<p>It is called a <a href="https://en.wikipedia.org/wiki/Subgame_perfect_equilibrium" rel="nofollow noreferrer">subgame perfect equilibrium.</a></p>

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2019-02-02T10:21:04.373Z
2019-02-02T10:21:04.373Z
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<p>Take a two-stage game with complete information and simultaneous actions in each state: </p>

<p>(1) Player 1 and 2 simultaneously choose action <span class="math-container">$a_1\in A_1$</span>  and
<span class="math-container">$a_2\in A_2$</span> respectively.</p>

<p>(2) Player 1 and 2 observe the outcome of the 1st stage
<span class="math-container">$(a_1, a_2)$</span>, then simultaneously choose action  <span class="math-container">$a_3\in A_3$</span> and
<span class="math-container">$a_4\in A_4$</span> respectively.</p>

<p>Payoffs are <span class="math-container">$u_i(a_1, a_2, a_3, a_4)$</span> for <span class="math-container">$i = 1,2$</span>. </p>

<hr>

<p>Suppose that I do the following (typically called backward induction):</p>

<p>(A) find the functions <span class="math-container">$a^*_3(a_1,a_2)$</span>  and <span class="math-container">$a^*_4(a_1,a_2)$</span> such that </p>

<p><span class="math-container">$$
\begin{cases}
a_3^*(a_1, a_2)\in argmax_{a_3(\cdot)}u_1(a_1, a_2, a_3(a_1, a_2), a_4(a_1, a_2))\\
a_4^*(a_1, a_2)\in argmax_{a_4(\cdot)}u_2(a_1, a_2, a_3(a_1, a_2), a_4(a_1, a_2))\\
\end{cases}
$$</span> </p>

<p>(B) find <span class="math-container">$a_1^*, a_2^*$</span> such that
<span class="math-container">$$
\begin{cases}
a_1^*\in argmax_{a_1}u_1(a_1, a_2, a^*_3(a_1, a_2), a^*_4(a_1, a_2))\\
a_2^*\in argmax_{a_2}u_2(a_1, a_2, a^*_3(a_1, a_2), a^*_4(a_1, a_2))\\
\end{cases}
$$</span> </p>

<p><strong>Which equilibrium notion I'm I applying?</strong> </p>

<p>Basically, in (A) I find the pure strategy Nash Equilibrium of stage 2, for each possible action played in stage 1; in (B) I find the pure strategy Nash Equilibrium of stage 1, given the solution from (A).</p>

<p>How is <span class="math-container">$\{a_1^*, a_2^*, a^*_3(a_1, a_2), a^*_4(a_1, a_2)\}$</span> called?</p>

<game-theory><nash-equilibrium>
Solving a two stage game by backward induction: which is the equilibrium notion?
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# Solving a two stage game by backward induction: which is the equilibrium notion?

(Asked by: username_0 on 2019-02-02)

Take a two-stage game with complete information and simultaneous actions in each state:

(1) Player 1 and 2 simultaneously choose action $a\_1\in A\_1$ and
$a\_2\in A\_2$ respectively.

(2) Player 1 and 2 observe the outcome of the 1st stage
$(a\_1, a\_2)$, then simultaneously choose action $a\_3\in A\_3$ and
$a\_4\in A\_4$ respectively.

Payoffs are $u\_i(a\_1, a\_2, a\_3, a\_4)$ for $i = 1,2$.

---

Suppose that I do the following (typically called backward induction):

(A) find the functions $a^\*\_3(a\_1,a\_2)$ and $a^\*\_4(a\_1,a\_2)$ such that

$$
\begin{cases}
a\_3^\*(a\_1, a\_2)\in argmax\_{a\_3(\cdot)}u\_1(a\_1, a\_2, a\_3(a\_1, a\_2), a\_4(a\_1, a\_2))\\
a\_4^\*(a\_1, a\_2)\in argmax\_{a\_4(\cdot)}u\_2(a\_1, a\_2, a\_3(a\_1, a\_2), a\_4(a\_1, a\_2))\\
\end{cases}
$$

(B) find $a\_1^\*, a\_2^\*$ such that
$$
\begin{cases}
a\_1^\*\in argmax\_{a\_1}u\_1(a\_1, a\_2, a^\*\_3(a\_1, a\_2), a^\*\_4(a\_1, a\_2))\\
a\_2^\*\in argmax\_{a\_2}u\_2(a\_1, a\_2, a^\*\_3(a\_1, a\_2), a^\*\_4(a\_1, a\_2))\\
\end{cases}
$$

**Which equilibrium notion I'm I applying?**

Basically, in (A) I find the pure strategy Nash Equilibrium of stage 2, for each possible action played in stage 1; in (B) I find the pure strategy Nash Equilibrium of stage 1, given the solution from (A).

How is $\{a\_1^\*, a\_2^\*, a^\*\_3(a\_1, a\_2), a^\*\_4(a\_1, a\_2)\}$ called?

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

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## 1 answers

### Answer by username_1 on 2019-02-02. Score: 3 | Accepted answer

It is called a [subgame perfect equilibrium.](https://en.wikipedia.org/wiki/Subgame_perfect_equilibrium)


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