<p>Adding to @soslow's answer: once you have an SPE, it should be easy to construct a non-subgame-perfect NE by modifying the off-equilibrium actions in such a way that 1) the players have no incentive to deviate to those actions and 2) the action profile is not a NE in any subgame.</p>
<p>For example, one SPE of the game is</p>
<blockquote>
<p>play <span class="math-container">$(A,A)$</span> in stage 1, and play <span class="math-container">$(A,A)$</span> in stage 2 regardless of the outcome in stage 1.</p>
</blockquote>
<p>We can modify this to</p>
<blockquote>
<p>play <span class="math-container">$(A,A)$</span> in stage 1, and play <span class="math-container">$(A,A)$</span> in stage 2 if the outcome in stage 1 is <span class="math-container">$(A,A)$</span>, otherwise play <span class="math-container">$(A,C)$</span>.</p>
</blockquote>
<p>This modified strategy profile is a NE, since players are still best responding to each other by playing <span class="math-container">$(A,A)$</span> in both stages (the off-equilibrium path outcome <span class="math-container">$(A,C)$</span> is Pareto dominated by <span class="math-container">$(A,A)$</span>). However, since <span class="math-container">$(A,C)$</span> is not a NE in any subgame, the modified strategy profile is not subgame perfect.</p>

CC BY-SA 4.0
2020-10-29T23:06:59.783Z
2020-10-29T23:06:59.783Z
Herr K.
<ol>
<li>Check for the Nash equilibria (pure or mixed) of the one-shot game.</li>
<li>Repetition of the strategy profile of the Nash equilibria of the one-shot version yields one set of subgame perfect equilibria: For instance, play <span class="math-container">$(A,A)$</span> in the first stage and for any action profile played at the first stage, play <span class="math-container">$(A,A)$</span> in the second stage. The same holds true for the mixed (complete or otherwise) as well. For instance, the totally mixed SPNE is :Play <span class="math-container">$\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$</span> in the first stage and for any action profile (assuming the randomisation is observable), play <span class="math-container">$\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$</span> in the second stage.</li>
<li>Combinations of the Nash equilibria of the one-shot game yield another set of SPNE: For instance, Play <span class="math-container">$(A,A)$</span> in the first stage and play <span class="math-container">$\left(0,\frac{1}{2},\frac{1}{2}\right)$</span> for any action profile in the second stage and so on.</li>
<li>Use credible threats to play a non-Nash action profile in the first period: Play <span class="math-container">$(B,B)$</span> in the first stage. At the second stage, if <span class="math-container">$(B,B)$</span> was played, play <span class="math-container">$\left(0,\frac{1}{2},\frac{1}{2}\right)$</span>. For any other action profile at the first stage, play the totally mixed Nash equilibrium of the one-shot game. Why does this work? There is no incentive for player 1 to deviate anyway because they're playing their best reply in the first stage. For player 2, if they were to deviate to <span class="math-container">$C$</span>. they'd receive <span class="math-container">$2$</span> in the first stage and <span class="math-container">$1/3$</span> in the second stage. Assuming no discounting, they get <span class="math-container">$7/3$</span> while if they obeyed the strategy, they would get <span class="math-container">$(1+3/2)$</span> which is strictly greater. Similarly <span class="math-container">$(C,C)$</span>, <span class="math-container">$(B,C)$</span> and <span class="math-container">$(C,B)$</span> can be played as first stage action profiles in a sub-game perfect equilibrium.</li>
</ol>

CC BY-SA 4.0
2020-10-29T20:13:08.287Z
Could you explain where the (0, 1/2 , 1/2) came from? And is there a nash equilibrium that's not spe?
Robin311
CC BY-SA 4.0
2020-10-29T20:15:28.560Z
That is a Nash equilibrium too: Suppose the probability of player 1 playing strategy $A$ is 0 and that of playing $B$ and $C$ equal, then player 2 gets 0 by playing $A$, 3/2 by playing $B$ and 3/2 by playing $C$. Hence, player 2 is indifferent between $B$ and $C$ and strictly prefers them to $A$. Hence any convex combination of $B$ and $C$ is a best reply. The same holds true for player 1.
soslow
CC BY-SA 4.0
2020-10-30T03:19:03.483Z
aha, I think I understand it now. Then can this be a subgame perfect equilibrium too? Play (A,A) on the first stage and then play (3/5, 1/5, 1/5) for any action in the second stage. You didn't mention this so I was wondering if this could be a subgame perfect equilibrium too.
Robin311
CC BY-SA 4.0
2020-10-29T16:24:11.617Z
2020-10-29T16:24:11.617Z
soslow
<p>In a game theory textbook there is something similar to the table below where there is one pure strategy nash equilibrium and multiple mixed strategy nash equilibria. It is a simultaneous game with the payoffs presented below.</p>
<p>If we assume that this game is played <strong>twice</strong>, How do I identify all subgame perfect equilibria for this game, as well as nash equilibrium that is not a subgame perfect equilibrium?</p>
<p>For a game with multiple pure strategy nash equilibrium I think I can find a solution by using backward induction, but for a game like this with just one pure strategy nash equilibrium and multiple mixed strategy nash equilibria, I have  no idea how to identify the subgame perfect equilibria and possibly a nash equilibrium that is not subgame perfect equilibrium, especially when there are mixed strategy equilibria included.</p>
<p>Any help in this would be appreciated.</p>
<p><span class="math-container">\begin{array}{|c|c|c|c|}
\hline
  &amp;  A    &amp;  B    &amp;  C \\\hline
A &amp; (1,1) &amp; (0,0) &amp; (0,0)\\\hline
B &amp; (0,0) &amp; (2,1) &amp; (1,2)\\\hline
C &amp; (0,0) &amp; (1,2) &amp; (2,1)\\\hline
\end{array}</span></p>

<game-theory><nash-equilibrium>
Subgame Perfect Equilibrium for Pure and Mixed strategy
Robin311
# Subgame Perfect Equilibrium for Pure and Mixed strategy

(Asked by: username_0 on 2020-10-29)

In a game theory textbook there is something similar to the table below where there is one pure strategy nash equilibrium and multiple mixed strategy nash equilibria. It is a simultaneous game with the payoffs presented below.

If we assume that this game is played **twice**, How do I identify all subgame perfect equilibria for this game, as well as nash equilibrium that is not a subgame perfect equilibrium?

For a game with multiple pure strategy nash equilibrium I think I can find a solution by using backward induction, but for a game like this with just one pure strategy nash equilibrium and multiple mixed strategy nash equilibria, I have no idea how to identify the subgame perfect equilibria and possibly a nash equilibrium that is not subgame perfect equilibrium, especially when there are mixed strategy equilibria included.

Any help in this would be appreciated.

\begin{array}{|c|c|c|c|}
\hline
& A & B & C \\\hline
A & (1,1) & (0,0) & (0,0)\\\hline
B & (0,0) & (2,1) & (1,2)\\\hline
C & (0,0) & (1,2) & (2,1)\\\hline
\end{array}

**Tags:** game-theory, nash-equilibrium

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## 2 answers

### Answer by username_2 on 2020-10-29. Score: 4 | Accepted answer

1. Check for the Nash equilibria (pure or mixed) of the one-shot game.
2. Repetition of the strategy profile of the Nash equilibria of the one-shot version yields one set of subgame perfect equilibria: For instance, play $(A,A)$ in the first stage and for any action profile played at the first stage, play $(A,A)$ in the second stage. The same holds true for the mixed (complete or otherwise) as well. For instance, the totally mixed SPNE is :Play $\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$ in the first stage and for any action profile (assuming the randomisation is observable), play $\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$ in the second stage.
3. Combinations of the Nash equilibria of the one-shot game yield another set of SPNE: For instance, Play $(A,A)$ in the first stage and play $\left(0,\frac{1}{2},\frac{1}{2}\right)$ for any action profile in the second stage and so on.
4. Use credible threats to play a non-Nash action profile in the first period: Play $(B,B)$ in the first stage. At the second stage, if $(B,B)$ was played, play $\left(0,\frac{1}{2},\frac{1}{2}\right)$. For any other action profile at the first stage, play the totally mixed Nash equilibrium of the one-shot game. Why does this work? There is no incentive for player 1 to deviate anyway because they're playing their best reply in the first stage. For player 2, if they were to deviate to $C$. they'd receive $2$ in the first stage and $1/3$ in the second stage. Assuming no discounting, they get $7/3$ while if they obeyed the strategy, they would get $(1+3/2)$ which is strictly greater. Similarly $(C,C)$, $(B,C)$ and $(C,B)$ can be played as first stage action profiles in a sub-game perfect equilibrium.

> **Comments:**
> **username_0**: Could you explain where the (0, 1/2 , 1/2) came from? And is there a nash equilibrium that's not spe?  
> **username_2**: That is a Nash equilibrium too: Suppose the probability of player 1 playing strategy $A$ is 0 and that of playing $B$ and $C$ equal, then player 2 gets 0 by playing $A$, 3/2 by playing $B$ and 3/2 by playing $C$. Hence, player 2 is indifferent between $B$ and $C$ and strictly prefers them to $A$. Hence any convex combination of $B$ and $C$ is a best reply. The same holds true for player 1.  
> **username_0**: aha, I think I understand it now. Then can this be a subgame perfect equilibrium too? Play (A,A) on the first stage and then play (3/5, 1/5, 1/5) for any action in the second stage. You didn't mention this so I was wondering if this could be a subgame perfect equilibrium too.  

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### Answer by username_1 on 2020-10-29. Score: 3

Adding to @soslow's answer: once you have an SPE, it should be easy to construct a non-subgame-perfect NE by modifying the off-equilibrium actions in such a way that 1) the players have no incentive to deviate to those actions and 2) the action profile is not a NE in any subgame.

For example, one SPE of the game is

> play $(A,A)$ in stage 1, and play $(A,A)$ in stage 2 regardless of the outcome in stage 1.

We can modify this to

> play $(A,A)$ in stage 1, and play $(A,A)$ in stage 2 if the outcome in stage 1 is $(A,A)$, otherwise play $(A,C)$.

This modified strategy profile is a NE, since players are still best responding to each other by playing $(A,A)$ in both stages (the off-equilibrium path outcome $(A,C)$ is Pareto dominated by $(A,A)$). However, since $(A,C)$ is not a NE in any subgame, the modified strategy profile is not subgame perfect.


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