<p>You're right. Let $s^*=(s_1^*,\dots,s_N^*)$ be the equilibrium of a strictly dominance solvable game. Then by definition, 
$$
u_i(s_i^*,s_{-i})&gt;u_i(s_i,s_{-i})
$$
for all $i$, all $s_i\ne s_i^*$ and all $s_{-i}$. This implies that 
$$
u_i(s_i^*,s_{-i})\ge u_i(s_i,s_{-i})
$$
for all $i$, all $s_i\ne s_i^*$, all $s_{-i}$ and with strict inequality for at least some $s_i$ (in fact, for all $s_i\ne s_i^*$). This makes $s^*$ an equilibrium satisfying the weak dominance solvability criterion.</p>

<hr>

<p>Your quoted definition of <em>strict/weak dominance solvability</em> makes me a little uncomfortable though. I'd say </p>

<blockquote>
  <p>A game is strict (or weak) dominance solvable if the process of iteratively removing strictly (or weakly) dominated strategies leads to a unique outcome (i.e. only one strategy for each player survives).</p>
</blockquote>

CC BY-SA 3.0
2017-02-21T20:15:27.733Z
You are right to be uneasy because iteratively removing weakly dominated strategies can lead to *different* unique outcomes depending on the order of the removal of the strategies. I would say that these are in some sense strange solutions.
Giskard
CC BY-SA 3.0
2017-02-21T20:20:25.753Z
An example: Both players have to choose between strategy $A$ and $B$. If your opponent choses $A$ your payoff is 0, otherwise it is 1. In these games both $A$ and $B$ are weakly dominant strategies. You can eliminate either one and have whatever you want as a solution: All strategy profiles are weakly dominant equilibria.
Giskard
CC BY-SA 3.0
2017-02-21T21:07:54.637Z
@denesp: Iteratively deleting weakly dominated strategies has long been known to be problematic. Aside from the point you mentioned, the fact that the procedure can remove Nash equilibria from a game makes it an undesirable method for analyzing games.
Herr K.
CC BY-SA 3.0
2017-02-21T22:56:42.647Z
I think that is included in my argument (you can remove weakly dominant equilibria and all w.d.equilibra are NE) but yes, we are in agreement. That is why the phrase 'solvable' is somewhat problematic in this context in my opinion.
Giskard
CC BY-SA 3.0
2017-02-22T23:41:16.397Z
@denesp in your example, for the player with payoffs specified, A and B are payoff equivalent and thus neither weakly dominates the other, no? A proper example of order dependence for iterated elimination of weakly dominated strategies necessarily needs more than one step.
Rahul Savani
CC BY-SA 3.0
2017-02-23T05:09:18.320Z
@RahulSavani You are correct sir. I knew the definition wrong (this is actually a major shock to me).
Giskard
CC BY-SA 3.0
2017-02-23T08:08:08.257Z
@denesp OK, still the spirit of the comments is quite right. BTW, I can give you a nice example of order dependence of for iterated elimination of weakly dominated strategies in case you are interested.
Rahul Savani
CC BY-SA 3.0
2017-02-21T19:33:04.177Z
2017-02-21T19:40:31.050Z
2017-02-21T19:40:31.050Z
Herr K.
<p>Okay consider a game $G$ if a strategy $s_i$ has the following property we call $s_i$ the strictly dominant strategy</p>

<p>$$u_i(s_i,s_{-i})&gt;u_i(s_i',s_{-i}) \\ \forall s_{-i} \ \forall s_i' \epsilon S'_i$$</p>

<p>Where $s_{i}$ indicates the strategies of players other then $i$ in the game and $S'_i$ is the set for strategies of player $i$ except the specific strategy $s_i$</p>

<p>Now let's look at the definition of a weakly dominant strategy</p>

<p>if a strategy $s_i$ has the following property we call $s_i$ the weakly dominant strategy</p>

<p>$$u_i(s_i,s_{-i})≥u_i(s_i',s_{-i}) \\ \forall s_{-i} \ \forall s_i' \epsilon S'_i \ and \\ \exists s_i' \epsilon S'_i \ such \ that: \ u_i(s_i,s_{-i})&gt;u_i(s_i',s_{-i})$$</p>

<p>Okay I believe from these two definitions we can derive that any strictly dominant strategy $s_i$ is also a weakly dominant strategy</p>

<p>Definition of strict dominance solvable is as follows :</p>

<p>A strict dominance solvable game is a game where the equilibrium outcome is strict dominance equilibrium.
A weakly dominance solvable games is a game where the equilibrium outcome is weakly dominance equilibrium.</p>

<p>So It seems like then a strictly dominance solvable game is always a weakly dominance solvable game. Am I wrong?</p>

<microeconomics><game-theory>
Are strict dominance solvable games weakly dominance solvable?
Kuantew
# Are strict dominance solvable games weakly dominance solvable?

(Asked by: username_0 on 2017-02-21)

Okay consider a game $G$ if a strategy $s\_i$ has the following property we call $s\_i$ the strictly dominant strategy

$$u\_i(s\_i,s\_{-i})>u\_i(s\_i',s\_{-i}) \\ \forall s\_{-i} \ \forall s\_i' \epsilon S'\_i$$

Where $s\_{i}$ indicates the strategies of players other then $i$ in the game and $S'\_i$ is the set for strategies of player $i$ except the specific strategy $s\_i$

Now let's look at the definition of a weakly dominant strategy

if a strategy $s\_i$ has the following property we call $s\_i$ the weakly dominant strategy

$$u\_i(s\_i,s\_{-i})≥u\_i(s\_i',s\_{-i}) \\ \forall s\_{-i} \ \forall s\_i' \epsilon S'\_i \ and \\ \exists s\_i' \epsilon S'\_i \ such \ that: \ u\_i(s\_i,s\_{-i})>u\_i(s\_i',s\_{-i})$$

Okay I believe from these two definitions we can derive that any strictly dominant strategy $s\_i$ is also a weakly dominant strategy

Definition of strict dominance solvable is as follows :

A strict dominance solvable game is a game where the equilibrium outcome is strict dominance equilibrium.
A weakly dominance solvable games is a game where the equilibrium outcome is weakly dominance equilibrium.

So It seems like then a strictly dominance solvable game is always a weakly dominance solvable game. Am I wrong?

**Tags:** microeconomics, game-theory

---

## 1 answers

### Answer by username_1 on 2017-02-21. Score: 1 | Accepted answer

You're right. Let $s^\*=(s\_1^\*,\dots,s\_N^\*)$ be the equilibrium of a strictly dominance solvable game. Then by definition,
$$
u\_i(s\_i^\*,s\_{-i})>u\_i(s\_i,s\_{-i})
$$
for all $i$, all $s\_i\ne s\_i^\*$ and all $s\_{-i}$. This implies that
$$
u\_i(s\_i^\*,s\_{-i})\ge u\_i(s\_i,s\_{-i})
$$
for all $i$, all $s\_i\ne s\_i^\*$, all $s\_{-i}$ and with strict inequality for at least some $s\_i$ (in fact, for all $s\_i\ne s\_i^\*$). This makes $s^\*$ an equilibrium satisfying the weak dominance solvability criterion.

---

Your quoted definition of *strict/weak dominance solvability* makes me a little uncomfortable though. I'd say

> A game is strict (or weak) dominance solvable if the process of iteratively removing strictly (or weakly) dominated strategies leads to a unique outcome (i.e. only one strategy for each player survives).

> **Comments:**
> **username_2**: You are right to be uneasy because iteratively removing weakly dominated strategies can lead to *different* unique outcomes depending on the order of the removal of the strategies. I would say that these are in some sense strange solutions.  
> **username_2**: An example: Both players have to choose between strategy $A$ and $B$. If your opponent choses $A$ your payoff is 0, otherwise it is 1. In these games both $A$ and $B$ are weakly dominant strategies. You can eliminate either one and have whatever you want as a solution: All strategy profiles are weakly dominant equilibria.  
> **username_1**: @denesp: Iteratively deleting weakly dominated strategies has long been known to be problematic. Aside from the point you mentioned, the fact that the procedure can remove Nash equilibria from a game makes it an undesirable method for analyzing games.  
> **username_2**: I think that is included in my argument (you can remove weakly dominant equilibria and all w.d.equilibra are NE) but yes, we are in agreement. That is why the phrase 'solvable' is somewhat problematic in this context in my opinion.  
> **username_3**: @denesp in your example, for the player with payoffs specified, A and B are payoff equivalent and thus neither weakly dominates the other, no? A proper example of order dependence for iterated elimination of weakly dominated strategies necessarily needs more than one step.  
> **username_2**: @RahulSavani You are correct sir. I knew the definition wrong (this is actually a major shock to me).  
> **username_3**: @denesp OK, still the spirit of the comments is quite right. BTW, I can give you a nice example of order dependence of for iterated elimination of weakly dominated strategies in case you are interested.  

---

