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

<p>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:</p>

<p>$\begin{bmatrix} -1,1 &amp; 3,-3 \\ 1,-1 &amp; -3,3 \end{bmatrix}$</p>

<p>Consider two versions of this game.</p>

<p>Version 1, $\Gamma'$
<br>  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.</p>

<p>Version 2, $\Gamma$
<br>  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.</p>

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

CC BY-SA 3.0
2016-04-07T00:32:32.943Z
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.
Martin Van der Linden
CC BY-SA 3.0
2016-04-07T04:19:48.523Z
@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 $\Gamma'$ .
Giskard
CC BY-SA 3.0
2016-04-07T14:29:38.717Z
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.
Martin Van der Linden
CC BY-SA 3.0
2016-04-07T16:22:42.750Z
@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.
Giskard
CC BY-SA 3.0
2016-04-06T21:10:12.617Z
2016-04-06T21:17:45.993Z
2016-04-06T21:17:45.993Z
Giskard
<p>Would this (and why or why not) provide an answer to your question?<a href="https://i.sstatic.net/XfL2K.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XfL2K.jpg" alt="enter image description here"></a></p>

CC BY-SA 3.0
2015-10-05T06:33:10.400Z
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.
Giskard
CC BY-SA 3.0
2015-10-05T13:32:58.783Z
@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.
Martin Van der Linden
CC BY-SA 3.0
2015-10-05T00:56:25.433Z
2015-10-05T00:56:25.433Z
ramazan
<p>Below is a lemma which I believe to be true and which I would like to use to derive other results</p>

<blockquote>
  <p>Take any two games in extensive form (complete information) $\Gamma$ and $\Gamma'$ differing <em>only</em> 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 $H'(x)$.</p>
  
  <p>Then for ever subgame perfect equilibrium <em>outcome</em> of $\Gamma$, there exists a subgame perfect equilibrium of $\Gamma'$ with the same outcome (but the converse does not need to be true).</p>
</blockquote>

<p>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:</p>

<ul>
<li>Do you know of any reference where that result is proven (or maybe disproven in case I got something wrong)?</li>
</ul>

<p><strong>Edit :</strong> I am only interested in pure strategies equilibria.</p>

CC BY-SA 3.0
2015-10-05T00:28:59.330Z
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.
ramazan
CC BY-SA 3.0
2015-10-05T00:42:18.607Z
@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.
Martin Van der Linden
CC BY-SA 3.0
2015-10-05T06:40:30.123Z
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.
Giskard
CC BY-SA 3.0
2015-10-05T07:09:19.530Z
When you write "every subgame perfect equilibrium of $\Gamma$ is also a subgame perfect equilibrium of $\Gamma$", presumably one should be a $\Gamma'$?
Ubiquitous
CC BY-SA 3.0
2015-10-05T13:26:41.973Z
@Ubiquitous: thanks, did not catch that one, sorry for the confusion.
Martin Van der Linden
CC BY-SA 3.0
2015-10-05T13:36:20.600Z
@denesp:  that's right, thanks for pointing it out. I believe what I had in mind actually was "Then for every subgame perfect equilibrium *outcome* of $\Gamma$, there exists a subgame perfect equilibrium with the same outcome in $\Gamma'$". I'll edit my question. Thanks again to all of you for keeping up with the question and helping me clarify/improve it.
Martin Van der Linden
<game-theory><extensive-games><proof-reference>
Proof of nestedness of subgame perfect equilibria
Martin Van der Linden
# Proof of nestedness of subgame perfect equilibria

(Asked by: username_0 on 2015-10-04)

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.

**Tags:** game-theory, extensive-games, proof-reference
> **Comments:**
> **username_1**: 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.  
> **username_0**: @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.  
> **username_2**: 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.  
> **username_3**: When you write "every subgame perfect equilibrium of $\Gamma$ is also a subgame perfect equilibrium of $\Gamma$", presumably one should be a $\Gamma'$?  
> **username_0**: @Ubiquitous: thanks, did not catch that one, sorry for the confusion.  
> **username_0**: @denesp:  that's right, thanks for pointing it out. I believe what I had in mind actually was "Then for every subgame perfect equilibrium *outcome* of $\Gamma$, there exists a subgame perfect equilibrium with the same outcome in $\Gamma'$". I'll edit my question. Thanks again to all of you for keeping up with the question and helping me clarify/improve it.  

---

## 2 answers

### Answer by username_2 on 2016-04-06. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: 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.  
> **username_2**: @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 $\Gamma'$ .  
> **username_0**: 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.  
> **username_2**: @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.  

---

### Answer by username_1 on 2015-10-05. Score: 2

Would this (and why or why not) provide an answer to your question?[![enter image description here](https://i.sstatic.net/XfL2K.jpg)](https://i.sstatic.net/XfL2K.jpg)

> **Comments:**
> **username_2**: 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.  
> **username_0**: @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.  

---

