<p>Mary's marginal cost is zero so she can sell at that price when Jim enters the market. Thus Jim needs to have deep pockets to survive a price war.If both Jim and Mary's financial resources and opportunity cost (i.e what they can earn elsewhere) are 'common knowledge' then, in an indefinitely repeated game, there is a 'Muth Rational' solution such that price is set where the demand curve has elasticity of one- i.e. Marginal Revenue is zero- and the market is shared according to a formula which takes account of what resources each can commit and what opportunity cost this entails. In your example it appears Jim has 'money lying around'- i.e. no opportunity cost as such while Mary has sunk costs only and no recurring or depreciation costs. Jim may as well produce widgets. If there is no cost of disposing of widgets and assuming both Jim and Mary are rational then the Muth rational solution obtains. However it is not stable, one defection is likely to collapse the market. However, the defector realizes that their greed destroyed future profit so they will set the Muth rational price twice- allowing the other to recoup lost profit- after which the Muth rational price obtains till another defection occurs.
It is rational to enter this market for Jim because he seems to have zero opportunity cost for the investment and there is a positive expected value for the investment- provided common knowledge and Muth rationality obtain.</p>

CC BY-SA 3.0
2015-09-25T07:15:46.523Z
2015-09-25T07:22:48.053Z
2015-09-25T07:22:48.053Z
Vivek Iyer
<p><strong>Answer to question</strong></p>

<p>If we take your assumptions literally, Jim will decide not to enter the widget business. For suppose he did incur the cost of entry and that Mary is selling at price $p_m$. Jim can only sell to consumers if his price $p_j\leq p_m$. The best price for Jim is $p_m-\epsilon$ (where $\epsilon$ is some very small, positive amount). But this would leave Mary with no sales, so she will have an incentive to reduce her price to $p_j-\epsilon$. This is the price war you describe and it will result in both parties reducing their price to marginal cost (i.e. zero—this is known as <a href="https://en.wikipedia.org/wiki/Bertrand_competition" rel="nofollow">Bertrand competition</a>).</p>

<p>Since there are no frictions in your model, this will all happen very quickly. Moreover, because a price of zero implies zero profits, Jim has no incentive to incur the cost of entry in the first place, so he will instead opt to stay out of the market.</p>

<hr>

<p><strong>Effect of relaxing assumptions</strong></p>

<p>This is obviously quite a stylised result owing to the stark assumptions of the model. But it makes a good foundation for thinking about some more realistic settings. For example:</p>

<ul>
<li><p>If Mary's factory has a marginal cost $c_m$ and Jim has a patent on a new technology that implies cost $c_j&lt;c_m$ then Jim can profitably set a price below the lowest price Mary is willing to set. Good news: if someone invents a more efficient technology they can enter the market and displace the older, less efficient technology.</p></li>
<li><p>Suppose that the factory has a monthly fixed maintenance cost. If Mary is a small, credit-constrained independent firm and Jim is a huge conglomerate with large reserves of cash then Jim can enter and practice <a href="https://en.wikipedia.org/wiki/Predatory_pricing" rel="nofollow">predatory pricing</a>. If he is patient enough, he can enter the market, set $p_j=0$, and wait for Mary to run out of money. She will then leave the market and Jim becomes a profitable monopoly. This kind of anti-competitive behaviour is illegal in most jurisdictions (yes, there are laws against setting too low a price!).</p></li>
<li><p>Suppose that Mary's widgets are blue and Jim's are pink. Consumers have an idiosyncratic preference for either blue or pink widgets. Then both firms can set a positive price and sell to the consumers who like their colour more. The more are products <em>differentiated</em> the more there is scope for both firms to exist profitably in the industry and Jim may find it worthwhile to enter. This is why firms talk so much about differentiation and unique selling points. There are various ways to model this in economics. <a href="http://www.people.fas.harvard.edu/~zlai/teaching/AppliedIO_Note3_HotellingSalop.pdf" rel="nofollow">Here is an example</a>.</p></li>
</ul>

<hr>

<p><strong>More formally, $p_j=p_m=0$ is the unique pure strategy Nash equilibrium</strong></p>

<p>To be slightly more formal, let us check that $p_m=p_j=0$ is indeed a (Nash) equilibrium of the subgame in which both firms simultaneously set $p\in\mathbb{R}$ with the lowest priced firm capturing the entire (finite) demand. Neither party can profit by deviating to $p&lt;0$ as this yields negative profits. A deviation to any $p&gt;0$ results in the rival firm having a lower price so demand (and profits) are zero---again not profitable. Thus, there is no profitable deviation and $p_i=p_m=0$ is <em>an</em> equilibrium.</p>

<p>Is there another pure strategy equilibrium with some $p&gt;0$? The answer is no. Consider the three possibilities and observe how a profitable deviation can be constructed for each:</p>

<ul>
<li>$p_m&gt;p_j$. In this case $j$ could increase his price to some $p_j'\in(p_j,p_m)$ without losing any demand. Such a $p_j'$ exists by the connectedness of the real line.</li>
<li>$p_m=p_j=p$. In this case at least one of the firms must be capturing less than all of the consumers. But by deviating to $p'=p-\epsilon$ it can capture all consumers. If we write $D$ for demand before the deviation and $D'&gt;D$ for demand afterwards then the change in profits is $D'(p-\epsilon)-Dp$. By the connectedness of the real line, there exists an $\epsilon$ sufficiently small that this is positive.</li>
<li>$p_m&lt;p_j$. This case is symmetric to $p_m&gt;p_j$.</li>
</ul>

<p>Thus, the only pure strategy equilibrium of this one-shot game is $p_j=p_m=0$.</p>

<hr>

<p><strong>If we repeat the game then we can sustain other (collusive) equilibria</strong></p>

<p>What if we repeat the game infinitely many times? Suppose the two firms have an implicit understanding (an explicit agreement would be illegal) that they will both set $p=p^*$ for some $p^*&gt;0$. Moreover, it is understood that if one firm deviates from this behaviour today, then both firms will revert to playing the static equilibrium ($p_j=p_m=0$) forever. Firms discount the future at rate $\delta$. For simplicity, suppose that demand is constant: $D(p)=1$ (not crucial).</p>

<p>If a firm plays in accordance with this understanding (and expects its rival to do likewise) then its profit is</p>

<p>$$\sum_{t=0}^\infty\frac{1}{2}\delta^t p^*=\frac{p^*}{2(1-\delta)}.$$</p>

<p>If a firm cheats and sets $p^*-\epsilon$ ($\epsilon$ small) then it captures the whole demand today, but expects to be punished forever thereafter, so profit is $p^*$. Thus, both firms wish to comply with their implicit understanding if
$$\frac{p^*}{2(1-\delta)}&gt;p^*\iff \delta&gt;\frac{1}{2}$$
(i.e. if they are sufficiently patient).</p>

<p>So, if we repeat the game and firms are very patient, <em>we can sustain any price in equilibrium</em>. But note that these equilibria require not only that firms are patient, but also that a) there is no threat of future entry that could destabilise the equilibrium; b) firms are able to coordinate on a $p^*$ without ever making an (illegal) explicit agreement; and c) firms are able to constantly and accurately monitor rivals behaviour in order to detect cheating.</p>

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2015-09-25T10:37:43.423Z
I picked the assumptions I picked because I was covertly talking about the pharmaceutical industry. There was a story recently about a firm realizing that they are in Mary's position and they just raised the price of their generic malaria resistance drug from 15 to 700 dollars a pill.
ZKe
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2015-09-25T18:30:08.497Z
@Zackkenyon Yes, I think in the particular case you are referring to the assumptions make a lot of sense. Indeed, it was not my intention to criticise your assumptions. But I find that thinking about what happens when we tweak the important assumptions of a model is very useful for understanding how robust and general the model's conclusions are.
Ubiquitous
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2015-09-27T01:49:13.230Z
So, I guess before I accept this answer, I would need you to show that the series of epsilons does, in fact, converge to $p_0-marginal cost$. Seems like they could easily set the same price in a prisoner's dilemma situation.
ZKe
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2015-09-27T12:48:41.683Z
@Zackkenyon I have added two sections. The first provides a more formal argument that the only pure strategy Nash equilibrium of the price-setting subgame is the zero price equilibrium. The second show how, with a bit of trickery, one can get other equilibria when the game is repeated (and explains why these equilibria might not be empirically plausible).
Ubiquitous
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2015-09-28T19:35:00.733Z
sorry I didn't get back to you earlier, but this argument is wrong. There is no non-iterative version of this game. Nash equilibrium is at setting your price exactly equal to your opponents at every "time step". You cannot do better against that strategy than to adopt exactly the same strategy.
ZKe
CC BY-SA 3.0
2015-09-28T20:18:00.457Z
@Zackkenyon No, I'm sorry but that's wrong. Here's a numerical example. Let market demand be $D(p)=100-p$. Suppose your strategy is to set price 50 in every period. Your claim is that the best I can do is to also set my price at 50. If I do that then my profit in each period is $50D(p)/2=[(100-50)50]/2=1250$ (division by 2 because if we set the same price then we have to share the market). But suppose I set a price of 49 in every period. Since my price is lower, I get the entire demand and my profit is $(100-49)49=2499$ per period. Higher profit, so your proposed play is not equilibrium
Ubiquitous
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2015-09-29T21:28:55.957Z
yes, and if you always choose rock, I will always choose paper... I'm not sure we are agreeing on the definition of Equilibrium here. my strategy does not have to be the best against ALL strategies in order to be a Nash equilibrium strategy, it just must be better against ITSELF than all other strategies. In that case it's very clear that when you are playing against an opponent who always sets their price value exactly equal to yours at every time step, the way to maximize your profit is to do the same... This is a very fundamental misunderstanding that I feel the need to resolve.
ZKe
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2015-09-29T23:02:52.040Z
I updated the question somewhat for clarity.
ZKe
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2015-09-25T07:06:06.400Z
2015-09-30T19:56:42.010Z
2015-09-30T19:56:42.010Z
Ubiquitous
<p>Mary is making a hefty profit manufacturing and selling widgets. Jim has some money laying around and he is trying to figure out if he shouldn't start manufacturing some widgets too. </p>

<p>In this example assume that the marginal cost to produce a widget is zero (time, money, etc are all close enough to zero to be indistinguishable), but the cost of making a widget manufacturing plant is quite high. </p>

<p>Also assume that the market for widgets is fairly centralized. There are two bins with widgets in them and consumers can purchase their widgets from either bin. They cannot be resold, however. There are laws against that for whatever reason, and they are ruthlessly enforced. You may only sell widgets that you manufacture yourself.</p>

<p>What does Jim decide? And if he decides to enter the widget business, what is the eventual price per widget and how long does it take to reach that price?</p>

<p><strong>More (but not too) formally:</strong> 
Our situation can be modeled by a few different games,</p>

<p>Case 1: assume the rules of the price war are</p>

<p>players take turns setting a new, lower price from $\mathbb R$ or passing. When both players pass, the prices are locked and the market is allowed to run for t.</p>

<p>in this case the optimal strategy to set your price to whatever price your opponent sets, if the prices are already equal, then pass.</p>

<p>Case 2: </p>

<p>The market is allowed to run for time t after each player sets a (lesser or equal) price, they take turns setting prices, and they choose prices from $\Bbb Z$.</p>

<p>Here the optimal strategy is in fact no different. if the price starts at $p1$, and the sharing strategy is $S$, then there exists some $t$ such that for all $t'$>$t$ $$EV(S,S,t)=\frac{t*p}{2} &gt;tk(p-1) &gt; EV(S,S',t)$$ for all $p,k,S'$</p>

<p>Case 3: </p>

<p>Both players set prices in each timestep t without the information of the price their opponent sets in that time step, and the market is run. </p>

<p>In this case, no pure strategy is guaranteed to exist, because the information is not perfect. I don't know what the Nash equilibrium would end up being, nor do I know that there is any reason to suspect that it converges (as t -> 0) to the same price as the previous two cases, despite the game seeming to do so.  </p>

<p>So I guess the question becomes, why does competition even exist at all in this market, and since all of these scenarios seem to converge towards the reality. ?</p>

CC BY-SA 3.0
2015-09-30T10:51:58.863Z
The answer I provided below gives the (correct) Nash equilibrium solution to the Bertrand (1883) game of price competition (the standard way to analyse oligopoly price competition). If you don't think this solution addresses your question then it must be either that (a) you want some equilibrium concept other than Nash equilibrium, or (b) you want some game other than the Bertrand game. But if you have a specific game/solution concept in mind then you need to provide a complete and precise description of them (strategies, payoffs, timing, and solution concept) so we can write a solution.
Ubiquitous
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2015-09-30T16:55:48.637Z
@ubiquitous The information in the bertrand model is not perfect, and you have failed to prove that there exist no mixed strategy equilibria, which are obviously relevant to my model. If the game requires that the players take turns, then equilibrium is defined at the cooperation point, so simply removing the assumption that the players move simultaneously will not suffice. I hope my argument is clear?
ZKe
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2015-09-30T18:48:02.950Z
I think I start to understand the game you have in mind. I created a new question (http://economics.stackexchange.com/questions/8473/dynamic-bertrand-competition-when-players-take-turns) with a formalised description of the game (as I understand it). I think having a separate question with a precise specification of the problem might attract more precisely targeted set of answers.
Ubiquitous
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2015-09-30T18:50:18.883Z
@ubiquitous, I still don't think you have proven the non-existence of non-trivial mixed strategy equilibria for the Bertrand model.
ZKe
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2015-09-30T18:53:30.923Z
I didn't because there *can* be a mixed strategy equilibrium. But Kapplan and Wettstein (2000) showed that a mixed strategy equilibrium of the Bertrand game exists only if revenue can be infiinite (which is not empirically plausible). The paper is here: http://people.exeter.ac.uk/trkaplan/papers/serkaplan.pdf
Ubiquitous
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2015-09-30T18:54:39.293Z
See Klemperer (2001) for a more succinct argument for why there can be a mixed strategy equilibium: http://www.nuff.ox.ac.uk/users/klemperer/WhyEveryEconomist.pdf
Ubiquitous
CC BY-SA 3.0
2015-09-30T18:55:54.097Z
@ubiquitous I have not read the paper yet, but I will conjecture that it relies on the fact that the marginal cost of production is not zero, which it *is* in my model.
ZKe
CC BY-SA 3.0
2015-09-30T18:58:27.077Z
No, it does not rely on that assumption. Note that the condition is that revenues have to be unbounded, which has nothing to do with costs. For revenue to be unbounded, we would need a very special demand function (such as constant-elasticity demand) for which demand remains positive even as prices become infinite.
Ubiquitous
CC BY-SA 3.0
2015-09-30T19:01:37.217Z
In fact, as early as section two we reach the statement "There may exist a mixed-strategy equilibrium of this model where prices announced
always exceed marginal cost." Which is later strengthened to exceed or equal to. Since it is an assumption of my model that marginal cost is zero, this statement holds, and does *not* depend on the infinite revenue scenario.
ZKe
CC BY-SA 3.0
2015-09-30T19:05:17.543Z
You have misunderstood. The statement that you quoted only says that the equilibrium they identify is such that $p>c$ (and does not exclude the $c=0$ case). It does not say anything about when that equilibrium exists. The relevant condition for existence is on p69: "There exist equilibria other than marginal-cost pricing if and only if $\lim_{p\rightarrow\infty} pD(p) = \infty$". Exactly the infinite revenue result I mentioned.
Ubiquitous
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2015-09-30T19:28:34.833Z
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/29731/discussion-between-zackkenyon-and-ubiquitous).
ZKe
<game-theory><elasticity><price-theory><competitive-equilibrium>
What determines the outcome of a price war, and why isn't that outcome reached instantaneously?
ZKe
# What determines the outcome of a price war, and why isn't that outcome reached instantaneously?

(Asked by: username_0 on 2015-09-25)

Mary is making a hefty profit manufacturing and selling widgets. Jim has some money laying around and he is trying to figure out if he shouldn't start manufacturing some widgets too.

In this example assume that the marginal cost to produce a widget is zero (time, money, etc are all close enough to zero to be indistinguishable), but the cost of making a widget manufacturing plant is quite high.

Also assume that the market for widgets is fairly centralized. There are two bins with widgets in them and consumers can purchase their widgets from either bin. They cannot be resold, however. There are laws against that for whatever reason, and they are ruthlessly enforced. You may only sell widgets that you manufacture yourself.

What does Jim decide? And if he decides to enter the widget business, what is the eventual price per widget and how long does it take to reach that price?

**More (but not too) formally:**
Our situation can be modeled by a few different games,

Case 1: assume the rules of the price war are

players take turns setting a new, lower price from $\mathbb R$ or passing. When both players pass, the prices are locked and the market is allowed to run for t.

in this case the optimal strategy to set your price to whatever price your opponent sets, if the prices are already equal, then pass.

Case 2:

The market is allowed to run for time t after each player sets a (lesser or equal) price, they take turns setting prices, and they choose prices from $\Bbb Z$.

Here the optimal strategy is in fact no different. if the price starts at $p1$, and the sharing strategy is $S$, then there exists some $t$ such that for all $t'$>$t$ $$EV(S,S,t)=\frac{t\*p}{2} >tk(p-1) > EV(S,S',t)$$ for all $p,k,S'$

Case 3:

Both players set prices in each timestep t without the information of the price their opponent sets in that time step, and the market is run.

In this case, no pure strategy is guaranteed to exist, because the information is not perfect. I don't know what the Nash equilibrium would end up being, nor do I know that there is any reason to suspect that it converges (as t -> 0) to the same price as the previous two cases, despite the game seeming to do so.

So I guess the question becomes, why does competition even exist at all in this market, and since all of these scenarios seem to converge towards the reality. ?

**Tags:** game-theory, elasticity, price-theory, competitive-equilibrium

> **Comments:**
> **username_1**: The answer I provided below gives the (correct) Nash equilibrium solution to the Bertrand (1883) game of price competition (the standard way to analyse oligopoly price competition). If you don't think this solution addresses your question then it must be either that (a) you want some equilibrium concept other than Nash equilibrium, or (b) you want some game other than the Bertrand game. But if you have a specific game/solution concept in mind then you need to provide a complete and precise description of them (strategies, payoffs, timing, and solution concept) so we can write a solution.  
> **username_0**: @ubiquitous The information in the bertrand model is not perfect, and you have failed to prove that there exist no mixed strategy equilibria, which are obviously relevant to my model. If the game requires that the players take turns, then equilibrium is defined at the cooperation point, so simply removing the assumption that the players move simultaneously will not suffice. I hope my argument is clear?  
> **username_1**: I think I start to understand the game you have in mind. I created a new question (http://economics.stackexchange.com/questions/8473/dynamic-bertrand-competition-when-players-take-turns) with a formalised description of the game (as I understand it). I think having a separate question with a precise specification of the problem might attract more precisely targeted set of answers.  
> **username_0**: @ubiquitous, I still don't think you have proven the non-existence of non-trivial mixed strategy equilibria for the Bertrand model.  
> **username_1**: I didn't because there *can* be a mixed strategy equilibrium. But Kapplan and Wettstein (2000) showed that a mixed strategy equilibrium of the Bertrand game exists only if revenue can be infiinite (which is not empirically plausible). The paper is here: http://people.exeter.ac.uk/trkaplan/papers/serkaplan.pdf  
> **username_1**: See Klemperer (2001) for a more succinct argument for why there can be a mixed strategy equilibium: http://www.nuff.ox.ac.uk/users/klemperer/WhyEveryEconomist.pdf  
> **username_0**: @ubiquitous I have not read the paper yet, but I will conjecture that it relies on the fact that the marginal cost of production is not zero, which it *is* in my model.  
> **username_1**: No, it does not rely on that assumption. Note that the condition is that revenues have to be unbounded, which has nothing to do with costs. For revenue to be unbounded, we would need a very special demand function (such as constant-elasticity demand) for which demand remains positive even as prices become infinite.  
> **username_0**: In fact, as early as section two we reach the statement "There may exist a mixed-strategy equilibrium of this model where prices announced always exceed marginal cost." Which is later strengthened to exceed or equal to. Since it is an assumption of my model that marginal cost is zero, this statement holds, and does *not* depend on the infinite revenue scenario.  
> **username_1**: You have misunderstood. The statement that you quoted only says that the equilibrium they identify is such that $p>c$ (and does not exclude the $c=0$ case). It does not say anything about when that equilibrium exists. The relevant condition for existence is on p69: "There exist equilibria other than marginal-cost pricing if and only if $\lim_{p\rightarrow\infty} pD(p) = \infty$". Exactly the infinite revenue result I mentioned.  
> **username_0**: Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/29731/discussion-between-zackkenyon-and-ubiquitous).  

---

## 2 answers

### Answer by username_1 on 2015-09-25. Score: 7 | Accepted answer

**Answer to question**

If we take your assumptions literally, Jim will decide not to enter the widget business. For suppose he did incur the cost of entry and that Mary is selling at price $p\_m$. Jim can only sell to consumers if his price $p\_j\leq p\_m$. The best price for Jim is $p\_m-\epsilon$ (where $\epsilon$ is some very small, positive amount). But this would leave Mary with no sales, so she will have an incentive to reduce her price to $p\_j-\epsilon$. This is the price war you describe and it will result in both parties reducing their price to marginal cost (i.e. zero—this is known as [Bertrand competition](https://en.wikipedia.org/wiki/Bertrand_competition)).

Since there are no frictions in your model, this will all happen very quickly. Moreover, because a price of zero implies zero profits, Jim has no incentive to incur the cost of entry in the first place, so he will instead opt to stay out of the market.

---

**Effect of relaxing assumptions**

This is obviously quite a stylised result owing to the stark assumptions of the model. But it makes a good foundation for thinking about some more realistic settings. For example:

* If Mary's factory has a marginal cost $c\_m$ and Jim has a patent on a new technology that implies cost $c\_j
* Suppose that the factory has a monthly fixed maintenance cost. If Mary is a small, credit-constrained independent firm and Jim is a huge conglomerate with large reserves of cash then Jim can enter and practice [predatory pricing](https://en.wikipedia.org/wiki/Predatory_pricing). If he is patient enough, he can enter the market, set $p\_j=0$, and wait for Mary to run out of money. She will then leave the market and Jim becomes a profitable monopoly. This kind of anti-competitive behaviour is illegal in most jurisdictions (yes, there are laws against setting too low a price!).
* Suppose that Mary's widgets are blue and Jim's are pink. Consumers have an idiosyncratic preference for either blue or pink widgets. Then both firms can set a positive price and sell to the consumers who like their colour more. The more are products *differentiated* the more there is scope for both firms to exist profitably in the industry and Jim may find it worthwhile to enter. This is why firms talk so much about differentiation and unique selling points. There are various ways to model this in economics. [Here is an example](http://www.people.fas.harvard.edu/~zlai/teaching/AppliedIO_Note3_HotellingSalop.pdf).

---

**More formally, $p\_j=p\_m=0$ is the unique pure strategy Nash equilibrium**

To be slightly more formal, let us check that $p\_m=p\_j=0$ is indeed a (Nash) equilibrium of the subgame in which both firms simultaneously set $p\in\mathbb{R}$ with the lowest priced firm capturing the entire (finite) demand. Neither party can profit by deviating to $p<0$ as this yields negative profits. A deviation to any $p>0$ results in the rival firm having a lower price so demand (and profits) are zero---again not profitable. Thus, there is no profitable deviation and $p\_i=p\_m=0$ is *an* equilibrium.

Is there another pure strategy equilibrium with some $p>0$? The answer is no. Consider the three possibilities and observe how a profitable deviation can be constructed for each:

* $p\_m>p\_j$. In this case $j$ could increase his price to some $p\_j'\in(p\_j,p\_m)$ without losing any demand. Such a $p\_j'$ exists by the connectedness of the real line.
* $p\_m=p\_j=p$. In this case at least one of the firms must be capturing less than all of the consumers. But by deviating to $p'=p-\epsilon$ it can capture all consumers. If we write $D$ for demand before the deviation and $D'>D$ for demand afterwards then the change in profits is $D'(p-\epsilon)-Dp$. By the connectedness of the real line, there exists an $\epsilon$ sufficiently small that this is positive.
* $p\_mp\_j$.

Thus, the only pure strategy equilibrium of this one-shot game is $p\_j=p\_m=0$.

---

**If we repeat the game then we can sustain other (collusive) equilibria**

What if we repeat the game infinitely many times? Suppose the two firms have an implicit understanding (an explicit agreement would be illegal) that they will both set $p=p^\*$ for some $p^\*>0$. Moreover, it is understood that if one firm deviates from this behaviour today, then both firms will revert to playing the static equilibrium ($p\_j=p\_m=0$) forever. Firms discount the future at rate $\delta$. For simplicity, suppose that demand is constant: $D(p)=1$ (not crucial).

If a firm plays in accordance with this understanding (and expects its rival to do likewise) then its profit is

$$\sum\_{t=0}^\infty\frac{1}{2}\delta^t p^\*=\frac{p^\*}{2(1-\delta)}.$$

If a firm cheats and sets $p^\*-\epsilon$ ($\epsilon$ small) then it captures the whole demand today, but expects to be punished forever thereafter, so profit is $p^\*$. Thus, both firms wish to comply with their implicit understanding if
$$\frac{p^\*}{2(1-\delta)}>p^\*\iff \delta>\frac{1}{2}$$
(i.e. if they are sufficiently patient).

So, if we repeat the game and firms are very patient, *we can sustain any price in equilibrium*. But note that these equilibria require not only that firms are patient, but also that a) there is no threat of future entry that could destabilise the equilibrium; b) firms are able to coordinate on a $p^\*$ without ever making an (illegal) explicit agreement; and c) firms are able to constantly and accurately monitor rivals behaviour in order to detect cheating.

> **Comments:**
> **username_0**: I picked the assumptions I picked because I was covertly talking about the pharmaceutical industry. There was a story recently about a firm realizing that they are in Mary's position and they just raised the price of their generic malaria resistance drug from 15 to 700 dollars a pill.  
> **username_1**: @Zackkenyon Yes, I think in the particular case you are referring to the assumptions make a lot of sense. Indeed, it was not my intention to criticise your assumptions. But I find that thinking about what happens when we tweak the important assumptions of a model is very useful for understanding how robust and general the model's conclusions are.  
> **username_0**: So, I guess before I accept this answer, I would need you to show that the series of epsilons does, in fact, converge to $p_0-marginal cost$. Seems like they could easily set the same price in a prisoner's dilemma situation.  
> **username_1**: @Zackkenyon I have added two sections. The first provides a more formal argument that the only pure strategy Nash equilibrium of the price-setting subgame is the zero price equilibrium. The second show how, with a bit of trickery, one can get other equilibria when the game is repeated (and explains why these equilibria might not be empirically plausible).  
> **username_0**: sorry I didn't get back to you earlier, but this argument is wrong. There is no non-iterative version of this game. Nash equilibrium is at setting your price exactly equal to your opponents at every "time step". You cannot do better against that strategy than to adopt exactly the same strategy.  
> **username_1**: @Zackkenyon No, I'm sorry but that's wrong. Here's a numerical example. Let market demand be $D(p)=100-p$. Suppose your strategy is to set price 50 in every period. Your claim is that the best I can do is to also set my price at 50. If I do that then my profit in each period is $50D(p)/2=[(100-50)50]/2=1250$ (division by 2 because if we set the same price then we have to share the market). But suppose I set a price of 49 in every period. Since my price is lower, I get the entire demand and my profit is $(100-49)49=2499$ per period. Higher profit, so your proposed play is not equilibrium  
> **username_0**: yes, and if you always choose rock, I will always choose paper... I'm not sure we are agreeing on the definition of Equilibrium here. my strategy does not have to be the best against ALL strategies in order to be a Nash equilibrium strategy, it just must be better against ITSELF than all other strategies. In that case it's very clear that when you are playing against an opponent who always sets their price value exactly equal to yours at every time step, the way to maximize your profit is to do the same... This is a very fundamental misunderstanding that I feel the need to resolve.  
> **username_0**: I updated the question somewhat for clarity.  

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### Answer by username_2 on 2015-09-25. Score: 0

Mary's marginal cost is zero so she can sell at that price when Jim enters the market. Thus Jim needs to have deep pockets to survive a price war.If both Jim and Mary's financial resources and opportunity cost (i.e what they can earn elsewhere) are 'common knowledge' then, in an indefinitely repeated game, there is a 'Muth Rational' solution such that price is set where the demand curve has elasticity of one- i.e. Marginal Revenue is zero- and the market is shared according to a formula which takes account of what resources each can commit and what opportunity cost this entails. In your example it appears Jim has 'money lying around'- i.e. no opportunity cost as such while Mary has sunk costs only and no recurring or depreciation costs. Jim may as well produce widgets. If there is no cost of disposing of widgets and assuming both Jim and Mary are rational then the Muth rational solution obtains. However it is not stable, one defection is likely to collapse the market. However, the defector realizes that their greed destroyed future profit so they will set the Muth rational price twice- allowing the other to recoup lost profit- after which the Muth rational price obtains till another defection occurs.
It is rational to enter this market for Jim because he seems to have zero opportunity cost for the investment and there is a positive expected value for the investment- provided common knowledge and Muth rationality obtain.


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