Behavior and deliberation in perfect-information games: Nash equilibrium and backward induction

Abstract

Doxastic characterizations of the set of Nash equilibrium outcomes and of the set of backward-induction outcomes are provided for general perfect-information games (where there may be multiple backward-induction solutions). We use models that are behavioral, rather than strategy-based, where a state only specifies the actual play of the game and not the hypothetical choices of the players at nodes that are not reached by the actual play. The analysis is completely free of counterfactuals and no belief revision theory is required, since only the beliefs at reached histories are specified.

A Proofs

Before proving Proposition 1 we introduce some notation and a definition.

Let G be a perfect-information game and $\sigma$ a pure-strategy profile of G. Let $f_{\sigma } : H \rightarrow Z$ (recall that H is the set of histories and Z is the set of terminal histories) be defined as follows: if $z\in Z$ then $f_{\sigma }(z)=z$ and if $h\in D$ (recall that D is the set of decision histories) then $f_{\sigma }(h)$ is the terminal history reached from h by following the choices prescribed by $\sigma$.

Definition 11

Let G be a perfect-information game and $\sigma$ a pure-strategy profile of G. The model of G generated by $\sigma$ is the following model:

• $\varOmega = Z$.

• $\zeta : Z \rightarrow Z$ is the identity function: $\zeta (z) = z, \forall z\in Z$.

• For every $h\in D$, ${\mathcal {B}}_h \subseteq Z\times Z$ is defined as follows: ${\mathcal {B}}_h (z)\ne \varnothing$ if and only if $h\prec z$ and $z^{\prime }\in {\mathcal {B}}_h (z)$ if and only if $z^{\prime }=f_{\sigma }(ha)$ for some $a\in A(h)$ (recall that A(h) is the set of actions available at h). That is, the active player at decision history h believes that if she takes action a then the outcome will be the terminal history reached from ha by $\sigma$.

Figure 8 shows an extensive form with perfect information and the model generated by the strategy profile $\sigma = (a_1,b_1,c_1,d_1)$ ($\sigma$ is highlighted by double edges).

Remark 7

Let G be a perfect-information game and ${\mathcal {M}}$ the model generated by a pure-strategy profile $\sigma$ of G. Then the no-uncertainty condition (Definition 7) is satisfied at every state, that is, ${\mathbf {C}} = Z$. Furthermore, if $z^{*}$ is the play generated by $\sigma$ (that is, $z^{*} = f_{\sigma }(\emptyset )$), then $z^{*}\in {\mathcal {B}}_h (z^{*})$ for all $h\in D$ such that $h\prec z^{*}$; that is, $z^{*}\in {\mathbf {T}}$.

Fig. 8

A game and the model generated by the strategy profile $\sigma = (a_1,b_1,c_1,d_1)$

Proof of Proposition 1

(A) Fix a perfect-information game G (not necessarily one that satisfies the no-consecutive-moves condition) and let $\sigma$ be a pure-strategy Nash equilibrium of G. If h is a decision history, to simplify the notation we shall write $\sigma (h)$ instead of $\sigma _{\iota (h)}(h)$ to denote the choice selected by $\sigma$ at h. Consider the model generated by $\sigma$ (Definition 11). Let $z^{*}$ be the play generated by $\sigma$, that is, $z^{*} = f_{\sigma }(\emptyset )$. By Remark 7, $z^{*}\in {\mathbf {C}}\cap {\mathbf {T}}$. Thus it only remains to show that $z^{*}\in {\mathbf {R}}$, that is, that $z^{*}\in {\mathbf {R}}_h$, for all $h\in D$ such that $h\prec z^{*}$. Fix an arbitrary $h\in D$ such that $h\prec z^{*}$ and let a be the action at h such that $ha\preceq z^{*}$, that is, $\sigma (h)=a$; then $f_{\sigma }(ha)=f_{\sigma }(\emptyset )= z^{*}$. Suppose that $z^{*}\notin {\mathbf {R}}_h$. Then there is an action $b\in A(h){\setminus } \{a\}$ that guarantees a higher utility to player $\iota (h)$, that is, if $z^{\prime }\in {\mathcal {B}}_h (z^{*})$ is such that $hb\preceq z^{\prime }$, then $u_{\iota (h)}(z^{\prime })>u_{\iota (h)}(z^{*})$. By Definition 11, $z^{\prime }=f_{\sigma }(hb)$ and thus $u_{\iota (h)}(f_{\sigma }(hb))>u_{\iota (h)}(f_{\sigma }(ha))$ so that by unilaterally changing his strategy at h from a to b (and leaving the rest of his strategy unchanged), player $\iota (h)$ can increase his payoff, contradicting the assumption that $\sigma$ is a Nash equilibrium.

(B) Let G be a perfect-information game that satisfies the no-consecutive-moves condition (Definition 8) and consider a model of it where there is a state $\alpha$ such that $\alpha \in {\mathbf {T}}\cap {\mathbf {R}}\cap {\mathbf {C}}$. We need to construct a pure-strategy Nash equilibrium $\sigma$ of G such that $f_{\sigma }(\emptyset )=\zeta (\alpha )$.

Step 1. For every $h\in D$ such that $h\prec \zeta (\alpha )$, let $\sigma (h)=a$ where $a\in A(h)$ is the action at h such that $ha\preceq \zeta (\alpha )$.

Step 2. Fix an arbitrary $h\in D$ such that $h\prec \zeta (\alpha )$ and an arbitrary $b\in A(h)$ such that $b\ne \sigma (h)$ ($\sigma (h)$ was defined in Step 1). Since $\alpha \in {\mathbf {C}}$, for every $\omega ,\omega ^{\prime }\in {\mathcal {B}}_h(\alpha )$ such that $hb\preceq \zeta (\omega )$ and $hb\preceq \zeta (\omega ^{\prime })$, $\zeta (\omega )=\zeta (\omega ^{\prime })$. Select an arbitrary $\omega \in {\mathcal {B}}_h(\alpha )$ such that $hb\preceq \zeta (\omega )$ and define, for every $h^{\prime }\in D$ such that $hb\preceq h^{\prime }\prec \zeta (\omega )$, $\sigma (h^{\prime })=c$ where $c\in A(h^{\prime })$ is the action at $h^{\prime }$ such that $h^{\prime } c\preceq \zeta (\omega )$.

So far we have defined the choices prescribed by $\sigma$ along the play to $\zeta (\alpha )$ and for paths at one-step deviations from this play. This is illustrated in Fig. 9, where ${\mathbf {T}}\cap {\mathbf {R}}\cap {\mathbf {C}}=\{\delta \}$.⁴⁷ Focusing on state $\delta$, the above two steps yield the following partial strategy profile (which is highlighted by double edges). By Step 1, $\sigma (\emptyset )=a_1, \sigma (a_1)=b_2$ and, by Step 2, $\sigma (a_2)=d_1, \sigma (a_1b_1)=c_1, \sigma (a_1b_1c_1)=e_1$, while $\sigma (a_2d_2)$ and $\sigma (a_1b_1c_2)$ are left undefined by Steps 1 and 2.

Fig. 9

A game and a model of it

Step 3. Complete $\sigma$ in an arbitrary way.⁴⁸

Because of Step 1, $\zeta (\alpha )=f_{\sigma }(h)$, for every $h\preceq \zeta (\alpha )$. We want to show that $\sigma$ is a Nash equilibrium. Suppose not. Then there is a decision history h with $h\prec \zeta (\alpha )$ such that, by changing her choice at h from $\sigma (h)$ to a different choice, player $\iota (h)$ can increase her payoff (recall the assumption that the game satisfies the no-consecutive-moves assumption and thus there are no successors of h that belong to player $\iota (h)$). Let $\sigma (h)=a$ (thus $ha\preceq \zeta (\alpha )$) and let b be the choice at h that yields a higher payoff to player $\iota (h)$; that is,

$$\begin{aligned} u_{\iota (h)}(f_{\sigma }(hb))>u_{\iota (h)}(\zeta (\alpha )). \end{aligned}$$

(2)

Let $\omega \in {\mathcal {B}}_h(\alpha )$ be such that $hb\preceq \zeta (\omega )$ (such an $\omega$ exists by Point 4 of Definition 2). Since $\alpha \in {\mathbf {C}}$, for every $\omega ^{\prime }\in {\mathcal {B}}_h(\alpha )$ such that $hb\preceq \zeta (\omega ^{\prime })$, $\zeta (\omega )=\zeta (\omega ^{\prime })$. By Step 2 above,

$$\begin{aligned} \zeta (\omega )=f_{\sigma }(hb). \end{aligned}$$

(3)

It follows from (3) that, at state $\alpha$ and history h, player $\iota (h)$ believes that if she plays b her payoff will be $u_{\iota (h)}(f_{\sigma }(hb))$. Since $\alpha \in {\mathbf {T}}$, $\alpha \in {\mathcal {B}}_h(\alpha )$, and since $\alpha \in {\mathbf {C}}$, for every $\omega ^{\prime }\in {\mathcal {B}}_h(\alpha )$ such that $ha\preceq \zeta (\omega ^{\prime })$, $\zeta (\omega ^{\prime })=\zeta (\alpha )$. Thus, at state $\alpha$ and history h, player $\iota (h)$ believes that if she plays a her payoff will be $u_{\iota (h)}(\zeta (\alpha ))$. It follows from this and (2) that at $\alpha$ and h player $\iota (h)$ believes that action b is better than action a, which implies that $\alpha \notin {\mathbf {R}}_h$, contradicting the assumption that $\alpha \in {\mathbf {R}}\subseteq {\mathbf {R}}_h$. $\square$

Before proving Proposition 2 we need to define the length of a game.

Definition 12

The length of a history h, denoted by L(h), is defined recursively as follows: $L(\emptyset )=0$ and, for every $a\in A(h)$, $L(ha)=L(h)+1$; thus the length of history h is the number of actions in h. The length of a game, denoted by $\ell$, is the length of a longest history in the game: $\ell = \mathop {\max }\limits _{h \in H} \{ L(h)\}$.

Proof of Proposition 2

(A) Fix a perfect-information game G and let the pure-strategy profile $\sigma$ be a backward-induction solution of G (that is, a possible output of the backward-induction algorithm). Consider the model generated by $\sigma$ (Definition 11); then—by construction—for every terminal history z and every decision history h such that $h\prec z$,

$$\begin{aligned} \forall z^{\prime }\in Z,\ z^{\prime }\in {\mathcal {B}}_{h}(z) \text { if and only if }\ z^{\prime }=f_{\sigma }(ha) \ \text {for some } \ a\in A(h). \end{aligned}$$

(4)

It follows from this that⁴⁹

$$\begin{aligned} \forall h\in D,\ \text { if } z=f_{\sigma }(h)\ \text { then } z\in {\mathbf {T}}_h \end{aligned}$$

(5)

and⁵⁰

$$\begin{aligned} \forall z\in Z \quad \text { and }\quad \forall h\in D \quad \text {such that }\quad h\prec z, \ z\in {\mathbf {C}}_h. \end{aligned}$$

(6)

Let $z^{*}$ be the play generated by $\sigma$, that is, $z^{*} = f_{\sigma }(\emptyset )$. Since every backward-induction solution is a Nash equilibrium, it follows from Part A of Proposition 1 that $z^{*}\in {\mathbf {T}}\cap {\mathbf {R}}\cap {\mathbf {C}}$. Let $\ell$ be the length of the game. If $\ell = 1$ there is nothing further to prove. Assume, therefore, that $\ell \ge 2$. We need to show that $z^{*}\in {\mathbf {I}}_{TRC}$, that is, that, for every decision history $h=a_1 \ldots a_m$ ($m \ge 1$) and for every sequence $\langle z_0,z_1,\ldots ,z_m\rangle$ that leads from $z^{*}$ to h (see Definition 9), $z_m\in {\mathbf {T}}_h\cap {\mathbf {R}}_h\cap {\mathbf {C}}_h$; however, by (6), we only need to show that $z_m\in {\mathbf {T}}_h\cap {\mathbf {R}}_h$. Let $h=a_1 \ldots a_m$ ($m\ge 1$) be a decision history and let $\langle z_0,z_1,\ldots ,z_m\rangle$ be a sequence that leads from $z^{*}$ to h (such a sequence exists: see Remark 3). By Definition 9, $z_m\in {\mathcal {B}}_{a_1 \ldots a_{m -1}}(z_{m-1})$, so that, by (4), $z_m=f_{\sigma }(a_1 \ldots a_{m -1}b)$ for some $b\in A(a_1 \ldots a_{m -1})$; hence $a_1 \ldots a_{m -1}b\preceq z_m$. Again by Definition 9, $a_1 \ldots a_{m -1}a_m=h\preceq z_m$ and thus $b=a_m$ so that

$$\begin{aligned} z_m=f_{\sigma }(h). \end{aligned}$$

(7)

Hence, by (5), $z_m\in {\mathbf {T}}_h$.

Let $a\in A(h)$ be the action taken at h at state $z_m$ (that is, $ha\preceq z_m$). It follows from (7) that $a=\sigma (h)$. Furthermore, by (4), for every $z^{\prime }\in {\mathcal {B}}_h(z_m)$, $z^{\prime }=f_{\sigma }(ha^{\prime })$ for some $a^{\prime }\in A(h)$. Hence at state $z_m$ and history h player $\iota (h)$ believes that after any choice $a^{\prime }$ at h the outcome will the one generated by $\sigma$ starting from $ha^{\prime }$ (that is, the backward-induction outcome induced by $\sigma$ in the subtree that starts at history $ha^{\prime }$). Furthermore, by (7), the action that she takes at h is $\sigma (h)$, the backward-induction choice prescribed by $\sigma$. Hence player $\iota (h)$ is rational at h and $z_m$, that is, $z_m\in {\mathbf {R}}_h$.

(B) Let G be a perfect-information game. Consider a model of G and a state $\alpha$ in that model such that $\alpha \in ({\mathbf {T}}\cap {\mathbf {R}}\cap {\mathbf {C}})\cap {\mathbf {I}}_{TRC}$. We want to show that $\zeta (\alpha )$ is a backward-induction outcome. Let $\ell$ be the length of the game. If $\ell = 1$ then every successor of $\emptyset$ (the root of the tree) is a terminal history. Hence, since $\alpha \in {\mathbf {R}}\subseteq {\mathbf {R}}_{\emptyset }$, the action chosen at $\emptyset$ at state $\alpha$ maximizes player $\iota (\emptyset )$’s payoff and thus is a backward-induction choice. Assume, therefore, that $\ell \ge 2$.

Step 1. First we show that, at every decision history of length $\ell - 1$ that is reachable from $\alpha$, the action chosen there is a backward-induction choice. Fix an arbitrary decision history $h=a_1 \ldots a_{\ell -1}$ of length $\ell -1$ (thus every successor of h is a terminal history) and let $\langle \omega _0,\omega _1, \ldots ,\omega _{\ell -1}\rangle$ be a sequence in $\varOmega$ that leads from $\alpha$ to h (such a sequence exists: see Remark 3), that is, (1) $\omega _0=\alpha$, (2) for every $i=1,\ldots ,\ell -1$, $a_1 \ldots a_i \prec \zeta (\omega _i)$, (3) $\omega _1\in {\mathcal {B}}_{\emptyset }(\alpha )$ and, for every $i=2,\ldots ,\ell -1$, $\omega _i\in {\mathcal {B}}_{a_1 \ldots a_{i-1}}(\omega _{i-1})$. Since, by hypothesis, $\alpha \in {\mathbf {I}}_{TRC}$, $\omega _{\ell -1}\in {\mathbf {R}}_{h}$ and thus if b is the action taken at history h at state $\omega _{\ell -1}$ (that is, $\zeta (\omega _{\ell -1})=hb$), then b maximizes the payoff of player $\iota (h)$, that is, b is a backward-induction choice at h.

Step 2. Next we show that, at every decision history of length $\ell - 2$, the active player believes that, for every $a\in A(h)$, if ha is a decision history then the action chosen at ha is a backward-induction action. Fix an arbitrary decision history $h=a_1 \ldots a_{\ell -2}$ of length $\ell -2$ and let $\langle \omega _0,\omega _1, \ldots ,\omega _{\ell -2}\rangle$ be a sequence in $\varOmega$ that leads from $\alpha$ to h (see Remark 3). Let $a\in A(h)$ be such that ha is a decision history and let $\omega \in {\mathcal {B}}_h(\omega _{\ell -2})$ be such that $ha\preceq \zeta (\omega )$ (such an $\omega$ exists by Point 4 of Definition 2). Then the sequence $\langle \omega _0,\omega _1, \ldots ,\omega _{\ell -2},\omega \rangle$ reaches ha from $\alpha$ and thus, by Step 1, the action chosen by the active player at ha is a backward-induction action (that is, if $\zeta (\omega )=hab$, with $b\in A(ha)$, then b is a backward-induction choice at ha). Furthermore, since $\alpha \in {\mathbf {I}}_{TRC}$, $\omega _{\ell -2}\in {\mathbf {C}}_{h}$ and thus, for every other $\omega ^{\prime }\in {\mathcal {B}}_h(\omega _{\ell -2})$ such that $ha\preceq \zeta (\omega ^{\prime })$, $\zeta (\omega ^{\prime })=\zeta (\omega )$ and thus, at h and $\omega _{\ell -2}$, player $\iota (h)$ believes that if she takes action a at h then the ensuing outcome is backward-induction outcome $\zeta (\omega )$. From $\alpha \in {\mathbf {I}}_{TRC}$ it also follows that $\omega _{\ell -2}\in {\mathbf {R}}_{h}$ and thus the action chosen by player $\iota (h)$ at h at state $\omega _{\ell -2}$ is optimal given her beliefs that after every choice a at h the outcome following ha is a backward-induction outcome. Finally, from $\alpha \in {\mathbf {I}}_{TRC}$ it follows that $\omega _{\ell -2}\in {\mathbf {T}}_{h}$ so that $\omega _{\ell -2}\in {\mathcal {B}}_h(\omega _{\ell -2})$ and thus player $\iota (h)$ has correct beliefs at h and at state $\omega _{\ell -2}$ about the outcome following the action actually taken at h and at $\omega _{\ell -2}$ (that is, if $\hat{a}$ is such that $h\hat{a}\preceq \zeta (\omega _{\ell -2})$ then player $\iota (h)$ believes that if she takes action $\hat{a}$ then the outcome will be the backward-induction outcome $\zeta (\omega _{\ell -2})$). Thus $\zeta (\omega _{\ell -2})$ is a backward-induction outcome in the subtree that starts at history h.

Step 3. Iterate the argument of Step 2 backwards to conclude that if $a\in A(\emptyset )$ is decision history of length 1 that is reachable from $\alpha$ via a sequence of the form $\langle \alpha ,\beta \rangle$, then $\zeta (\beta )$ is a backward-induction outcome in the subtree that starts at history a.

Step 4. Use the fact that $\alpha \in {\mathbf {T}}_{\emptyset }\cap {\mathbf {C}}_{\emptyset }$ to conclude that at state $\alpha$ and history $\emptyset$ player $\iota (\emptyset )$ has correct and certain beliefs about the outcomes following decision histories in $A(\emptyset )$ and thus, using the fact that $\alpha \in {\mathbf {R}}_{\emptyset }$ and the conclusion of Step 3, deduce that the action taken at state $\alpha$ by $\iota (\emptyset )$ is a backward induction action, so that $\zeta (\alpha )$ is a backward-induction outcome. $\square$