Trembling hand perfection and weakly dominated strategies

Question

It is well known that players cannot use weakly dominated strategies in a trembling hand perfect equilibrium. My question, however, is a little different: does iterated deletion of weakly dominated strategies preserve trembling hand perfect equilibria? In other words, if $\sigma$ is a trembling hand perfect equilibrium of a finite game $G$, is $\sigma$ also a trembling hand perfect equilibrium of the game $G'$ obtained by iteratively deleting weakly dominated strategies from $G$?

Apologies in advance if the answer is obvious!

Answer 1

A trembling hand perfect (THP) equilibrium may not survive iterated deletion of weakly dominated strategies, as is shown by the following example. Let game $G$ be

\begin{array}{|c|c|c|c|}\hline &A&B&C\\\hline A&0,0&0,0&0,-2\\\hline B&0,0&1,1&-1,-2\\\hline C&-2,0&-2,-1&-2,-2\\\hline \end{array}

There are two pure strategy NEs: $(A,A)$ and $(B,B)$, with the former involving iteratively weakly dominated strategies. And yet $(A,A)$ is a THP equilibrium. To see this, consider the following sequence of totally mixed strategies \begin{equation} \sigma_i^k=\left(1-\epsilon^{-2k}-\epsilon^{-k},\epsilon^{-2k},\epsilon^{-k}\right),\qquad i\in\{1,2\} \text{ and }\epsilon>2, \end{equation} which converges to $(1,0,0)=A$ as $k\to\infty$. It is easy to verify that, for any $k$, the best response of player $j\ne i$ to $\sigma_i^k$ is $A$: \begin{align} u_j(A,\sigma_i^k)&=0 \\ u_j(B,\sigma_i^k)&=\epsilon^{-2k}-\epsilon^{-k}<0 \\ u_j(C,\sigma_i^k)&=-2<0. \end{align} Therefore, $(A,A)$ is THP.

However, the game $G'$ that survives iterated deletion of weakly dominated strategies does not contain the strategy $A$ for either player. Hence, the THP equilibrium $(A,A)$ of the original game $G$ is not preserved in game $G'$.