Proposition. Every finite extensive form game is associated with a unique strategic form representation.

I think this proposition is true. But how do we prove it rigorously?

  • $\begingroup$ This naturally depends on the definition of an extensive form you use. $\endgroup$ Dec 15 '14 at 0:42
  • $\begingroup$ And on what you mean by "is associated with". $\endgroup$ Jan 28 '15 at 4:24

I rely on the definitions from Chapter 2 of the Handbook of Game Theory, Volume 1, by Sergiu Hart.

If I understand you correctly, the proposition can be re-written as

Proposition. For, every finite extensive form game $\Gamma^E$, there exists a single strategic form representation $\Gamma^N =[I^N,\{S^N_i\},\{u^N_i(\cdot)\}]$ (up to relabeling of the agents) such that

  1. $I^N = I^E$,

and for all $i\in I^N = I^E$,

  1. $S^N_i = \{$ pure strategies of $i$ in $\Gamma^E$ $\}$ ,

  2. and $u^N_i(s) = u^E_i(c(s))$, where $c(s) $ associates every profile of pure strategy in $\Gamma^E$ with a terminal node of $\Gamma^E$ resulting from the pure strategy profile $s$.

I think 1. and 2. are obvious. There only remains to show 3, which is equivalent to proving that $c(s)$ is a function, i.e. every profile of pure strategies is associated with one and only one terminal node in $\Gamma^E$. This follows directly from the fact that the pure strategy of some player $i$ is a function selecting one and only one possible action from every information set.

  • Start from the root node, $r_0$.
  • By definition of a game in extensive form, the nodes are partitioned between the agents.
  • Thus $r_0$ belongs to some agent $i_0$.
  • By definition of the game in extensive form, the nodes of $i_0$ are partitioned into information sets.
  • Thus $r_0$ belongs to some information set of $i_0$, say $H_i^0$.
  • By definition of the pure strategy $s$, for any node in $H_i^0$, $i_0$ always choses a single successor, among the possible successors at $H_i^0$.
  • Let this successor be $r_1$.
  • Then $r_1$ must be the successor of $r_0$, and the next node on the path.

  • Now consider $r_1$.

  • By definition of agame in extensive form, the nodes are partitioned between the agents.
  • Thus $r_1$ belong to some agent $i_1$ (where $i_1 = i_0$ is allowed).
  • $\vdots$
  • Repeating the argument as many time as needed, because the game is finite, we necessarily reach a single terminal node $r_* = c(s)$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.