How can we prove that an equilibrium in dominant strategies is a Nash equilibrium?

Question

So far I have that an equilibrium in dominant strategies is one where two strictly dominant strategy profiles meet (e.g. to report the other prisoner in the Prisoner's Dilemma). Going by the definition of a dominant strategy, it is one where the strategy profile is the best no matter what the opponent chooses, and hence would also fit the Nash equilibrium definition where neither player has incentive to change their strategy since by definition the dominant strategy is already the best one.

However, this seems too simple - am I missing anything?

Thanks!

Answer 1

Your intuition about the proof is correct indeed. A more formal proof would involve examining the definitions of dominant strategy and Nash equilibrium:

A strategy $s_i^d$ is a dominant strategy for player $i$ if \begin{equation} u_i(s_i^d,s_{-i})> u_i(s_i,s_{-i}),\quad \forall s_i\in S_i\setminus\{s_i^d\},\;\forall s_{-i}\in S_{-i} \tag{1} \end{equation}

A strategy profile $(s_1^*,\dots,s_n^*)$ constitutes a Nash equilibrium if for every player $i=1,\dots,n$, \begin{equation} u_i(s_i^*,s_{-i}^*)\ge u_i(s_i,s_{-i}^*), \quad\forall s_i\in S_i \tag{2} \end{equation}

In a game where every player has a dominant strategy $s_i^d$, i.e. a strategy that satisfies condition $(1)$ above, then condition $(2)$ must also be satisfied by the strategy profile $(s_1^d,\dots,s_n^d)$, hence a Nash equilibrium.