0
$\begingroup$

I am confused between the difference of what a solutions that a Nash equilibrium method provides versus an IESDS method. Let me explain through an example:

We have player 1 (P1) with strategies, $s_1$ = $\{ X, Y,Z \}$ and player 2 (P2) with strategies, $s_2$ = $\{ A, B, C \}$ $$ \begin{array}{c|lcr} & \text{A} & \text{B} & \text{C} \\ \hline X & 0,2 & 1,0 & 0,0 \\ Y & 0,1 & 0,0 & 1,1 \\ Z & 0,1 & 0,1 & 0,1 \end{array} $$ Now, using the method of Nash equilbrium, for every column, which is strategy for P2, we identify the highest payoff for P1. Let's Underline these payoffs. This gives: $$ \begin{array}{c|lcr} & \text{A} & \text{B} & \text{C} \\ \hline X & \underline{0,2} & \underline{1,0} & 0,0 \\ Y & \underline{0,1} & 0,0 & \underline{1,1} \\ Z & \underline{0,1} & 0,1 & 0,1 \end{array} $$ The second step of this method is, for every row, which is a strategy for P1, identify the highest payoff for P2. Let's Over-line these payoffs that satisfy this condition. This yields: $$ \begin{array}{c|lcr} & \text{A} & \text{B} & \text{C} \\ \hline X & \overline{\underline{0,2}} & \underline{1,0} & 0,0 \\ Y & \overline{\underline{0,1}} & 0,0 & \overline{\underline{1,1}} \\ Z & \overline{\underline{0,1}} & \overline{0,1} & \overline{0,1} \end{array} $$ All entries that are left with both an underline and an overline are Nash equilibria, correct? So we have four in this case: $(X,A),(Y,A),(Z,A),(Y,C)$.

I am confused because if we go with the IESDS (by weak dominance), then the only strategy that remains is $(Y,A)$, so are they both right?

$\endgroup$
2
$\begingroup$

For the record: this game has the four Nash equilibria in pure strategies that you have found above. However, contrary to your statement above, under IEWDS (iterated elimination of weakly dominated strategies) three of them survive: $(X,A),(Y,A),(Z,A)$.

It is generally known that IESDS never eliminates NE, while IEWDS may rule out some NE. The game presented in your question illustrates this fact, with the additional twist that the eliminated NE achieves the (unique) first-best for Player~1. So, here IEWDS eliminates the equilibrium uniquely preferred by one player.

This is why we say (tongue-in-cheek) that IEWDS should be applied with caution.

$\endgroup$
  • $\begingroup$ This interesting because I never realized that IEWDS could rule out a NE since the definition of a NE does not have a strict equality. $\endgroup$ – guy Apr 1 '17 at 20:27
1
$\begingroup$

I am confused as to what you are doing here. First, IESDS is iterated elimination of strictly dominated actions? If so, then no action can be eliminated since the definition of dominated action is: a pure action $a_{i}$ of player $i$ is strictly dominated if there exists a mixed action of player $i$, $\alpha_i$, that strictly dominates $a_{i}$ for any pure action of the other players.

To see that nothing can be eliminated notice that the row player has payoff $0$ if the column player plays $A$ and the column player has payoff $1$ if the row player plays $Z$.

I guess I am not sure what you mean by 'IESDS (weak dominance)'.

Likewise, what do you mean by 'right' in 'so are they both right'. Nash equilibrium, IESDS are just two examples of 'solution concepts' and solution concepts cannot be right or wrong. They just give you predictions. Whether they give you predictions that match those actually observed is a separate issue.

$\endgroup$
  • $\begingroup$ Thanks Jan for the explanation. I meant IESDS by weak dominance as in perform IESDS but rather by doing it strictly, a strategy may be equal to it as well. Meaning if we look at P1's remaining strategies $\{ X, Y\}$, we see that $X$ is weakly dominated by $Y$ since $v_1(X, s_{-i}) \leq v_1(Y, s_{-i})$ and therefore, we can eliminate strategy $X$. At the same time, we can also eliminate strategy $C$ for P2 since $v_2( s_{-i}, C) \leq v_2( s_{-i}, A)$ and therefore $C$ is weakly dominated by strategy $A$. Is that clear now? Is my logic for finding Nash equilibria sound? $\endgroup$ – guy Feb 6 '17 at 17:04
  • $\begingroup$ @tbone I believe by doing iterated elimination of weakly dominated strategies you might eliminate NE. Also, I am not sure about your elimination. I can see how you eliminate $Z$ in the first step and $C$ in the second, or vice versa. Then you can eliminate $Y$ or $B$ and if you eliminate $B$, you might eliminate $Y$ or $X$ in the next step and if you eliminate $Y$ you eliminate $B$ in the next step. The bottom line is i) that I am not sure how you eliminate $X$ in the first step and ii) iterated elimination of weakly dominated strategies might give different results depending on order. $\endgroup$ – Jan Feb 6 '17 at 19:11
  • $\begingroup$ yes the order is important I suppose. I just don't get why IESDS by weak dominance and Nash equilibrium solution concepts yield different results since they appear the same mathematically. $\endgroup$ – guy Feb 6 '17 at 22:26
  • $\begingroup$ @tbone Well, appear is a soft concept. To me those are very different. NE is a fixed point. IESDS is about common knowledge of rationality. $\endgroup$ – Jan Feb 6 '17 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.