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Below is a lemma which I believe to be true and which I would like to use to derive other results

Take any two games in extensive form (complete information) $\Gamma$ and $\Gamma'$ differing only through their collections of information sets $\mathcal{H}$ and $\mathcal{H'}$. Suppose that for each decision node $x$ in the set of decisions nodes $\mathcal{X} = \mathcal{X}'$, $H(x)$ is a finer partition of the set of actions $c(x) = c'(x)$ than $H'(x)$.

Then for ever subgame perfect equilibrium outcome of $\Gamma$, there exists a subgame perfect equilibrium of $\Gamma'$ with the same outcome (but the converse does not need to be true).

I could try to write a complete proof but this seems to be basic enough a statement and I don't want to re-invent the wheel (or fail to pay tribute to former proofs). So my question is:

  • Do you know of any reference where that result is proven (or maybe disproven in case I got something wrong)?

Edit : I am only interested in pure strategies equilibria.

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    $\begingroup$ There seems to be many typos (and/or unfamiliar convention) in your post that prevents it from being clearly understood. Please proof-read your question. $\endgroup$
    – ramazan
    Commented Oct 5, 2015 at 0:28
  • $\begingroup$ @ramazan : I changed a couple of things but could not find major typos. Could you point at specific parts of the questions which make it unclear? I'd be happy to try to clarify. The notational conventions and terminology I use are standard, straight out of Mas-collel, Whinston and Green. $\endgroup$
    – Martin Van der Linden
    Commented Oct 5, 2015 at 0:42
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    $\begingroup$ Technically the strategy profiles would not match for such games. E.g. in @ramazan's answer a strategy of player 2 consists of one action in $\Gamma'$ and two actions in $\Gamma$, so the profiles would be $(O,l)$ versus $(O,(l,l))$. Technically these are not the same. $\endgroup$
    – Giskard
    Commented Oct 5, 2015 at 6:40
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    $\begingroup$ When you write "every subgame perfect equilibrium of $\Gamma$ is also a subgame perfect equilibrium of $\Gamma$", presumably one should be a $\Gamma'$? $\endgroup$
    – Ubiquitous
    Commented Oct 5, 2015 at 7:09
  • $\begingroup$ @Ubiquitous: thanks, did not catch that one, sorry for the confusion. $\endgroup$
    – Martin Van der Linden
    Commented Oct 5, 2015 at 13:26

2 Answers 2

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Perhaps I misunderstand something, maybe you do not allow mixed equilibria. That may be strange in games of imperfect information.

Consider an asymmetric matching pennies game. Both players show either Heads or Tails. I forgot how to type game matrices in mathjax, but the payoffs look something like this:

$\begin{bmatrix} -1,1 & 3,-3 \\ 1,-1 & -3,3 \end{bmatrix}$

Consider two versions of this game.

Version 1, $\Gamma'$
The moves are simultaneous so player 2 does not observe the move of player 1 and hence has two decision nodes in her information set. In the unique equilibrium player 1 mixes Heads and Tails with 50%-50% probability. Player 2 mixes Heads and Tails with 75%-25% probability respectively. Expected payoff is 0 for both.

Version 2, $\Gamma$
Sequential matching pennies. Player 1 moves first and his move is observed by player 2. In the unique equilibrium player 1 shows Heads, as does player 2. Payoff is -1 for player 1 and 1 for player 2.

Information-wise $\Gamma$ is a refinement of $\Gamma'$ but the equilibrium outcomes do not match.

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  • $\begingroup$ Thanks @denesp. Sorry, yet another thing I forgot to include in my question : I am interested in pure strategies only. But I like you're example for mixed strategies. $\endgroup$
    – Martin Van der Linden
    Commented Apr 7, 2016 at 0:32
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    $\begingroup$ @MartinVanderLinden The counterexample still holds. Information-wise $\Gamma$ is a refinement of $\Gamma'$. There is a pure strategy equilibrium in $\Gamma$ and no pure strategy equilibrium with the same payoff exists in $\Gamma'$ because there are no pure strategy equilibria in $\Gamma'$ . $\endgroup$
    – Giskard
    Commented Apr 7, 2016 at 4:19
  • $\begingroup$ That's right. Thanks a lot for keeping up with the question and getting back to it after a while. Your example helped me realize the above "conjecture" was based on very wrong intuition. $\endgroup$
    – Martin Van der Linden
    Commented Apr 7, 2016 at 14:29
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    $\begingroup$ @MartinVanderLinden The conjecture did seem logical. I did not post an answer first because I thought it was trivially true but would require a very technical proof. $\endgroup$
    – Giskard
    Commented Apr 7, 2016 at 16:22
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Would this (and why or why not) provide an answer to your question?enter image description here

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  • $\begingroup$ Although @Martin Van der Linden wrote $\Gamma$ twice I am pretty sure he meant that an SPE of the refined game $\Gamma$ would be an SPE of the coarse game $\Gamma'$, not the other way around. $\endgroup$
    – Giskard
    Commented Oct 5, 2015 at 6:33
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    $\begingroup$ @ramazan: Thanks for your efforts, and sorry for the confusion. denesp comment is correct. Again, sorry, it was impossible to guess from the former version of my question. $\endgroup$
    – Martin Van der Linden
    Commented Oct 5, 2015 at 13:32

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