<p>Your intuition about the proof is correct indeed. A more formal proof would involve examining the definitions of dominant strategy and Nash equilibrium:</p>
<blockquote>
<p>A strategy <span class="math-container">$s_i^d$</span> is a <em><strong>dominant strategy for player <span class="math-container">$i$</span></strong></em> if
<span class="math-container">\begin{equation}
u_i(s_i^d,s_{-i})&gt; 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}</span></p>
</blockquote>
<blockquote>
<p>A strategy profile <span class="math-container">$(s_1^*,\dots,s_n^*)$</span> constitutes a <em><strong>Nash equilibrium</strong></em> if for every player <span class="math-container">$i=1,\dots,n$</span>,
<span class="math-container">\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}</span></p>
</blockquote>
<p>In a game where every player has a dominant strategy <span class="math-container">$s_i^d$</span>, i.e. a strategy that satisfies condition <span class="math-container">$(1)$</span> above, then condition <span class="math-container">$(2)$</span> must also be satisfied by the strategy profile <span class="math-container">$(s_1^d,\dots,s_n^d)$</span>, hence a Nash equilibrium.</p>

CC BY-SA 4.0
2022-07-18T17:46:51.247Z
2022-07-18T17:46:51.247Z
Herr K.
<p>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 <em>no matter what the opponent chooses</em>, 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.</p>
<p>However, this seems too simple - am I missing anything?</p>
<p>Thanks!</p>

<game-theory><nash-equilibrium><dominant-strategy>
How can we prove that an equilibrium in dominant strategies is a Nash equilibrium?
user41781
# How can we prove that an equilibrium in dominant strategies is a Nash equilibrium?

(Asked by: username_0 on 2022-07-18)

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!

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


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## 1 answers

### Answer by username_1 on 2022-07-18. Score: 2

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.


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