<p>You are along the right track but are making certain errors in your proof,</p>

<ol>
<li><p>Your notation is confusing. You haven't defined what <span class="math-container">$S^{'D}$</span> is. I would suggest sticking with the defined strategy space and not creating unnecessary notation in the proof. </p></li>
<li><p>When you write <span class="math-container">$v(s'_{i},s'_{-i}) \geq v(s_{i},s_{-i})$</span>, you are not only changing player i's strategy but also other players' strategies. However, you need to keep other players' strategies constant and then compare player i's options to choose the best strategy. Also, note which player's payoff function is being compared. So, you should instead write <span class="math-container">$v_{i}(s'_{i},s'_{-i}) \geq v_{i}(s_{i},s'_{-i})$</span>. </p></li>
</ol>

<p>Now, let's move on to how a proof by contradiction could look like.</p>

<ol>
<li><p>Suppose there exists another dominant strategy equilibrium, <span class="math-container">$s^{A} \in S$</span>. (Note that we are talking about both weakly and strictly dominant strategy equilibria here)</p></li>
<li><p>This means that <span class="math-container">$\exists i \in n$</span> such that <span class="math-container">$s^{A}_{i} \neq  s^{D}_{i}$</span> and <span class="math-container">$v_{i}(s^{A}_{i}, s^{A}_{-i}) \geq v_{i}(s^{D}_{i}, s^{A}_{-i})$</span>. </p></li>
<li><p>However, <span class="math-container">$s^{D}$</span> is a strictly dominant strategy equilibrium, and so, <span class="math-container">$s^{D}_{i}$</span> is a strictly dominant strategy for player i. Inter alia, this implies that <span class="math-container">$v_{i}(s^{D}_{i}, s^{A}_{-i}) &gt; v_{i}(s^{A}_{i}, s^{A}_{-i})$</span>. </p></li>
<li><p>The inequalities in (2) and (3) contradict each other. Hence, the strictly dominant strategy equilibrium must be unique.</p></li>
</ol>

CC BY-SA 4.0
2020-05-07T22:25:54.760Z
I could also see that you've previously asked a similar question here: https://math.stackexchange.com/questions/3398749/strictly-dominant-strategy-equilibrium-uniqueness-proof but did not get a reply. The basic logic there is sound except that you don't consider another weakly dominant strategy equilibrium.
Amol Singh Raswan
CC BY-SA 4.0
2022-10-20T16:26:32.263Z
Really very explanatory and clear proof! Thank you!:)
studentp
CC BY-SA 4.0
2020-05-07T22:23:13.943Z
2020-05-07T22:23:13.943Z
Amol Singh Raswan
<p>This exercise is from book: Steve Tadelis, Intro to game theory, chapter 4 exercises.</p>

<p>Prove: If the game <span class="math-container">$\Gamma= \{N,\{S_i\}^n_{i=1},\{v\}^n_{i=1}\}$</span> has a strictly dominant strategy equilibrium <span class="math-container">$s^D$</span>, then <span class="math-container">$s^D$</span> is the unique dominant strategy equilibrium.</p>

<p>My attempt:</p>

<p>Proof by contradiction: Suppose <span class="math-container">$s^D$</span> were not the unique DSE. Then <span class="math-container">$\exists s'^D\in S'^D$</span> such that <span class="math-container">$v(s'_i,s'_{-i})\geq v(s_i,s_{-i})$</span> where <span class="math-container">$s'_i,s'_{-i}\in S^D$</span> and <span class="math-container">$s_is_{-i}\in S^d$</span>. But this contradicts the assumption that <span class="math-container">$s^D$</span> is the DSE.</p>

<p>Would this be correct?</p>

<game-theory>
proof of uniqueness of Dominant strategy equilibrium
Slim Shady
# proof of uniqueness of Dominant strategy equilibrium

(Asked by: username_0 on 2020-05-07)

This exercise is from book: Steve Tadelis, Intro to game theory, chapter 4 exercises.

Prove: If the game $\Gamma= \{N,\{S\_i\}^n\_{i=1},\{v\}^n\_{i=1}\}$ has a strictly dominant strategy equilibrium $s^D$, then $s^D$ is the unique dominant strategy equilibrium.

My attempt:

Proof by contradiction: Suppose $s^D$ were not the unique DSE. Then $\exists s'^D\in S'^D$ such that $v(s'\_i,s'\_{-i})\geq v(s\_i,s\_{-i})$ where $s'\_i,s'\_{-i}\in S^D$ and $s\_is\_{-i}\in S^d$. But this contradicts the assumption that $s^D$ is the DSE.

Would this be correct?

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2020-05-07. Score: 3 | Accepted answer

You are along the right track but are making certain errors in your proof,

1. Your notation is confusing. You haven't defined what $S^{'D}$ is. I would suggest sticking with the defined strategy space and not creating unnecessary notation in the proof.
2. When you write $v(s'\_{i},s'\_{-i}) \geq v(s\_{i},s\_{-i})$, you are not only changing player i's strategy but also other players' strategies. However, you need to keep other players' strategies constant and then compare player i's options to choose the best strategy. Also, note which player's payoff function is being compared. So, you should instead write $v\_{i}(s'\_{i},s'\_{-i}) \geq v\_{i}(s\_{i},s'\_{-i})$.

Now, let's move on to how a proof by contradiction could look like.

1. Suppose there exists another dominant strategy equilibrium, $s^{A} \in S$. (Note that we are talking about both weakly and strictly dominant strategy equilibria here)
2. This means that $\exists i \in n$ such that $s^{A}\_{i} \neq s^{D}\_{i}$ and $v\_{i}(s^{A}\_{i}, s^{A}\_{-i}) \geq v\_{i}(s^{D}\_{i}, s^{A}\_{-i})$.
3. However, $s^{D}$ is a strictly dominant strategy equilibrium, and so, $s^{D}\_{i}$ is a strictly dominant strategy for player i. Inter alia, this implies that $v\_{i}(s^{D}\_{i}, s^{A}\_{-i}) > v\_{i}(s^{A}\_{i}, s^{A}\_{-i})$.
4. The inequalities in (2) and (3) contradict each other. Hence, the strictly dominant strategy equilibrium must be unique.

> **Comments:**
> **username_1**: I could also see that you've previously asked a similar question here: https://math.stackexchange.com/questions/3398749/strictly-dominant-strategy-equilibrium-uniqueness-proof but did not get a reply. The basic logic there is sound except that you don't consider another weakly dominant strategy equilibrium.  
> **username_2**: Really very explanatory and clear proof! Thank you!:)  

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