Bayesian games with unawareness and unawareness perfection

Abstract

Applying unawareness belief structures introduced in Heifetz et al. (Games Econ Behav 77:100–121, 2013a), we develop Bayesian games with unawareness, define equilibrium, and prove existence. We show how equilibria are extended naturally from lower to higher awareness levels and restricted from higher to lower awareness levels. We apply Bayesian games with unawareness to investigate the robustness of equilibria to uncertainty about opponents’ awareness of actions. We show that a Nash equilibrium of a strategic game is robust to unawareness of actions if and only if it is not weakly dominated. Finally, we discuss the relationship between standard Bayesian games and Bayesian games with unawareness.

Appendices

Appendix A: Bayesian games with unawareness: allowing for unawareness of players

So far, we did not allow for unawareness of players. In standard Bayesian game theory, ignorance of players is modeled by dummy players. This is distinct from being unable to conceive of a player at all. In this subsection, we allow for unawareness of players. This requires that we generalize our interactive unawareness belief structure such that a player may exist only at some states but not at others. Such a generalization may be useful to extensions to psychological games with unawareness (see Nielsen and Sebald 2011). For instance, being aware of an observer may affect behavior even if the observer has no active role in the game. This is known in psychology as the observer effect.

Definition 8

A Bayesian game with unawareness (that also allows for unawareness of players) is a tuple

$$\begin{aligned} \Gamma (\underline{\mathcal {S}}) := \left\langle \mathcal {S}, \left( r_{S_{\beta }}^{S_{\alpha }} \right) _{S_{\beta } \preceq S_{\alpha }}, \mathcal {E}, \left( t_i \right) _{i \in I}, \left( M_i \right) _{i \in I}, (\mathcal {M}_i)_{i \in I}, \left( u_i \right) _{i \in I} \right\rangle \end{aligned}$$

defined as follows:

• (0) $\mathcal {S} = \{S_{\alpha }\}_{\alpha \in \mathcal {A}}$ is as before a complete lattice of spaces with surjective and commuting projections $(r_{S_\beta }^{S_\alpha })$, for $S_{\beta } \preceq S_{\alpha }$ (see Sect. 2).

• (i) $\mathcal {E}: I \longrightarrow \Sigma$ is the “existence” correspondence that assigns to each player $i \in I$ an event in which she exists. Moreover, $\mathcal {S}_i := \{ S \in \mathcal {S} : \mathcal {E} (i) \cap S \ne \emptyset \}$ is the complete sublattice of spaces with states in which player $i$ exists.

• (ii) For every player $i \in I$, $t_i: \mathcal {E}(i) \longrightarrow \bigcup _{S \in \mathcal {S}_i} \Delta \left( S \right)$ is a type mapping that satisfies Properties (i)–(iii) (see Sect. 2) such that for every $\omega \in \mathcal {E}(i)$, $t_i(\omega )(\mathcal {E}(i)) = 1$.

• (iii) For every $i \in I$, $M_{i}$ is a nonempty finite set of actions. $\mathcal {M}_i: \mathcal {E}(i) \longrightarrow 2^{M_i} {\setminus } \{\emptyset \}$ is a correspondence, for $i \in I$, with the following properties:

  • a. For every $M'_i$ with $\emptyset \ne M_i' \subseteq M_i$: If there is a state $\omega$ with $\omega \in \mathcal {E}(i)$, then the set $\{\omega ' \in \Omega : M'_i \subseteq \mathcal {M}_i(\omega ') \}$ is an event.

  • b. If $\omega \in \mathcal {E}(i)$ and $\omega ', \omega '' \in S_{t_i(\omega )} \cap [t_i(\omega )]$, then $\mathcal {M}_i(\omega ') = \mathcal {M}_i(\omega '')$.

• (iv) Further, we impose introspection as follows: For $\omega \in \mathcal {E}(i)$,

  $$\begin{aligned} t_i(\omega )\left( \{\omega ' \in \mathcal {E}(i): t_i(\omega ')_{|S_{t_i(\omega )}} = t_i(\omega ) \}\right) = 1. \end{aligned}$$

• (v) For $i \in I$, $u_{i}:\bigcup _{\omega \in \mathcal {E}(i)} \left( \left( \prod _{j \in I(\omega )} \mathcal {M}_j(\omega ) \right) \times \{\omega \} \right) \longrightarrow \mathbb {R}$ is the utility function of $i$, where $I(\omega ) := \{i \in I : \omega \in \mathcal {E}(i)\}$.

This game allows for unawareness of events, actions, outcomes, and players. For every player $i \in I$, the “existence” correspondence $\mathcal {E}$ assigns to $i$ the event in which she exists. Consequently, we restrict player $i$’s type mapping to states at which she exists. Moreover, player $i$’s type is concentrated only on states in which she exists. A player cannot assign strict positive probability to states at which she does not exist. The correspondence $\mathcal {M}_i$ assigns a nonempty set of actions for player $i$ only to the set of states in which player $i$ exists. The dimension of the domain of a utility function may vary from state to state, since players may exist in some states but not in others, and each players utility at a state depends on the actions of all the players that exist in that state.

Note that if $\mathcal {E}(i) = \Omega$ for all $i \in I$, then we obtain an unawareness belief structure and a Bayesian game with unawareness as defined before.

Note further that if $\omega \in \mathcal {E}(i)$, then $[t_i(\omega )] := \{\omega ^{\prime }\in \Omega : t_i(\omega ^{\prime }) = t_i(\omega ) \} \subseteq \mathcal {E}(i)$.

A strategy of player $i$ is now adapted to the event in which she exists.

Definition 9

A strategy of player $i$ in a Bayesian game with unawareness is a function $\sigma _{i}: \mathcal {E}(i) \longrightarrow \Delta (M_{i})$ such that for all $\omega \in \mathcal {E}(i)$,

• (i) $\sigma _i(\omega ) \in \Delta \left( \mathcal {M}_i(\omega _{S_{t_i}(\omega )})\right)$.

• (ii) $t_{i}(\omega ^{\prime })=t_{i}(\omega )$ implies $\sigma _{i}(\omega ^{\prime })=\sigma _{i}(\omega )$.

Denote $\sigma _{S_{t_{i}(\omega )}} := ( ( \sigma _{j}(\omega ^{\prime }) )_{j \in I(\omega ^{\prime })} )_{\omega ^{\prime }\in S_{t_{i}(\omega )}}$. The expected utility of player-type $(i,t_{i}(\omega ))$ from the strategy profile $\sigma _{S_{t_{i}(\omega )}}$ is given by

$$\begin{aligned}&U_{(i, t_{i}(\omega ))}(\sigma _{S_{t_{i}(\omega )}}) := \nonumber \\&\quad \sum _{\omega ^{\prime } \in S_{t_{i}(\omega )}}\,\,\sum _{m \in \prod _{j \in I(\omega ')} \mathcal {M}_j \left( \omega _{S_{t_{j} (\omega ^{\prime })}}^{\prime }\right) } \left( \,\,\prod _{j\in I(\omega ^{\prime })} \sigma _{j}(\omega ^{\prime }) \left( \{m_{j}\} \right) \right) \cdot u_{i} \left( (m_j)_{j \in I(\omega ')}, \omega ^{\prime }\right) t_{i}(\omega )(\{\omega ^{\prime }\}).\nonumber \\ \end{aligned}$$

(2)

This expression is analogous to Eq. (1) except that the set of players $I$ is now replaced by $I(\omega ')$.

Definition 5 and Proposition 1 are now generalized to the following:

Definition 10

(Equilibrium) Given a Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}})$, define the associated strategic game by

• (i) $\{(i,t_i(\omega )): \omega \in \Omega \text{ and } i \in I(\omega )\}$ is the set of players, and for each player $(i, t_i(\omega ))$,

• (ii) the set of mixed strategies is $\Delta (\mathcal {M}_i(\omega _{S_{t_i}(\omega )})$, and

• (iii) the utility function is given by Eq. (2).

A profile $(\sigma _i)_{i \in I}$ is an equilibrium of the Bayesian game with unawareness if and only if the following is an equilibrium of the associated strategic game: $\left( i, t_i(\omega )\right)$ plays $\sigma _i(\omega )$, for all $i \in I(\omega )$ and $\omega \in \Omega$.

Proposition 3

(Existence) Let $\Gamma (\underline{\mathcal {S}}) = \big \langle \mathcal {S},(r_{S_{\beta }}^{S_{\alpha }} )_{S_{\beta } \preceq S_{\alpha }}, \mathcal {E}, (t_i )_{i \in I}, (M_i)_{i \in I}, (\mathcal {M}_i)_{i \in I}, (u_i)_{i \in I} \big \rangle$ be a Bayesian game with unawareness. If $I$, $\Omega$, and $(M_{i})_{i \in I}$ are finite, then there exists an equilibrium.

Proof

By Nash’s (1950) theorem.

Recall $l(S) := \{S^{\prime }\in \mathcal {S} : S^{\prime }\preceq S\}$. Definition 6, Proposition 2, and Remark 1 are now generalized, respectively, as follows:

Definition 11

Given a Bayesian game with unawareness

$$\begin{aligned} \Gamma (\underline{\mathcal {S}}) = \left\langle \mathcal {S},\left( r_{S_{\beta }}^{S_{\alpha }}\right) _{S_{\beta }\preceq S_{\alpha }},\mathcal {E},\left( t_{i}\right) _{i\in I},\left( M_{i}\right) _{i\in I}, (\mathcal {M}_i)_{i \in I},\left( u_{i}\right) _{i\in I}\right\rangle , \end{aligned}$$

we can define for any $S' \in \mathcal {S}$ an $S^{\prime }$-partial Bayesian game with unawareness

$$\begin{aligned}&\Gamma (\underline{l(S')}) \\&\quad = \left\langle l(S^{\prime }),\left( r_{S_{\beta }}^{S_{\alpha }}\right) _{S_{\beta }\preceq S_{\alpha } \preceq S'}, \mathcal {E}^{\prime },\left( t_{i}\right) _{i\in I(\Omega ')},\left( M_{i}\right) _{i\in I(\Omega ')}, (\mathcal {M}'_i)_{i \in I(\Omega ')},\left( u_{i}\right) _{i\in I(\Omega ')}\right\rangle \!, \end{aligned}$$

in which $\mathcal {E}^{\prime }(i)=\mathcal {E}(i)\cap \Omega ^{\prime }$, where $\Omega ^{\prime } = \bigcup _{S^{''} \in l(S^{\prime })} S^{''}$, and for any $i \in I(\Omega ') := \bigcup _{\omega \in \Omega '} I(\omega )$, $\mathcal {M}'_i$ is $\mathcal {M}_i$ restricted to $\mathcal {E}^{\prime }(i)$.

Proposition 4

(“Upwards Induction”) Given a Bayesian game with unawareness $\big \langle \mathcal {S},( r_{S_{\beta }}^{S_{\alpha }}) _{S_{\beta }\preceq S_{\alpha }},\mathcal {E},( t_{i}) _{i\in I},( M_{i}) _{i\in I}, (\mathcal {M}_i)_{i \in I},( u_{i}) _{i\in I}\big \rangle$, consider for $S^{\prime }, S^{\prime \prime }\in \mathcal {S}$ with $S^{\prime } \preceq S^{\prime \prime }$ the $S^{\prime }$-partial (resp. $S^{\prime \prime }$-partial) Bayesian game with unawareness. If $I$, $\Omega$, and $(M_{i})_{i \in I}$ are finite, then for every equilibrium of the $S^{\prime }$-partial Bayesian game, there is an equilibrium of the $S^{\prime \prime }$-partial Bayesian game in which equilibrium strategies of player-types in $\{(i, t_i(\omega )) : \omega \in \Omega ' = \bigcup _{S \in l(S^{\prime })} S \text{ and } i \in I(\Omega ') \}$ are identical with the equilibrium strategies in the $S^{\prime }$-partial Bayesian game.

Proof

Let $\sigma ^*_{| \Omega ^{\prime }}$ be an equilibrium in the $S^{\prime }$-partial Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^{\prime })$. For $S^{\prime \prime }\succeq S^{\prime }$, we define a strategic form game with

• $I(\Omega ^{\prime \prime }{\setminus } \Omega ^{\prime }) := \{(i, t_i(\omega )): \omega \in \Omega ^{\prime \prime }, i \in I(\omega ) \} {\setminus } \{(i, t_i(\omega )): \omega \in \Omega ^{\prime }, i \in I(\omega ) \}$ being the set of players,

• the set of strategies of player $(i, t_i(\omega )) \in I(\Omega ^{\prime \prime } {\setminus } \Omega ^{\prime })$ is $\Delta (\mathcal {M}_i(\omega _{S_{t_i(\omega )}}))$,¹³

• the payoff function of player $(i, t_i(\omega ))$ is given by Eq. (1) but fixing the strategy of each (dummy) player in $\{(i, t_i(\omega ^{\prime })): \omega ^{\prime }\in \Omega ^{\prime }, i \in I(\omega ^{\prime }) \}$ to her respective equilibrium strategy $\sigma ^*_i(\omega )$ of the $S^{\prime }$-partial Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^{\prime })$.

Since $I$, $\Omega$, and $(M_{i})_{i \in I}$ are finite, this strategic game has an equilibrium by Nash’s (1950) theorem. Fix one equilibrium of this game.

Consider now the strategy profile $\sigma ^*_{| \Omega ^{\prime \prime }}$ in which players in $\{(i, t_i(\omega )): \omega \in \Omega ^{\prime }, i \in I(\omega ) \}$ play their component of the profile $\sigma ^*_{| \Omega ^{\prime }}$ and players in $I(\Omega ^{\prime \prime } {\setminus } \Omega ^{\prime })$ play the equilibrium strategies of the equilibrium in the above defined strategic game.

We need to show that $\sigma ^*_{| \Omega ^{\prime \prime }}$ is an equilibrium of the $S^{\prime \prime }$-partial Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^{\prime \prime })$. Suppose not, then for some player $(i, t_i(\omega )) \in I(\Omega ^{\prime \prime }) = \{(i, t_i(\omega ^{\prime })) : \omega ^{\prime }\in \Omega ^{\prime \prime }, i \in I(\omega ^{\prime }) \}$ there exists $\sigma _i(\omega ) \in \Delta (\mathcal {M}_i(\omega _{S_{t_i(\omega )}}))$ with $\sigma _i(\omega ) \ne \sigma _i^*(\omega )$ such that for $\sigma _{S_{t_i( \omega )}} := (\sigma _i(\omega ), (\sigma _{j}^*(\omega ^{\prime }))_{\omega ^{\prime }\in S_{t_i(\omega )}, j \in I(\omega ) {\setminus } \{i\}})$ we have

$$\begin{aligned} U_{(i, t_i(\omega ))}\left( \sigma _{S_{t_i(\omega )}}\right) > U_{(i, t_i(\omega ))}\left( \sigma ^*_{S_{t_i(\omega )}}\right) , \end{aligned}$$

i.e., there exists a profitable deviation from $\sigma ^*_{| \Omega ^{\prime \prime }}$ for some player-type $(i, t_i(\omega ))$ with $\omega \in \Omega ^{\prime \prime }$ and $i \in I(\omega )$ given that all other player-types in $I(\Omega ^{\prime \prime })$ play their equilibrium strategy.

If $(i, t_i(\omega )) \in I(\Omega ^{\prime \prime }{\setminus } \Omega ^{\prime })$, then her strategy is not an equilibrium strategy in the above defined strategic game, a contradiction. If $(i, t_i(\omega )) \in \{(i, t_i(\omega )): \omega ^{\prime }\in \Omega ^{\prime }, i \in I(\omega ^{\prime })\}$, then since her payoffs are identical in both games, and her strategy is not an equilibrium strategy in the $S^{\prime }$-partial Bayesian game with unawareness $\Gamma ( \underline{\mathcal {S}}')$, a contradiction. Hence, $\sigma ^*_{|\Omega ^{\prime \prime }}$ must be an equilibrium of the $S^{\prime \prime }$-partial Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^{\prime \prime })$. $\square$

It is easy to construct examples of Bayesian games with unawareness in which unaware players may have commitment power and the “value of awareness” may be negative. For instance, in a simultaneous-move linear Cournot duopoly, a player who is unaware of his opponent can obtain the Stackelberg leader profit if the opponent knows that the first player is unaware of him.¹⁴

The converse to Proposition 4 follows from the consistency of Nash equilibrium (see Peleg and Tijs 1996; Peleg et al. 1996):

Remark 5

Let $\big \langle \mathcal {S},( r_{S_{\beta }}^{S_{\alpha }}) _{S_{\beta }\preceq S_{\alpha }},\mathcal {E},( t_{i}) _{i\in I},( M_{i})_{i\in I},(\mathcal {M}_i)_{i \in I},( u_{i}) _{i\in I}\big \rangle$ be a Bayesian game with unawareness. Consider for $S^{\prime }, S^{\prime \prime }\in \mathcal {S}$ with $S^{\prime }\preceq S^{\prime \prime }$ the $S^{\prime }$-partial (resp. $S^{\prime \prime }$-partial) Bayesian game with unawareness. Then for every equilibrium of the $S^{\prime \prime }$-partial Bayesian game there is a unique equilibrium of the $S^{\prime }$-partial Bayesian game in which the equilibrium strategies of player-types in $\{(i, t_i(\omega )) : \omega \in \Omega ' = \bigcup _{S \in l(S^{\prime })} S \text{ and } i \in I(\Omega ') \}$ are identical to the equilibrium strategies of the $S^{\prime \prime }$-partial Bayesian game.

Appendix B: Connection to standard Bayesian games

In this section, we compare Bayesian games with unawareness to standard Bayesian games. In particular, we show how to derive a standard type-space with zero probability from our unawareness structure by “flattening” our lattice of spaces. “Flattening” the game is a purely technical procedure. While we can show a correspondence between equilibria in a Bayesian game with unawareness and equilibria in a standard Bayesian game, the equilibrium in the standard Bayesian game cannot be interpreted anymore under unawareness because the “language” required to identify events of which a player could be unaware is essentially “erased.” Since a flattened structure is a standard type-space, the “Dekel-Lipman-Modica-Rustichini critique” applies (Modica and Rustichini 1994; Dekel et al. 1998). Hence, unawareness is trivial in the flattened game.

Let $\underline{\mathcal {S}}$ be an unawareness belief structure. We define the flattened type-space associated with the unawareness belief structure $\underline{\mathcal {S}}$ by

$$\begin{aligned} F(\underline{\mathcal {S}}) := \langle \Omega , (t^F_i)_{i \in I} \rangle , \end{aligned}$$

where $\Omega$ is the union of all state-spaces in the unawareness belief structure $\underline{\mathcal {S}}$ and $t^F_i: \Omega \longrightarrow \Delta (\Omega )$ is defined by,

$$\begin{aligned} t^F_i(\omega )(E) := \left\{ \begin{array}{ll} t_i(\omega )(E \cap S_{t_i(\omega )}) &{}\quad \text{ if } E \cap S_{t_i(\omega )} \ne \emptyset \\ 0 &{} \quad \text{ otherwise. } \end{array} \right. \end{aligned}$$

Note that when flattening an unawareness belief structure, we lose the lattice structure of spaces and thus the event structure. For instance, recall that in an unawareness belief structure, the negation of the event $(D^{\uparrow }, S)$ with $D \subseteq S$ is defined by $\lnot (D^{\uparrow }, S) = ((S{\setminus } D)^{\uparrow },S)$. This is typically a proper subset of the complement $\Omega {\setminus } D^{^{\uparrow }}$. That is, $\left( S{\setminus } D\right) ^{\uparrow }\subsetneqq \Omega {\setminus } D^{^{\uparrow }}$. But it is precisely our lattice structure that allows us to circumvent the impossibility result by Dekel et al. (1998).

A standard type-space on $Y$ for the player set $I$ is a tuple

$$\begin{aligned} \underline{Y}:=\left\langle Y, \left( t_{i}\right) _{i \in I} \right\rangle , \end{aligned}$$

where $Y$ is a nonempty set and for $i \in I$, $t_{i}$ is a function from $Y$ to $\Delta (Y)$, the space of countably additive probability measures on $Y$, such that for all $\omega \in Y$, $t_{i}\left( \omega \right) \left( \left[ t_{i}\left( \omega \right) \right] \right) = 1$ with $\left[ t_{i}\left( \omega \right) \right] : = \left\{ \omega ^{\prime } \in Y : t_{i}\left( \omega ^{\prime }\right) = t_{i}\left( \omega \right) \right\}$ (i.e., introspection).

The properties of the type mapping in the unawareness belief structure $\underline{\mathcal {S}}$ implies immediately the following:

Proposition 5

If $\underline{\mathcal {S}}$ is an unawareness belief structure, then $F(\underline{\mathcal {S}})$ is a standard type-space. Moreover, it has the following property: For every $p > 0$, $E \subseteq \Omega$, and $i \in I$: $\{\omega \in \Omega : t_i(\omega )(E \cap S_{t_i(\omega )}) \ge p\} = \{\omega \in \Omega : t_i^F(\omega )(E) \ge p\}$.

Proof

We only have to show introspection. That is, for all $\omega \in \Omega$, $i \in I$, and $E \in \mathcal {F}$, $t_i^F(\omega )([t_i^F(\omega )]) = 1$ with $[t_i^F(\omega )] = \{\omega ^{\prime }\in \Omega : t_i^F(\omega ^{\prime }) = t_i^F(\omega ) \}$. But this follow directly from property (iv) of the type mapping in the unawareness belief structure $\underline{\mathcal {S}}$. $\square$

A flattened unawareness structure is just a standard type-space. To derive such a type-space, one extends a player’s type mapping by assigning probability zero to sets for which the player’s belief was previously undefined. Of course, once an unawareness structure is flattened, there is no way to analyze reasoning about unawareness anymore since by Dekel et al. (1998), unawareness is trivial.

Note that the converse to Proposition 5 is not true. That is, given a standard type-space, it is not always possible to find some unawareness structure with nontrivial unawareness. For instance, let $X = \{\omega _1, \omega _2, \omega _3\}$ with $t_i(\omega _1) = t_i(\omega _2) = t_i(\omega _3) = \tau _i$ and $\tau _i(\{\omega _1\}) = \tau _i(\{\omega _2\}) = \frac{1}{2}$ and $\tau _i(\{\omega _3\}) = 0$. If $\Omega = S = X$, then by Dekel et al. (1998), the unawareness structure has trivial unawareness only. Any nontrivial partition of $X$ into separate spaces yields either no projections or violates properties (i)–(iii). We conclude that not every standard types-space with zero probability can be used to model unawareness. We understand the contribution of our work as making restrictions required for modeling unawareness precise in unawareness belief structures.

Definition 12

(Flattened game) Given a Bayesian game with unawareness of events and (possibly) actions $\Gamma (\underline{\mathcal {S}})$, we can associate a standard Bayesian game $F(\Gamma (\underline{\mathcal {S}}))$ played on a standard type-space (with possibly allowing for varying action sets of the players across different types) in the following manner:

If $\Gamma (\underline{\mathcal {S}}) = \langle \underline{\mathcal {S}}, (M_i)_{i \in I}, (\mathcal {M}_i)_{i \in I}, (u_i)_{i \in I} \rangle$, where $\underline{\mathcal {S}} = \langle \mathcal {S}, (r_{S_\beta }^{S_\alpha })_{S_{\beta } \preceq S_{\alpha }}, (t_i)_{i \in I} \rangle$ is a unawareness belief structure, then set $F(\Gamma (\underline{\mathcal {S}})) \!:=\! \langle F(\underline{\mathcal {S}}), (M_i)_{i \in I}, (\mathcal {M}_i)_{i \in I}, (u_i)_{i \in I} \rangle$, where $F(\underline{\mathcal {S}})$ is the flattened structure associated with $\underline{\mathcal {S}}$, and $(M_i)_{i \in I}$, $(\mathcal {M}_i)_{i \in I}$, and $(u_i)_{i \in I}$ remain unchanged.

The flattened game is a standard Bayesian game (apart from explicitly modeling uncertainty about opponents’ action sets).

Proposition 6

Since the strategy sets and the utility functions remain unchanged, we have that any strategy profile is a Bayesian equilibrium in $\Gamma (\underline{\mathcal {S}})$ if and only if it is a Bayesian equilibrium in $F(\Gamma (\underline{\mathcal {S}}))$.

The interpretation of a flattened game may be flawed in several ways. For instance, we can have types of players who are certain of their set of actions, but consider it possible that they have a larger set of actions even though they do not have a larger set of actions. This leads to serious conceptual problems, if a player were to choose such an action. A player could then “test” his own beliefs by trying to choose such actions. Consider as a simple example an unawareness belief structure with two disjoint spaces, $S_1 = \{\omega _1\} \succeq S_2 = \{\omega _2\}$ and one player only. Let the set of the player’s available actions at $\omega _1$ be $\{a, b\}$, while it is just $\{a\}$ at $\omega _2$. The type mapping is defined by $t(\omega )(\{\omega _2\}) = 1$ for all $\omega$. That is, although at $\omega _1$ the player has actions $a$ and $b$ available, he is unaware of $b$. In the flattened game, he is aware of $b$ but is certain at $\omega _2$ that he has just action $a$ available. But because he is aware of $b$ at $\omega _2$, he could try to test his belief by trying to choose $b$. The flattened game is not well-defined because it is not specified what happens if a player tries to take an action that is not available to him.¹⁵

The fact that an unawareness beliefs structure can be “flattened” into a standard type-spaces does not mean that unawareness has no behavior implications. Unawareness has very different properties from probability zero belief. For instance, one property that is satisfied by unawareness is symmetry (see Heifetz et al. 2013a, Proposition 5). An agent is unaware of an event if and only if she is unaware of its negation. Clearly, such a property cannot be satisfied by probability zero belief because if an agent assigns probability zero to an event, then she must assign probability one to its complement. Schipper (2013) shows that this feature captures also behavioral differences between unawareness and probability zero belief. Let us say that a decision-maker chooses among different contracts for buying a firm. A second contract may differ from a first contract only in a consequence for an event $E$ that is disadvantageous to the buyer. If the decision-maker is indifferent between both contracts, then this is consistent with $E$ being Savage null. Yet, if the decision-maker is also indifferent between the first and a third contract that differs from the first only in assigning this disadvantageous consequence to the negation of the event $E$ instead the event $E$ itself, then this behavior is inconsistent with the negation of the event $E$ or the event $E$ itself being Savage null. The decision-maker behaves as if both the event $E$ and its negation are Savage null, which is impossible with probability zero belief but consistent with unawareness of the $E$ and of its negation. Thus, in models with primitives that are sufficiently rich to study decisions under unawareness, unawareness can have behavioral implications distinct from zero probability.

Appendix C: Further proofs

1.1 C1: Proof of Remark 2

Let $\omega = \left( \alpha _0, (\alpha _i)_{i \in I}\right)$ with $\alpha _0 = \alpha$, and let $\alpha \succeq \beta \succeq \delta$. We have to show for every $i \in I^0$ that $\inf \{\alpha _i, \delta \} = \inf \{\delta , \inf \{\alpha _i, \beta \}\}$. Since $\delta \preceq \beta$, we have $\inf \{\alpha _i, \delta \} \preceq \inf \{\alpha _i, \beta \}$ and hence $\inf \{\alpha _i, \delta \} \preceq \inf \{\delta , \inf \{\alpha _i, \beta \}\}$. We have $\inf \{\delta , \inf \{\alpha _i, \beta \}\} \preceq \delta$, by definition, but since $\inf \{\alpha _i, \beta \} \preceq \alpha _i$, we also have $\inf \{\delta , \inf \{\alpha _i, \beta \}\} \preceq \alpha _i$. This implies $\inf \{\delta , \inf \{\alpha _i, \beta \}\} \preceq \inf \{\alpha _i, \delta \}$. $\square$

1.2 C2: Proof of Lemma 1

We prove only the last claim, since the rest is obvious.

Let $\alpha \succeq \beta \succeq \delta$, $\omega = (\alpha _j)_{j \in I^0} \in S_{\alpha }$ and $t_i^k(\omega _{\beta }) \in \Delta (S_{\delta })$. By definition $t_i^k(\omega ) \in \Delta (S_{\alpha _i})$. We have to show that $\alpha _i \succeq \delta$. Since $t_i^k(\omega _{\beta }) \in \Delta (S_{\delta })$, we have by (i’) that $\omega _{\beta } = (\beta _j)_{j \in I^0}$ is such that $\beta _i = \delta$. By the definition of $r_{\beta }^{\alpha }$, we have $\beta _i = \inf \{\alpha _i, \beta \}$. Hence, $\delta \preceq \alpha _i$. $\square$

1.3 C3: Proof of Remark 3

We define $\alpha \cap \beta$ to be the restricted game such that $M_i^{\alpha \cap \beta } := M_i^{\alpha } \cap M_i^{\beta }$, for all $i \in I$. Note that if $\alpha , \beta \in G(\gamma )$, then $\alpha \cap \beta = \alpha \wedge \beta$ since $G(\gamma )$ is a meet-sublattice of the lattice of all restricted games.

Let $\emptyset \ne M_i' \subseteq M_i$, where $M_i$ is the action set of player $i$ in $\gamma$. We have to show that $\{\omega \in \Omega : \mathcal {M}_i(\omega ) \supseteq M_i'\}$ is an event.

By definition, $G(\gamma )$ is a finite meet-sublattice of the lattice of all restricted games given $\gamma$ ordered by set inclusion of $\prod _{i \in I} M_i$. Fix a player $i \in I$. Recall that $\mathcal {M}_i(\omega ) = M_i^{\alpha }$, for all $\omega \in S_{\alpha }$, where $M_i^{\alpha }$ is the action set of player $i$ in the restricted game $\alpha$.

Let $A = \{\alpha \in G(\gamma ) : M_i^{\alpha } \supseteq M_i'\}$. Since $G(\gamma )$ is a finite meet-sublattice, $\bigcap _{\alpha \in A} \alpha =: \alpha (M_i') \in G(\gamma )$. We have $M_i' \subseteq M_i^{\alpha (M_i')}$ and $M_i^{\alpha (M'_i)} \subseteq \mathcal {M}_i(\omega )$, for all $\omega$ such that $\mathcal {M}_i(\omega ) \supseteq M_i'$. Since for $\omega \in S_{\alpha }$, $\mathcal {M}_i(\omega ) = M_i^{\alpha }$, we have that $[M_i'] = \{\omega \in \Omega : \mathcal {M}_i(\omega ) \supseteq M_i'\} = (S_{\alpha (M_i')})^{\uparrow }$, which is an event. $\square$

1.4 C4: Proof of Lemma 2

Suppose for some $i \in I$, $\nu _i$ would not be a best reply to $\nu _{-i}$. Then, there exists $m_i \in M_i$ such that

$$\begin{aligned} \varepsilon \le v_i(m_i, \nu _{-i}) - v_i(\nu _i, \nu _{-i}), \end{aligned}$$

for some $\varepsilon > 0$. By continuity of the utility functions in mixed strategies and in beliefs on types, there exists a $k_1$ such that

$$\begin{aligned} \left| v_i(\nu _i, \nu _{-i}) - U_{(i, t^k_i(\bar{\gamma }))} (\sigma _i^k, \sigma _{-i}^k) \right| < \frac{\varepsilon }{3}, \end{aligned}$$

for all $k \ge k_1$. And likewise there exits $k_2$ such that

$$\begin{aligned} \left| U_{(i, t^k_i(\bar{\gamma }))}(m_i, \sigma _{-i}^k) - v_i(m_i, \nu _{-i}) \right| < \frac{\varepsilon }{3}, \end{aligned}$$

for all $k \ge k_2$.

Let $k \ge \max \{k_1, k_2\}$, then

$$\begin{aligned}&U_{(i, t^k_i(\bar{\gamma }))}(m_i, \sigma _{-i}^k) - U_{(i, t^k_i(\bar{\gamma }))}(\sigma _i^k, \sigma _{-i}^k) \\&\quad = \mathop {\underbrace{U_{(i, t^k_i(\bar{\gamma }))}(m_i, \sigma _{-i}^k) - v_i(m_i, \nu _{-i})}}\limits _{\ge - \frac{\varepsilon }{3}}+ \mathop {\underbrace{v_i(m_i, \nu _{-i}) - v_i(\nu _i, \nu _{-i})}}\limits _{\ge \varepsilon } \\&\qquad +\mathop {\underbrace{v_i(\nu _i, \nu _{-i}) - U_{(i, t^k_i(\bar{\gamma }))}(\sigma _i^k, \sigma _{-i}^k)}}\limits _{\ge - \frac{\varepsilon }{3}}\\&\quad \ge \frac{\varepsilon }{3}, \end{aligned}$$

that is, $\sigma _i^k$ is not a best reply to $\sigma _{-i}^k$, for sufficiently large $k$. This implies that $\sigma ^k$ is not an equilibrium of the game $\Gamma (\underline{\mathcal {S}}^k_\gamma )$, for sufficiently large $k$, a contradiction.$\square$

1.5 C5: Proof of Theorem 1

For every $k = 0, 1, \ldots$, the Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^{k}_{\gamma })$ has an equilibrium by Corollary 1. Since the set of mixed strategy combinations at $\bar{\gamma }$ is a closed and bounded subset of an Euclidean space, the sequence of equilibria $(\sigma ^{k}(\bar{\gamma }))_{k = 0}^{\infty }$ has a subsequence that converges to some $\nu$. By Lemma 2, $\nu$ is an unawareness perfect equilibrium of $\tilde{\gamma }$. $\square$

1.6 C6: Proof of Theorem 2

“$\Rightarrow$”: Let $\nu$ be an unawareness perfect equilibrium of $\tilde{\gamma }$. Then, there exists a sequence of Bayesian games with unawareness $\left( \Gamma (\underline{\mathcal {S}}^k_{\gamma })\right) _{k = 0}^{\infty }$ with corresponding Bayesian Nash equilibria $\sigma ^k \in E(\Gamma (\underline{\mathcal {S}}^k_{\gamma }))$, $k = 0, 1, \ldots ,$ for which $\sigma ^k(\bar{\gamma }) \rightarrow \nu$ as $k \rightarrow \infty$.

From Pearce (1984, Lemma 4) follows that a Nash equilibrium profile $\nu$ of the original game $\tilde{\gamma }$ is undominated if and only if for every $i \in N$, $\nu _i$ is a best response to a completely mixed strategy profile of opponents in the original game $\tilde{\gamma }$. Note that by Property (v), $\sum _{\omega \in S_{\gamma }} t_i^k(\bar{\gamma })(\{\omega \}) \sigma _{-i}^k(\omega )$ is equivalent to a completely mixed strategy profile of opponents for any $k = 0, 1, \ldots ,$ in the original game $\tilde{\gamma }$. Since $\sigma _i^k(\bar{\gamma }) \rightarrow \nu _i$, for any $m_i \in M_i$ with $\nu _i(m_i) > 0$, there exists a sufficiently large $k(m_i)$ such that for all $k \ge k(m_i)$, $\sigma _i^k(\bar{\gamma })(m_i) > 0$. Since $M_i$ is finite, there is a $k_{\max }$ such for all $k \ge k_{\max }$, $\sigma _i^k(\bar{\gamma })(m_i) > 0$ for all $m_i \in M_i$ with $\nu _i(m_i) > 0$. Thus, any such $m_i \in M_i$ is a best reply to $\sum _{\omega \in S_{\gamma }} t_i^k(\bar{\gamma })(\{\omega \}) \sigma _{-i}^k(\omega )$, for all $k \ge k_{\max }$. Hence, $\nu _i$ is a best reply to the completely mixed belief equivalent to $\sum _{\omega \in S_{\gamma }} t_i^{k_{\max }}(\bar{\gamma })(\{\omega \}) \sigma _{-i}^{k_{\max }}(\omega )$.

“$\Leftarrow$”: Recall that from Pearce (1984, Lemma 4) it follows that a Nash equilibrium profile $\nu$ is undominated in the original game $\tilde{\gamma }$ if and only if for every $i \in N$, $\nu _i$ is a best response to a completely mixed strategy profile of opponents $\tilde{\nu }_{-i}$ in $\tilde{\gamma }$. We will use these completely mixed strategy profiles $\tilde{\nu }_{-i}$, $i \in I$, to construct a sequence of Bayesian games with unawareness $\left( \Gamma (\underline{\mathcal {S}}^k_{\gamma })\right) _{k = 0}^{\infty }$ with Bayesian Nash equilibria $\sigma ^k \in E(\Gamma (\underline{\mathcal {S}}^k_{\gamma }))$, $k = 0, 1, \ldots$, for which $\sigma ^k(\bar{\gamma }) \rightarrow \nu$ as $k \rightarrow \infty$.

Given the augmented strategic game $\gamma$, consider any restricted strategic game $\alpha$ with $d_i \in M_i^{\alpha }$ for all $i \in I$ and $1 \le |M^\alpha _i| \le 2$. Note that by construction in any such restricted game, each player has at most one non-default action and if there is a non-default action for player $i$, then this non-default action is the strict dominant Nash equilibrium action of $\alpha$. Let $G(\gamma )$ be the set comprising of all such games, $\bot$, and $\gamma$ itself. Note that $G(\gamma )$ is rich (see page ) and a lattice.

We now construct a sequence of Bayesian games with unawareness of actions by defining for each player $i \in I$ the type mapping as follows: For $m_{-i} \in \tilde{M}_{-i}$, define $\ell (m_{-i})$ to be the number of profiles $(\alpha _j)_{j \in I {\setminus } \{i\}}$ with $\alpha _j \in G(\gamma ) {\setminus } \{\bot , \gamma \}$, $j \in I {\setminus }\{i\}$, for which $\{m_{-i}\} = \prod _{j \in I {\setminus } \{i\}} \left( M^{\alpha _j}_{j} {\setminus } \{d_{j}\}\right)$. (Note that in such a profile we have $| M_j^{\alpha _j} | = 2$, for all $j \in I {\setminus } \{i\}$).

Fix $\varepsilon \in (0, 1)$. If $\omega = (\alpha _j)_{j \in I^0}$ with $\alpha _j \in G(\gamma ) {\setminus } \{\bot , \gamma \}$ such that $| M_j^{\alpha _j} | = 2$, for $j \in I {\setminus } \{i\}$ and $\alpha _j = \gamma$ for $j \in \{0, i\}$, let $t_i^k(\bar{\gamma })(\{ \omega \}) = \frac{\varepsilon ^k}{\ell (m_{-i})} \tilde{\nu }_{-i}(m_{-i})$ with $\{m_{-i}\} = \prod _{j \in I {\setminus } \{i\}} (M^{\alpha _j}_{j} {\setminus } \{d_{j}\})$.

All the remaining probability mass of player $i$’s type at state $\bar{\gamma }$ at $k$, $1 - \varepsilon ^k$, is assigned to $t_i^k(\bar{\gamma })(\{\bar{\gamma }\})$. We impose properties (i’), (ii’), and (iv’). Property (iii’) is implied by Lemma 1. Properties (v) and (vi) are satisfied by construction. Beliefs of player $i$ in states different from $\bar{\gamma }$ are then either completely determined by beliefs in $\bar{\gamma }$ via properties (ii’) and (iv’) or can be assigned so as to satisfy properties (i’) to (vi).

Next, we construct a sequence of Bayesian Nash equilibria whose limit in state $\bar{\gamma }$ is the undominated Nash equilibrium $\nu$ of the original game $\tilde{\gamma }$. For any player $i$ and any $k$, let $\sigma _i^k(\omega ) = \nu _i$ for all $\omega \in S_{\gamma }$ with $\omega = (\beta _j)_{j \in I^0}$ such that $\beta _i = \gamma$. Moreover, set $\sigma _i^k(\omega )(d_i) = 1$ in any other state $\omega = (\beta _j)_{j \in I^0}$ with $M^{\beta _i} = \{d_i\}$. Finally, set $\sigma _i^k(\omega )(m_i) = 1$ with $m_i \ne d_i$ in any state $\omega = (\beta _j)_{j \in I^0}$ with $M^{\beta _i} = \{m_i, d_i\}$. (Recall that such a non-default action is strictly dominant.) Since in the latter states, $\sigma _{i}^k(\omega ) \in \Delta (M_{i}^\alpha )$, for $\alpha \prec \gamma$, we extend $\sigma _{i}^k(\omega )$ to $\tilde{\sigma }_{i}^k(\omega ) \in \Delta (M_{i})$ by setting $\tilde{\sigma }^k_{i}(\omega )(m_{i}) := \sigma ^k_{i}(\omega )(m_{i})$ for all $\omega \in \Omega _{\gamma }$ and $m_{i} \in M_{i}^\alpha$.

Note that $\sum _{\omega \in S_{\gamma }} t_i^{k}(\bar{\gamma })(\{\omega \}) \tilde{\sigma }^{k}_{-i}(\omega ) \rightarrow \nu _{-i}$ for $k \rightarrow \infty$. Moreover, for any $k$,

$\sum _{\omega \in S_{\gamma }} t_i^{k}(\bar{\gamma })(\{\omega \}) \tilde{\sigma }^{k}_{-i}(\omega )$ is a convex combination of $\tilde{\nu }_{-i}$ and $\nu _{-i}$.

To see that $\sigma ^k$ is a Bayesian Nash equilibrium for any $k = 0, 1, \ldots ,$ note that $\sigma _i^k(\omega ) = \nu _i$ with $\omega = (\beta _j)_{j \in I^0}$ such that $\beta _i = \gamma$ is a best reply to $\nu _{-i}$ since $\nu$ is Nash equilibrium of $\tilde{\gamma }$. Since $\nu$ is an undominated Nash equilibrium of $\tilde{\gamma }$, we also noted above that $\nu _i$ is a best reply against $\tilde{\nu }_{-i}$. Hence $\nu _i$ is also a best reply to any convex combination of $\nu _{-i}$ and $\tilde{\nu }_{-i}$. It follows that for any $k = 0, 1, \ldots$, $\sigma _i^k$ is a Bayesian Nash equilibrium mixture of player $i$ in the Bayesian game with unawareness $\Gamma (\underline{\mathcal {S}}^k_{\gamma })$. Since $\sigma _i^k(\bar{\gamma }) = \nu _i$ for all $k$ and $i \in I$, $\nu$ is an Unawareness perfect equilibrium of $\tilde{\gamma }$. $\square$