# Are all strategies that survive IESDS part of Nash equilibria?

I know that Iterated Elimination of Strictly Dominated Strategies (IESDS) never eliminates a strategy which is part of a Nash equilibrium. Is the reverse also true? And is there a proof somewhere? I only found this as a statement in a series of slides, but without proof. It seems like this should be true, but I can't prove it myself properly.

Q: If a strategy survives IESDS, is it part of a Nash equilibrium? (mixed strategies also allowed)

Maxim

No. Observe the following payoff matrix: $$\begin{bmatrix} 1,1 & 1,5 & 5,2 \\ 1,2 & 1,1 & 1,1 \\ 5,1 & 1,5 & 1,2 \\ \end{bmatrix}$$.

The row player's strategy space is $$(U,M,B)$$ and the column palyer's is $$(L,M,R)$$.

Ther is no pure Nash equilibrium if where the row player plays $$M$$, because column's best response is $$U$$, but to $$U$$ row's best response ins $$B$$.

There are also no mixed equilibria in which row plays $$B$$:

if column mixes over his entire strategy space - $$x = (a, b, 1-a-b)$$

$$u_1(U,x) = 5-4(a+b)$$, $$u_1(M,x) = 1$$, $$u_1(B,x) = 1+4a$$. I.e. $$u_1(U,x) > u_1(M,x) \wedge u_1(B,x) > u_1(M,x) \Rightarrow$$ if column plays x row plays $$M$$ with probability zero.

If column mixes over $$(L, M)$$ - $$x = (a, 1-a, 0)$$ $$u_1(U,x) = 1$$, $$u_1(M,x) = 1$$, $$u_1(B,x) = 1+4a$$. I.e. $$u_1(B,x) > u_1(U,x) \wedge u_1(B,x) > u_1(M,x) \Rightarrow$$ if column plays x row plays $$M$$ and $$U$$ with probability zero.

If column mixes over $$(L, R)$$ - $$x = (a, 0, 1-a)$$ $$u_1(U,x) = 5-4a$$, $$u_1(M,x) = 1$$, $$u_1(B,x) = 1+4a$$. I.e. $$u_1(U,x) > u_1(M,x) \wedge u_1(B,x) > u_1(M,x) \Rightarrow$$ if column plays x row plays $$M$$ with probability zero.

If column mixes over $$(M, R)$$ - $$x = (0, a, 1-a)$$ $$u_1(U,x) = 5-4a$$, $$u_1(M,x) = 1$$, $$u_1(B,x) = 1$$. I.e. $$u_1(U,x) > u_1(M,x) \wedge u_1(B,x) > u_1(M,x) \Rightarrow$$ if column plays x row plays $$M$$ and $$B$$ with probability zero.

• Wow, thanks a lot! This is a great example, and presented in a really nice way! (I briefly thought that maybe rows M could be dominated by a mixed strategy, but that is not the case. And for column nothing can be eliminate anyway.) Feb 19, 2020 at 15:17
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