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I have a confusion on how to define strategy sets;

I understand that in dynamic games strategy sets are defined as $\textbf{sets of functions}$ - see for example my previous post, link below:

Define and characterize equilibria of the following game

However I cannot square this with what would be a strategy set in a simultaneous move game; as an illustration consider the following simple game:

There are two players competing for a prize of $1$. Each player $i$ choses effort level $x_i \in \mathbb{R}$. When the efforts levels are $(x_1,x_2)$, the probability that player 1 wins the prize is given by $p(x_1,x_2)$ and the probability that player 2 wins the prize is given by $1-p(x_1,x_2)$. Player $i$'s cost of exerting effort $x_i$ is equal to $x_i$.

Players choose their effort levels simultaneously.

Now looking standard textbooks I suspect that the strategy sets of both players would be defined as: $S_1 = \mathbb{R}$, $S_2 = \mathbb{R}$ Since both players will choose an effort level $x\in \mathbb{R}$.

However given that a $\textbf{strategy}$ for player $i$ is defined as a complete contingent plan, it would be natural for me to define a strategy as a function:
$s_i: \mathbb{R} \to \mathbb{R}$,

that is for each possible effort level $x_j \in \mathbb{R}$ that player $j$ chooses, player $i$ will choose an effort level $x_i \in \mathbb{R}$. But then since $S_i$ is the $\textbf{set of}$ strategies it would appropriate to define the strategy set of this game for player i as:

$S_i = \{s_i : \mathbb{R} \to \mathbb{R}\}$ , that is the set of all possible functions $s_i$.

But then this definition would conflict with the original one I gave at the beginning.

So which of the two (if any) would be the correct "strategy set" of this simultaneous move game? Thank You!

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1 Answer 1

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There is no conflict. The strategy is a function which maps decision points to actions, that is it tells the player in each decision point (to be precise, in each information set) what she should do. In a dynamic game there are usually more decisions points, so this function nature of the strategy is more obvious.

In the simultaneous game you describe there is only one decision point (information set), players have to make a choice at the start of the game. A strategy maps an action to this one decision point (information set), so basically the strategy space is just the action space.

EDIT:
You write " it would be natural for me to define a strategy as a function: $$s_i: \mathbb{R} \to \mathbb{R}."$$ There is some murkiness here, because what is the first $\mathbb{R}$ supposed to be? The strategy set of the other player? But that is not observed by player $i$, so it cannot factor into her decision. As a function the strategy would look actually look like $$s_i: \left\{I_i\right\} \to \mathbb{R},$$ where $I_i$ is the single information set of player $i$, so $\left\{I_i\right\}$ is a singleton set. There is a bijective mapping between the vector space of these strategies and the space $\mathbb{R}$ so it does no harm to define $S_i$ as equal to $\mathbb{R}$.

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  • $\begingroup$ thank you, but I don't understand: if the "strategy space is just the action space" why is it valid to specify $S_i$ as a function? (I mean even if we can specify it as a function, then the domain of this function will be constant..and surely cannot be R as I specified above); since each player has only one information set then an example of a strategy for player i and j would simply be: si=1, si=1 "no matter what the other player strategy is" and the respective sets of strategies would be $S_i=\mathbb{R},S_j=\mathbb{R}$ $\endgroup$ Commented Oct 25, 2015 at 20:10
  • $\begingroup$ @night_owl89 Sorry, I missed a part of your question. I will edit my answer. $\endgroup$
    – Giskard
    Commented Oct 25, 2015 at 20:48
  • $\begingroup$ Ok thank You very much; now it is much clearer. So, now would you agree that $\textbf{an example of a strategy for player i}$ would be given by $s_i = k$ where $k \in \mathbb{R}$? $\endgroup$ Commented Oct 25, 2015 at 21:00
  • $\begingroup$ @night_owl89 Yes, I would. $\endgroup$
    – Giskard
    Commented Oct 25, 2015 at 21:36

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