<p>The following two claims hold in the general <span class="math-container">$n$</span>-shop case.</p>
<p><strong>Claim 1.</strong> In equilibrium a shop closest to an edge (0 or 1) cannot be alone.</p>
<p><strong>Proof.</strong> Such a shop could gain customers by moving slightly inward.</p>
<p><strong>Claim 2.</strong> In equilibrium at most two shops can be in any location.</p>
<p><strong>Proof.</strong> Assume there is an equilibrium where there are three or more shops in a location. Denote the number of customers coming to this location from the left and right by <span class="math-container">$c_l$</span> and <span class="math-container">$c_r$</span> respectively. (In the 3-shop case <span class="math-container">$c_l + c_r = 1$</span>.) All shops in the location have a payoff of <span class="math-container">$(c_l + c_r)/n$</span>. By moving slightly to the left, a shop can attain a payoff that is arbitrarily close to <span class="math-container">$c_l$</span>, and by moving slightly to the right a shop can attain a payoff that is arbitrarily close to <span class="math-container">$c_r$</span>, therefore it can attain (almost) <span class="math-container">$\max(c_l;c_r)$</span>. It is easy to show that if <span class="math-container">$n &gt;2$</span> then
<span class="math-container">$$
\frac{c_l + c_r}{n} &lt; \max(c_l;c_r). 
$$</span></p>
<hr />
<p>Using these two properties it is very easy to show that in the 3-shop case there is no pure Nash-equilibrium, so this is left as an exercise.</p>
<p>I happen to have a <a href="https://student.desmos.com/join/pu8zpg" rel="nofollow noreferrer">Desmos</a> for this, if you'd like you can use it to validate the claims and reason about the proof.</p>
<hr />
<p>Note: a mixed equilibrium does exist, see <a href="https://www.jstor.org/stable/2098006" rel="nofollow noreferrer">Shaked, A. (1982): <em>Existence and Computation of Mixed Strategy Nash Equilibrium for 3-Firms Location Problem</em>.</a></p>
<hr />
<p>Interestingly, <span class="math-container">$n = 3$</span> is the only case without a pure equilibrium, <span class="math-container">$n \in \left\{2,4,5\right\}$</span> all have one, and for other <span class="math-container">$n$</span> values there are infinitely many equilibria. For a formal characterization one needs some additional properties, but no advanced mathematics. For a detailed discussion see
<a href="https://academic.oup.com/restud/article-abstract/42/1/27/1538204?redirectedFrom=fulltext" rel="nofollow noreferrer">Eaton, B.C., and R.G. Lipsey (1975): <em>The Principle of Minimum Differentiation Reconsidered: Some New Developments in the Theory of Spatial Competition</em>.</a></p>

CC BY-SA 4.0
2020-07-12T08:04:49.240Z
2020-07-12T10:23:50.960Z
2020-07-12T10:23:50.960Z
Giskard
<p>So if two ice cream shops were to be placed in the location <span class="math-container">$[0,1]$</span>, inorder to maximize their own pay offs, they both would finally come to the location <span class="math-container">$[\frac{1}{2}, \frac{1}{2}]$</span>. This is also the Nash Equilibrium of the problem and does not require a lot of mathematics to understand.</p>
<p>Now my problem was when we are playing this game for 3 different shops, like there is no intuitive answer for the problem. Every possible answer I thought of (<span class="math-container">$[\frac{1}{2}, \frac{1}{2}, \frac{1}{2}]$</span>, <span class="math-container">$[\frac{1}{4}, \frac{1}{2}, \frac{3}{4}]$</span> etc.) is not a Nash Equilibrium.</p>
<p>So like is there a mathematical way that I can use to find the Nash equilibrium location for this game, or prove that it doesn't exist.</p>
<p>Also can this be generalized to <span class="math-container">$4,5 ...n$</span> games?</p>

<game-theory><nash-equilibrium>
Nash Equilibrium for n-shops Location Game
Harshit Joshi
# Nash Equilibrium for n-shops Location Game

(Asked by: username_0 on 2020-07-12)

So if two ice cream shops were to be placed in the location $[0,1]$, inorder to maximize their own pay offs, they both would finally come to the location $[\frac{1}{2}, \frac{1}{2}]$. This is also the Nash Equilibrium of the problem and does not require a lot of mathematics to understand.

Now my problem was when we are playing this game for 3 different shops, like there is no intuitive answer for the problem. Every possible answer I thought of ($[\frac{1}{2}, \frac{1}{2}, \frac{1}{2}]$, $[\frac{1}{4}, \frac{1}{2}, \frac{3}{4}]$ etc.) is not a Nash Equilibrium.

So like is there a mathematical way that I can use to find the Nash equilibrium location for this game, or prove that it doesn't exist.

Also can this be generalized to $4,5 ...n$ games?

**Tags:** game-theory, nash-equilibrium


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## 1 answers

### Answer by username_1 on 2020-07-12. Score: 5 | Accepted answer

The following two claims hold in the general $n$-shop case.

**Claim 1.** In equilibrium a shop closest to an edge (0 or 1) cannot be alone.

**Proof.** Such a shop could gain customers by moving slightly inward.

**Claim 2.** In equilibrium at most two shops can be in any location.

**Proof.** Assume there is an equilibrium where there are three or more shops in a location. Denote the number of customers coming to this location from the left and right by $c\_l$ and $c\_r$ respectively. (In the 3-shop case $c\_l + c\_r = 1$.) All shops in the location have a payoff of $(c\_l + c\_r)/n$. By moving slightly to the left, a shop can attain a payoff that is arbitrarily close to $c\_l$, and by moving slightly to the right a shop can attain a payoff that is arbitrarily close to $c\_r$, therefore it can attain (almost) $\max(c\_l;c\_r)$. It is easy to show that if $n >2$ then
$$
\frac{c\_l + c\_r}{n} < \max(c\_l;c\_r).
$$

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Using these two properties it is very easy to show that in the 3-shop case there is no pure Nash-equilibrium, so this is left as an exercise.

I happen to have a [Desmos](https://student.desmos.com/join/pu8zpg) for this, if you'd like you can use it to validate the claims and reason about the proof.

---

Note: a mixed equilibrium does exist, see [Shaked, A. (1982): *Existence and Computation of Mixed Strategy Nash Equilibrium for 3-Firms Location Problem*.](https://www.jstor.org/stable/2098006)

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Interestingly, $n = 3$ is the only case without a pure equilibrium, $n \in \left\{2,4,5\right\}$ all have one, and for other $n$ values there are infinitely many equilibria. For a formal characterization one needs some additional properties, but no advanced mathematics. For a detailed discussion see
[Eaton, B.C., and R.G. Lipsey (1975): *The Principle of Minimum Differentiation Reconsidered: Some New Developments in the Theory of Spatial Competition*.](https://academic.oup.com/restud/article-abstract/42/1/27/1538204?redirectedFrom=fulltext)


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