<p>I assume you know how the proof goes using Kakutani to prove existence for finite games.</p>
<p>As standard, let <span class="math-container">$X_{-i} = \times_{j \not = i} X_{j}$</span>.</p>
<p>Define the best response correspondence <span class="math-container">$BR_i: S_{-i} \rightrightarrows S_i$</span> by
<span class="math-container">$$
BR_i(s_{-i}) = \arg \max_{s_i \in S_i} u_i(s_i,s_{-i})
$$</span></p>
<p>Apart from the usual existence and upper hemicontinuity, quasiconcavity of <span class="math-container">$u_i$</span> implies the correspondence is convex valued.</p>
<p>As usual, stack them up to define the correspondence <span class="math-container">$BR: S \rightrightarrows S$</span>.</p>
<p>Apply Kakutani, the desired fixed point is an element of <span class="math-container">$s^* \in S$</span>, and is thus a pure strategy.</p>
<p>Compare this to the proof for finite games, where we define the correspondence on the space <span class="math-container">$\times_{i} \Delta (S_i)$</span>!</p>

CC BY-SA 4.0
2021-12-27T20:38:21.730Z
2021-12-27T20:38:21.730Z
Walrasian Auctioneer
<p>I have a question about <strong>Theorem 1.2</strong> in <em>Game Theory</em> by Fudenberg and Tirole.</p>
<blockquote>
<p><strong><strong>Theorem 1.2</strong> in FT(Debreu, Glicksberg, Fan)</strong> Consider a strategic-form game whose strategy spaces <span class="math-container">$S_{i}$</span> are nonempty compact convex subsets of an Euclidean space. If the payoff functions <span class="math-container">$u_{i}$</span> are continuous in <span class="math-container">$s$</span> and quasi-concave in <span class="math-container">$s_{i}$</span>, there exists a pure-strategy Nash Equilibrium.</p>
</blockquote>
<blockquote>
<p><strong>Theorem 1.1 in FT (Nash)</strong>: Every finite strategic-form game has a mixed strategy equilibrium.</p>
</blockquote>
<p>My question here is that, how do we know that the Nash Equilibrium in <strong>theorem 1.2</strong> is a pure-strategy one?</p>
<p>In the proof, they also use Kakutani's fixed point theorem, but I think it can only give us the existence of some Nash Equilibrium, and we know nothing about its characterizations. Additionally, Nash's existence theorem should be a special case of theorem 1.2 since the set of mixed strategies over finite actions is a simplex, which is compact. However, in Nash's theorem, we only know that a mixed-strategy equilibrium will exist.</p>
<p>My understanding of Debreu's theorem is that, the &quot;pure strategy&quot; in this theorem is with respect to the strategy spaces, not to action spaces. To be more specific, take the Nash Existence Theorem example above, the mixed strategy in the theorem is actually a &quot;pure strategy&quot; in the space of mixed strategies, so both theorems don't contradict with each other.</p>

CC BY-SA 4.0
2021-12-27T16:53:50.057Z
The usual proof of Theorem 1.2. uses a setting in which fixed points correspond to Nash equilibria in pure strategies. One can prove Theorem 1.1 from 1.2 by showing that the mixed extension of a finite game has a pure strategy equilibrium. This pure strategy equilibrium of the mixed extension is then a mixed equilibrium of the underlying finite game.
Michael Greinecker
CC BY-SA 4.0
2021-12-27T20:02:24.623Z
@MichaelGreinecker could you elaborate a bit more on the first sentence? i.e The proof uses a setting in which fixed points correspond to Nash equilibria in pure strategies? And what is the "mixed extension" of a finite game?
John
CC BY-SA 4.0
2021-12-27T22:20:51.003Z
In the mixed extension, you take your strategy space to be the space of probability distributions over the original strategies. Another way to see the connection: In Nash's proof, one uses the mixed strategy best reply correspondence, while can also work with the pure strategy best reply correspondence under the assumptions of Theorem 1.2.
Michael Greinecker
<game-theory>
Existence of some pure-strategy Nash Equilibrium
John
# Existence of some pure-strategy Nash Equilibrium

(Asked by: username_0 on 2021-12-27)

I have a question about **Theorem 1.2** in *Game Theory* by Fudenberg and Tirole.

> ****Theorem 1.2** in FT(Debreu, Glicksberg, Fan)** Consider a strategic-form game whose strategy spaces $S\_{i}$ are nonempty compact convex subsets of an Euclidean space. If the payoff functions $u\_{i}$ are continuous in $s$ and quasi-concave in $s\_{i}$, there exists a pure-strategy Nash Equilibrium.

> **Theorem 1.1 in FT (Nash)**: Every finite strategic-form game has a mixed strategy equilibrium.

My question here is that, how do we know that the Nash Equilibrium in **theorem 1.2** is a pure-strategy one?

In the proof, they also use Kakutani's fixed point theorem, but I think it can only give us the existence of some Nash Equilibrium, and we know nothing about its characterizations. Additionally, Nash's existence theorem should be a special case of theorem 1.2 since the set of mixed strategies over finite actions is a simplex, which is compact. However, in Nash's theorem, we only know that a mixed-strategy equilibrium will exist.

My understanding of Debreu's theorem is that, the "pure strategy" in this theorem is with respect to the strategy spaces, not to action spaces. To be more specific, take the Nash Existence Theorem example above, the mixed strategy in the theorem is actually a "pure strategy" in the space of mixed strategies, so both theorems don't contradict with each other.

**Tags:** game-theory
> **Comments:**
> **username_1**: The usual proof of Theorem 1.2. uses a setting in which fixed points correspond to Nash equilibria in pure strategies. One can prove Theorem 1.1 from 1.2 by showing that the mixed extension of a finite game has a pure strategy equilibrium. This pure strategy equilibrium of the mixed extension is then a mixed equilibrium of the underlying finite game.  
> **username_0**: @MichaelGreinecker could you elaborate a bit more on the first sentence? i.e The proof uses a setting in which fixed points correspond to Nash equilibria in pure strategies? And what is the "mixed extension" of a finite game?  
> **username_1**: In the mixed extension, you take your strategy space to be the space of probability distributions over the original strategies. Another way to see the connection: In Nash's proof, one uses the mixed strategy best reply correspondence, while can also work with the pure strategy best reply correspondence under the assumptions of Theorem 1.2.  

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## 1 answers

### Answer by username_2 on 2021-12-27. Score: 5 | Accepted answer

I assume you know how the proof goes using Kakutani to prove existence for finite games.

As standard, let $X\_{-i} = \times\_{j \not = i} X\_{j}$.

Define the best response correspondence $BR\_i: S\_{-i} \rightrightarrows S\_i$ by
$$
BR\_i(s\_{-i}) = \arg \max\_{s\_i \in S\_i} u\_i(s\_i,s\_{-i})
$$

Apart from the usual existence and upper hemicontinuity, quasiconcavity of $u\_i$ implies the correspondence is convex valued.

As usual, stack them up to define the correspondence $BR: S \rightrightarrows S$.

Apply Kakutani, the desired fixed point is an element of $s^\* \in S$, and is thus a pure strategy.

Compare this to the proof for finite games, where we define the correspondence on the space $\times\_{i} \Delta (S\_i)$!


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