proof of uniqueness of Dominant strategy equilibrium

Question

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?

Answer 1

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.