<p>The problem formulation admits the following Normal Form representation. We can reject any strategy involving price greater than 2, as demand falls to zero and such strategies are strictly dominated by those for which prices are either 1 or 2.</p>

<pre><code>    0         1         2

0 [0,0]     [0,0]     [0,0]    

1 [0,0]   [0.5,0.5]   [1,0]

2 [0,0]     [0,1]     [1,1]
</code></pre>

<p>With this setup you can see that there are three Nash equilibria: (0,0), (1,1) and (2,2). At (0,0), firms are indifferent to the outcome of a higher price, and will not deviate. At (1,1), Firm 1 will not deviate to 2, nor will Firm 2 deviate to 2, as this cedes the market to the opponent. At (2,2) firms are indifferent to reducing price to 1.  </p>

<p>(0,1), (0,2), (1,0), and (2,0) are strictly dominated by deviating to a price-matching strategy, splitting the market between firms.</p>

<p>Similarly, (1,2) and (2,1) fail to be Nash equilibria because whichever firm opens with a price of 2 can capture half the market by deviating to a price of 1.</p>

CC BY-SA 4.0
2019-09-16T19:01:44.013Z
What's wrong with weak dominance, why are you leaving out 0?
Giskard
CC BY-SA 4.0
2019-09-16T19:17:48.073Z
I edited my answer.  I initially left it out because I thought the result wasn't interesting - but then your comment helped me realize I was leaving out the classic Bertrand equilibrium result.
heh
CC BY-SA 4.0
2019-09-16T19:40:36.043Z
I think 1 is weakly dominating 2 (not the other way around), but otherwise you are spot on. Perhaps you mean 2 is weakly dominated?
Giskard
CC BY-SA 4.0
2019-09-16T19:59:09.970Z
Yep - that is indeed what I meant.  I literally joined and started posting because my game theory has been getting rusty - I had no idea by how much!
heh
CC BY-SA 4.0
2019-09-16T22:28:14.557Z
Pricing above 2 is only *weakly*, not strictly, dominated, since profit is zero if the other firm prices at zero, regardless of what your price is.
Herr K.
CC BY-SA 4.0
2019-09-17T14:33:51.207Z
I suppose you could add 3 as a choice to be robust, but any price above 3 is clearly degenerate, and unlike allowing for 0 - which can be justified by a desire to leave room for the Bertrand outcome - it is hard to see why either firm would price itself out of the market in this game.
heh
CC BY-SA 4.0
2019-09-16T18:52:54.270Z
2019-09-16T19:58:09.973Z
2019-09-16T19:58:09.973Z
heh
<p>Firstly, at reservation prices, it's not compulsory that demand will be 0. The consumers will be indifferent between buying and not buying the good.
Let q* be the demand by the mass when price is 2 or less.
If you take out the profit function of both the firms,  you will see that at p1=2, profit received by firm 2 will be the same if P2=1,2. Because costs are 0, it will share the market at p2=2 and gain profit equal to q* and at p2= 1 it will gain the entire market and make profit equal to q* again. 
When p1=1, firm 2's best response is p2=1. When p1=0, firm 2's best response is p2 is a member of N, where N is a natural number (0 is a natural number). This is because firm 2 will earn profits 0 for any price. The best response of firm 1 will be analogous because cost functions are the same. Therefore Nash Equilibriums are {(0,0),(1,1),(2,2)}.
For the reservation price P1=2, suppose the demand is 0, then the best response by firm 2 will be p2 is a member of N again, because it will earn 0 profits at all N. Also, when p1>2, best response of firm 2 will be p2=2, but there will be no intersection of best responses beyond (2,2).</p>

CC BY-SA 4.0
2019-09-16T18:53:39.357Z
Why do you think demand will be the same $q^*$ when price is 2 *or less*, given that consumers have inelastic unit demand?
Herr K.
CC BY-SA 4.0
2019-09-17T04:06:43.397Z
You're right, I mistakenly took it for perfectly inelastic demand under 2.
ZlatanIbrahimovic
CC BY-SA 4.0
2019-09-21T19:33:05.200Z
@Herr K. Don't you think that inelastic unit demand at reservation price 2 means that the consumers will demand 1 unit at price less than 2 and 0 at prices more than 2? Isn't that why 2,2 is also a Nash equilibrium because the firm will get same profits at p1=P2=2 and p1=2, P2=1?
ZlatanIbrahimovic
CC BY-SA 4.0
2019-09-21T21:29:27.240Z
Yes, (2,2) is a NE if we assume that *all* consumers (i.e. the entire unit mass of them) will purchase at price equal to 2. But when you use $q^*$, I thought you meant any $q^*\in[0,1]$, which is okay when consumers are indifferent when price is 2, but when price is 1 or 0, all consumers should make the purchase.
Herr K.
CC BY-SA 4.0
2019-09-16T16:43:34.633Z
2019-09-16T16:43:34.633Z
ZlatanIbrahimovic
<ul>
<li><em>There are two identical firms, <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, with zero marginal costs. They produce homogenous product, which is demanded by a unit mass of identical consumers, each of which has inelastic unit demand with a reservation price of <span class="math-container">$2$</span>. Prices are constrained to take only integer values. Using standard game theoretic reasoning, determine whether or not each possible pair <span class="math-container">$(p_1,p_2)$</span> can be regarded as Bertrand Nash equilibrium.</em></li>
</ul>

<p>Would the Nash equilibria be <span class="math-container">$(p_{1}^{*},p_{2}^{*})=$</span> <span class="math-container">$\{(0,0),(1,1)\}$</span>. 
For symmetric prices, <span class="math-container">$\{(2,2)\}$</span> since the demand would be <span class="math-container">$0$</span>, I can see how unilateral deviation to <span class="math-container">$1$</span> could increase profits for a firm. 
However for prices <span class="math-container">$\{(0,0),(1,1)\}$</span>, no unilateral profits are possible. For other non symmetric price pairings, unilateral deviations are possible. Therefore, the Nash equilirium prices are <span class="math-container">$\{(0,0),(1,1)\}$</span>. 
Is my reasoning correct? </p>

<microeconomics><game-theory><self-study><nash-equilibrium><industrial-organisation>
Bertrand Duopoly Equilibrium for Discrete Prices
S.Rana
# Bertrand Duopoly Equilibrium for Discrete Prices

(Asked by: username_0 on 2019-09-16)

* *There are two identical firms, $1$ and $2$, with zero marginal costs. They produce homogenous product, which is demanded by a unit mass of identical consumers, each of which has inelastic unit demand with a reservation price of $2$. Prices are constrained to take only integer values. Using standard game theoretic reasoning, determine whether or not each possible pair $(p\_1,p\_2)$ can be regarded as Bertrand Nash equilibrium.*

Would the Nash equilibria be $(p\_{1}^{\*},p\_{2}^{\*})=$ $\{(0,0),(1,1)\}$.
For symmetric prices, $\{(2,2)\}$ since the demand would be $0$, I can see how unilateral deviation to $1$ could increase profits for a firm.
However for prices $\{(0,0),(1,1)\}$, no unilateral profits are possible. For other non symmetric price pairings, unilateral deviations are possible. Therefore, the Nash equilirium prices are $\{(0,0),(1,1)\}$.
Is my reasoning correct?

**Tags:** microeconomics, game-theory, self-study, nash-equilibrium, industrial-organisation


---

## 2 answers

### Answer by username_1 on 2019-09-16. Score: 1 | Accepted answer

The problem formulation admits the following Normal Form representation. We can reject any strategy involving price greater than 2, as demand falls to zero and such strategies are strictly dominated by those for which prices are either 1 or 2.

```
    0         1         2

0 [0,0]     [0,0]     [0,0]    

1 [0,0]   [0.5,0.5]   [1,0]

2 [0,0]     [0,1]     [1,1]

```

With this setup you can see that there are three Nash equilibria: (0,0), (1,1) and (2,2). At (0,0), firms are indifferent to the outcome of a higher price, and will not deviate. At (1,1), Firm 1 will not deviate to 2, nor will Firm 2 deviate to 2, as this cedes the market to the opponent. At (2,2) firms are indifferent to reducing price to 1.

(0,1), (0,2), (1,0), and (2,0) are strictly dominated by deviating to a price-matching strategy, splitting the market between firms.

Similarly, (1,2) and (2,1) fail to be Nash equilibria because whichever firm opens with a price of 2 can capture half the market by deviating to a price of 1.

> **Comments:**
> **username_2**: What's wrong with weak dominance, why are you leaving out 0?  
> **username_1**: I edited my answer.  I initially left it out because I thought the result wasn't interesting - but then your comment helped me realize I was leaving out the classic Bertrand equilibrium result.  
> **username_2**: I think 1 is weakly dominating 2 (not the other way around), but otherwise you are spot on. Perhaps you mean 2 is weakly dominated?  
> **username_1**: Yep - that is indeed what I meant.  I literally joined and started posting because my game theory has been getting rusty - I had no idea by how much!  
> **username_3**: Pricing above 2 is only *weakly*, not strictly, dominated, since profit is zero if the other firm prices at zero, regardless of what your price is.  
> **username_1**: I suppose you could add 3 as a choice to be robust, but any price above 3 is clearly degenerate, and unlike allowing for 0 - which can be justified by a desire to leave room for the Bertrand outcome - it is hard to see why either firm would price itself out of the market in this game.  

---

### Answer by username_4 on 2019-09-16. Score: 0

Firstly, at reservation prices, it's not compulsory that demand will be 0. The consumers will be indifferent between buying and not buying the good.
Let q\* be the demand by the mass when price is 2 or less.
If you take out the profit function of both the firms, you will see that at p1=2, profit received by firm 2 will be the same if P2=1,2. Because costs are 0, it will share the market at p2=2 and gain profit equal to q\* and at p2= 1 it will gain the entire market and make profit equal to q\* again.
When p1=1, firm 2's best response is p2=1. When p1=0, firm 2's best response is p2 is a member of N, where N is a natural number (0 is a natural number). This is because firm 2 will earn profits 0 for any price. The best response of firm 1 will be analogous because cost functions are the same. Therefore Nash Equilibriums are {(0,0),(1,1),(2,2)}.
For the reservation price P1=2, suppose the demand is 0, then the best response by firm 2 will be p2 is a member of N again, because it will earn 0 profits at all N. Also, when p1>2, best response of firm 2 will be p2=2, but there will be no intersection of best responses beyond (2,2).

> **Comments:**
> **username_3**: Why do you think demand will be the same $q^*$ when price is 2 *or less*, given that consumers have inelastic unit demand?  
> **username_4**: You're right, I mistakenly took it for perfectly inelastic demand under 2.  
> **username_4**: @username_3. Don't you think that inelastic unit demand at reservation price 2 means that the consumers will demand 1 unit at price less than 2 and 0 at prices more than 2? Isn't that why 2,2 is also a Nash equilibrium because the firm will get same profits at p1=P2=2 and p1=2, P2=1?  
> **username_3**: Yes, (2,2) is a NE if we assume that *all* consumers (i.e. the entire unit mass of them) will purchase at price equal to 2. But when you use $q^*$, I thought you meant any $q^*\in[0,1]$, which is okay when consumers are indifferent when price is 2, but when price is 1 or 0, all consumers should make the purchase.  

---

