<p>If the stage game <span class="math-container">$G$</span> of a finitely repeated game has a unique Nash equilibrium (NE), then the finitely repeated game <span class="math-container">$G^T$</span> has a unique subgame perfect equilibrium (SPE).</p>
<p>To see why, simply observe that at the last stage <span class="math-container">$T$</span>, SPE requires that the stage game NE be played. This means in the second to last stage <span class="math-container">$T-1$</span>, the players cannot condition <span class="math-container">$T-1$</span> behavior on what they will do in <span class="math-container">$T$</span>. So they must play the stage NE again in <span class="math-container">$T-1$</span>. Continue this backward induction until stage <span class="math-container">$1$</span>, and we can conclude that the SPE of <span class="math-container">$G^T$</span> must be unique and it involves playing the stage NE regardless of history.</p>

CC BY-SA 4.0
2024-10-28T16:39:25.930Z
One lingering question: If the game is repeated infinitely, can then the SPE be something else? Not always, but sometimes (depending on the payoff matrix)!
madamsapple
CC BY-SA 4.0
2024-10-28T19:07:47.823Z
@madamsapple: Yes, infinity does open a whole new set of possibilities. The result is the celebrated [folk theorem](https://en.wikipedia.org/wiki/Folk_theorem_(game_theory)). It turns out that the existence of SPE depends not so much on the payoff matrix (though it still matters) but on players' discount factor(s).
Herr K.
CC BY-SA 4.0
2024-10-28T14:00:31.703Z
2024-10-28T14:00:31.703Z
Herr K.
<p>Consider a <span class="math-container">$2$</span>-player normal formal game that is being repeated twice. If there's only one pure strategy Nash-Equilibrium (call it <span class="math-container">$(X,x)$</span>), can a subgame perfect equilibrium exist where an action-profile <span class="math-container">$(Y,y) \neq (X,x)$</span> is played?</p>
<p>In other words, if there's only one pure strategy NE, will every round in a finitely repeated game see players playing the same strategy?</p>

CC BY-SA 4.0
2024-10-28T08:40:02.083Z
This looks like a homework question. Have you tried to solve it yourself?
Herr K.
CC BY-SA 4.0
2024-10-28T08:52:08.560Z
@HerrK. Not a HW question. I am studying subgame perfect equilibriums and found that there can be NEs defined by, let's say, finite state automata, that are not actually the pure NE that we get from the payoff matrix. That's where my doubt stems from - if for NEs, that is possible, what about SPEs? I think the answer to the question is in the positive (that is, the starting strategy may not necessarily be $(X,x)$ - and that can be done by defining an appropriate trigger strategy NE).
madamsapple
<game-theory><nash-equilibrium>
Subgame perfect equilibrium starting with a non-NE
madamsapple
# Subgame perfect equilibrium starting with a non-NE

(Asked by: username_0 on 2024-10-28)

Consider a $2$-player normal formal game that is being repeated twice. If there's only one pure strategy Nash-Equilibrium (call it $(X,x)$), can a subgame perfect equilibrium exist where an action-profile $(Y,y) \neq (X,x)$ is played?

In other words, if there's only one pure strategy NE, will every round in a finitely repeated game see players playing the same strategy?

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: This looks like a homework question. Have you tried to solve it yourself?  
> **username_0**: @HerrK. Not a HW question. I am studying subgame perfect equilibriums and found that there can be NEs defined by, let's say, finite state automata, that are not actually the pure NE that we get from the payoff matrix. That's where my doubt stems from - if for NEs, that is possible, what about SPEs? I think the answer to the question is in the positive (that is, the starting strategy may not necessarily be $(X,x)$ - and that can be done by defining an appropriate trigger strategy NE).  

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## 1 answers

### Answer by username_1 on 2024-10-28. Score: 3 | Accepted answer

If the stage game $G$ of a finitely repeated game has a unique Nash equilibrium (NE), then the finitely repeated game $G^T$ has a unique subgame perfect equilibrium (SPE).

To see why, simply observe that at the last stage $T$, SPE requires that the stage game NE be played. This means in the second to last stage $T-1$, the players cannot condition $T-1$ behavior on what they will do in $T$. So they must play the stage NE again in $T-1$. Continue this backward induction until stage $1$, and we can conclude that the SPE of $G^T$ must be unique and it involves playing the stage NE regardless of history.

> **Comments:**
> **username_0**: One lingering question: If the game is repeated infinitely, can then the SPE be something else? Not always, but sometimes (depending on the payoff matrix)!  
> **username_1**: @madamsapple: Yes, infinity does open a whole new set of possibilities. The result is the celebrated [folk theorem](https://en.wikipedia.org/wiki/Folk_theorem_(game_theory)). It turns out that the existence of SPE depends not so much on the payoff matrix (though it still matters) but on players' discount factor(s).  

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