;0.$$</p>

<p>Could A do better with a lower price? A price of $p=10$ is less than his cost, so this obviously can't be better. Likewise, a price of $p=20$ yields profit</p>

<p>$$D(20)(p-c)=0,$$</p>

<p>which is less than the profit (200/3) in the putative equilibrium. So, again, a deviation to $p_A=20$ is not profitable.</p>

<p>Can A do better by deviating to $p_A&gt;30$? The answer is clearly no because any $p_A&gt;p_B,p_c=30$ implies that A attracts no demand and therefore earns zero profit.</p>

<p>We thus observe that A cannot find a profitable way to deviate from the putative equilibrium $\{30,30,30\}$, so this is indeed a Nash equilibrium.</p>

<hr>

<p>To address your final question: there are two general approaches to problems of this type.</p>

<ol>
<li><p>For each player in the game, compute their <strong>best response</strong> (i.e. profit-maximising strategy) to each possible combination of strategies by the other players. Then look for all situations in which every player is playing a best response; that is a Nash equilibrium.</p></li>
<li><p>Guess a set of strategies that you think might be an equilibrium. Then verify that is is indeed an equilibrium by checking that no player could do better by changing their strategy.</p></li>
</ol>

CC BY-SA 3.0
2018-01-16T18:38:09.640Z
2018-01-18T07:44:01.790Z
2018-01-18T07:44:01.790Z
Ubiquitous
<p>Consider a market for a homogenous product with three producers, firms A, B and
C. The firms have constant marginal costs which are equal to $c = 20$ for each firm. Consumers always buy from the firrm with the lowest price. If the lowest price is set by two (three) firms, half (one third) of the consumers go to each of those firms.</p>

<p>Consumer demand is given by $D(p) = 50 - p$. Each firm can only choose prices from the set $\{10, 20, 30, 40, 50, 60\}$.</p>

<p>Determine all Bertrand-Nash equilibria.</p>

<p><em>Attempt:</em> First, we observe that the monopoly price is 35 (this is easy to determine). Also, profit is zero when the price of any firm is 20, and profit is negative for $p = 10$. I thought that this means that any price-set of the form $\{20, p_B, p_C\}$ would do (and similarly $\{p_A, 20, p_C\}$ and $\{p_A, p_B, 20\}$), as long as all other prices are at least 20. This is not the case, however, since $\{20, 30, 30\}$ will cause all customers to go to firm $A$, resulting in 0 profits for everyone, but if firm $A$ increases price to 30, all firms will make a positive profit, hence $\{20, 30, 30\}$ is not Pareto-efficient. Hence, the only Pareto-efficient price-sets are those where <em>two</em> firms have price 20, and the remaining firm has any price that is 20 or higher.</p>

<p>My mistake was that I missed the $\{30, 30, 30\}$ equilibrium. But how could I have seen this? I always thought that if all companies have constant and identical marginal costs (in this case 20), the only Nash-equilibrium is the one in which all companies have their price equal to the marginal costs. Apparently, this is not the case here. </p>

<p>What am I missing? What is the general approach to problems of this type? Any help is greatly appreciated.</p>

<nash-equilibrium><pareto-efficiency>
Bertrand-equilibrium with discrete price set
TheLimitIntegral
# Bertrand-equilibrium with discrete price set

(Asked by: username_0 on 2018-01-16)

Consider a market for a homogenous product with three producers, firms A, B and
C. The firms have constant marginal costs which are equal to $c = 20$ for each firm. Consumers always buy from the firrm with the lowest price. If the lowest price is set by two (three) firms, half (one third) of the consumers go to each of those firms.

Consumer demand is given by $D(p) = 50 - p$. Each firm can only choose prices from the set $\{10, 20, 30, 40, 50, 60\}$.

Determine all Bertrand-Nash equilibria.

*Attempt:* First, we observe that the monopoly price is 35 (this is easy to determine). Also, profit is zero when the price of any firm is 20, and profit is negative for $p = 10$. I thought that this means that any price-set of the form $\{20, p\_B, p\_C\}$ would do (and similarly $\{p\_A, 20, p\_C\}$ and $\{p\_A, p\_B, 20\}$), as long as all other prices are at least 20. This is not the case, however, since $\{20, 30, 30\}$ will cause all customers to go to firm $A$, resulting in 0 profits for everyone, but if firm $A$ increases price to 30, all firms will make a positive profit, hence $\{20, 30, 30\}$ is not Pareto-efficient. Hence, the only Pareto-efficient price-sets are those where *two* firms have price 20, and the remaining firm has any price that is 20 or higher.

My mistake was that I missed the $\{30, 30, 30\}$ equilibrium. But how could I have seen this? I always thought that if all companies have constant and identical marginal costs (in this case 20), the only Nash-equilibrium is the one in which all companies have their price equal to the marginal costs. Apparently, this is not the case here.

What am I missing? What is the general approach to problems of this type? Any help is greatly appreciated.

**Tags:** nash-equilibrium, pareto-efficiency

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## 1 answers

### Answer by username_1 on 2018-01-16. Score: 4

The key is the definition of the Nash equilibrium solution concept that you are applying to solve your game. In non-formal terms, a Nash equilibrium is a set of strategies such that no player can increase his or her payoff by deviating to some alternative strategy.

Let's consider $\{30,30,30\}$. In this 'putative' equilibrium, firm A's profit is

$$\frac{D(p)}{3}(p-c)=\frac{20}{3}(30-20)=\frac{200}{3}>0.$$

Could A do better with a lower price? A price of $p=10$ is less than his cost, so this obviously can't be better. Likewise, a price of $p=20$ yields profit

$$D(20)(p-c)=0,$$

which is less than the profit (200/3) in the putative equilibrium. So, again, a deviation to $p\_A=20$ is not profitable.

Can A do better by deviating to $p\_A>30$? The answer is clearly no because any $p\_A>p\_B,p\_c=30$ implies that A attracts no demand and therefore earns zero profit.

We thus observe that A cannot find a profitable way to deviate from the putative equilibrium $\{30,30,30\}$, so this is indeed a Nash equilibrium.

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To address your final question: there are two general approaches to problems of this type.

1. For each player in the game, compute their **best response** (i.e. profit-maximising strategy) to each possible combination of strategies by the other players. Then look for all situations in which every player is playing a best response; that is a Nash equilibrium.
2. Guess a set of strategies that you think might be an equilibrium. Then verify that is is indeed an equilibrium by checking that no player could do better by changing their strategy.


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