On the equivalence between (quasi-)perfect and sequential equilibria

Abstract

We prove the generic equivalence between quasi-perfect equilibrium and sequential equilibrium. Combining this result with Blume and Zame (Econometrica 62:783–794, 1994) shows that perfect, quasi-perfect and sequential equilibrium coincide in generic games.

Appendix

In what follows, if $c\in C(h)$ we denote $Z(h,c)$ the set of final nodes that come after some node in $\{(x,c):x\in h\}$. Recall that a system of beliefs $\mu$ assigns, for every player $n\in \mathcal N$ and every $h\in H_n$ a probability distribution on $h$.

Proof of Proposition 1

Let $s\in SE (u)$. We construct a sequence $\{(u^t,\varepsilon ^t,s^t)\}_{t=1}^{\infty }\subset W^{\circ }$ that converges to $(u,0,s)$. Since $s$ is a sequential equilibrium strategy, there is a system of beliefs $\mu$ and a sequence $\{(\mu ^t,s^t)\}_{t=1}^{\infty }$ converging to $(\mu ,s)$ such that $s^t\in S^{\!\circ }$ and $\mu ^t$ is derived from $s^t$ using Bayes rule for every $t$. Moreover, $(\mu ,s)$ is sequentially rational, i.e., for every player $n$ and information set $h\in H_n$, the strategy $s_n$ prescribes optimal behavior at $h$ given the value of $\mu$ at $h$ and the other players’ future behavior as described by $s_{-n}$. Fix an element $t$ of the sequence $\{(\mu ^t,s^t)\}_{t=1}^{\infty }$. Define $\varepsilon ^t=\max _{n,h,c}\left\{ s_n^t(c\mid h):\right. \left. s_n(c\mid h)=0\right\}$. We now construct a payoff vector $u^t$ such that $(u^t,\varepsilon ^t,s^t)\in W^{\circ }$.

Let $H^*_n\subset H_n$ be the collection of last information sets of player $n$. For each information set $h\in H_n^*$ and for each choice $c\in C(h)$ compute:

$$\begin{aligned} x^t_n(h,c)=\max _{i_n\in I_n(h)}{v_n^h(s_{-n}^t,i_n,u)}-\max _{j_n\in I_n(h,c)}{v_n^h(s_{-n}^t,j_n,u)}. \end{aligned}$$

Obtain the payoff vector $w^t$ from $u$ as follows:

$$\begin{aligned} w^t_n(z) = {\left\{ \begin{array}{ll} u_n(z)+x^t_n(h,c)\quad &{}\text{ if } z\in Z(h,c),\ h\in H^*_n,\ c\in C(h), \text{ and } s_n(c\mid h)>0;\\ u_n(z)&{}\text{ otherwise }. \end{array}\right. } \end{aligned}$$

Let $H^{\prime }_n$ be the collection of last information sets in $H_n\setminus H^*_n$. For each $h^{\prime }\in H^{\prime }_n$ and for each choice $c^{\prime }\in C(h^{\prime })$ let:

$$\begin{aligned} x^t_n(h^{\prime },c^{\prime })=\max _{i_n\in I_n(h^{\prime })}{v_n^h(s_{-n}^t,i_n,{w}^t)}-\max _{j_n\in I_n(h^{\prime },c^{\prime })}{v_n^h(s_{-n}^t,j_n,{w}^t)}. \end{aligned}$$

Obtain the utility vector $\tilde{w}^t$ from ${w}^t$ according to:

$$\begin{aligned} \tilde{w}^t_n(z) \!=\! {\left\{ \begin{array}{ll} {w}^t_n(z)\!+\!x^t_n(h^{\prime },c^{\prime })\quad &{}\text{ if } z\in Z(h^{\prime },c^{\prime }),\ h^{\prime }\in H^{\prime }_n,\ c^{\prime }\in C(h^{\prime }), \text{ and } s_n(c^{\prime }\mid h^{\prime })>0;\\ {w}^t_n(z) &{}\text{ otherwise }. \end{array}\right. } \end{aligned}$$

In particular, note that if $h\in H_n$ follows $c^{\prime }$ then $x^t_n(h^{\prime },c^{\prime })$ is added to player $n$’s utility value at every node in $Z(h)$. Hence, player $n$’s optimal set of continuation strategies at $h$ is the same under $\tilde{w}^t$ as it is under ${w}^t$. Continue with this procedure with the remaining of information sets in $H_n$. Repeat it, in the same fashion, for every other player $m\ne n$. Since the game is finite, we stop after a finite number of steps obtaining a payoff vector $u^t$ such that, by construction, $(u^t,\varepsilon ^t,s^t)\in W^{\circ }$.

To see that $\{u^t\}_{t=1}^{\infty }$ converges to $u$ note that, if it did not, there would be a player $n$, an information set $h\in H_n$ and a choice $c\in C(h)$ with $s_n(c\mid h)>0$ such that $\{x^t(c,h)\}_{t=1}^{\infty }$ does not converge to zero. Letting $h$ be a last information set in $H_n$ within those with such a property, for some $c\in C(h)$, the (bounded) sequence $\{x^t(c,h)\}_{t=1}^{\infty }$ (passing to a sub-sequence if necessary) converges to some strictly positive number. Continuity of the function $v^h_n$ implies that $c$ is not a sequentially rational choice at $h$ in the original game $\Gamma (u)$.

Conversely, let $\{(u^t,\varepsilon ^t,s^t)\}_{t=1}^{\infty }\subset W^{\circ }$ converge to $(u,0,s)$. Letting $\mu ^t$ be derived from $s^t$ using Bayes rule, we see that the limit point $(\mu ,s)$ of $\{(\mu ^t,s^t)\}^{\infty }_{t=1}$ is a consistent assessment. Sequential rationality of $(\mu ,s)$ in $\Gamma (u)$ follows again from the continuity of $v^h_n$. Hence, $s\in SE (u)$.