<p>There are a couple of imprecisions. First, there are 3 subgames: two that start with player 2 moving, and the complete game is also a subgame. </p>

<p>Second I think that A is a best response for player 1 iff <span class="math-container">$r\geq \frac{z-y}{x-y}$</span> otherwise, B is the best response (perhaps this was just a typo). </p>

<p>Third, note that the fraction is guaranteed to be positive and smaller than 1 given the restrictions on the parameters,  but it is not necessarily equal to <span class="math-container">$0.5$</span>, For example, let <span class="math-container">$x=5$</span>, <span class="math-container">$z=3$</span> and <span class="math-container">$y=2$</span>, then the fraction is equal to 1/3. </p>

<p>Fourth, you are missing other equilibria, when <span class="math-container">$r=\frac{z-y}{x-y}$</span>, then player 1 is indifferent between A and B, and any randomization of A and B also constitutes an equilibrium.</p>

<p>Lastly, when writing the equilibria, you must specify what player 2 does in each information set (or node in this game) </p>

<p>Thus: <span class="math-container">$$NE=\{(A, (ra+(1-r)b)b)|r\geq \frac{z-y}{x-y}\}\cup\{(B, (ra+(1-r)b)b)|r\leq \frac{z-y}{x-y}\}$$</span>
<span class="math-container">$$\cup \{((pA+(1-p)B), (ra+(1-r)b)b)|r=\frac{z-y}{x-y}, 0\leq p\leq 1\}$$</span></p>

<p>Where <span class="math-container">$(ra+(1-r)b)b$</span> specifies that player 2 randomizes in the first node and chooses b in the second node.</p>

CC BY-SA 4.0
2020-01-03T00:54:58.963Z
2020-01-03T00:54:58.963Z
Regio
<p>I have doubts about whether my solution is right. 
<a href="https://i.sstatic.net/kmQFP.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kmQFP.png" alt="game"></a></p>

<p>Let <span class="math-container">$x&gt;z&gt;y&gt;1$</span>.
Given the extended game above, we have two subgames and will find the Nash equilibria of these.</p>

<p>Observe that player 2 will play <span class="math-container">$b$</span> in the right subgame.
In the left subgame player 2 will be indifferent between playing <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.
Let <span class="math-container">$0 \le r \le 1$</span> such that <span class="math-container">$(r,1-r)$</span> is a mixed strategy for player 2 and such that the player plays this strategy in the left subgame. </p>

<p>Player 1 has expected payoffs
<span class="math-container">$$
E u(s) = 
(rx+(1-r)y) \cdot 1_{s=A} + z \cdot 1_{s=B}
$$</span>
Observe that player 1's best response is <span class="math-container">$A$</span> if <span class="math-container">$r \ge (z-1)/(x-y)$</span> and <span class="math-container">$B$</span> if <span class="math-container">$r \le (z-1)/(x-y)$</span>, hence SPNE will be 
<span class="math-container">$$
\{(A,(r,1-r))|r \ge (z-1)/(x-y) \} \cup \{(B,(r,1-r))|r \le (z-1)/(x-y)\}
$$</span>
Is this correct? </p>

<game-theory><extensive-games>
Extended game mixed SPNE
bubububub
# Extended game mixed SPNE

(Asked by: username_0 on 2020-01-02)

I have doubts about whether my solution is right.
[![game](https://i.sstatic.net/kmQFP.png)](https://i.sstatic.net/kmQFP.png)

Let $x>z>y>1$.
Given the extended game above, we have two subgames and will find the Nash equilibria of these.

Observe that player 2 will play $b$ in the right subgame.
In the left subgame player 2 will be indifferent between playing $a$ and $b$.
Let $0 \le r \le 1$ such that $(r,1-r)$ is a mixed strategy for player 2 and such that the player plays this strategy in the left subgame.

Player 1 has expected payoffs
$$
E u(s) =
(rx+(1-r)y) \cdot 1\_{s=A} + z \cdot 1\_{s=B}
$$
Observe that player 1's best response is $A$ if $r \ge (z-1)/(x-y)$ and $B$ if $r \le (z-1)/(x-y)$, hence SPNE will be
$$
\{(A,(r,1-r))|r \ge (z-1)/(x-y) \} \cup \{(B,(r,1-r))|r \le (z-1)/(x-y)\}
$$
Is this correct?

**Tags:** game-theory, extensive-games

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## 1 answers

### Answer by username_1 on 2020-01-03. Score: 1

There are a couple of imprecisions. First, there are 3 subgames: two that start with player 2 moving, and the complete game is also a subgame.

Second I think that A is a best response for player 1 iff $r\geq \frac{z-y}{x-y}$ otherwise, B is the best response (perhaps this was just a typo).

Third, note that the fraction is guaranteed to be positive and smaller than 1 given the restrictions on the parameters, but it is not necessarily equal to $0.5$, For example, let $x=5$, $z=3$ and $y=2$, then the fraction is equal to 1/3.

Fourth, you are missing other equilibria, when $r=\frac{z-y}{x-y}$, then player 1 is indifferent between A and B, and any randomization of A and B also constitutes an equilibrium.

Lastly, when writing the equilibria, you must specify what player 2 does in each information set (or node in this game)

Thus: $$NE=\{(A, (ra+(1-r)b)b)|r\geq \frac{z-y}{x-y}\}\cup\{(B, (ra+(1-r)b)b)|r\leq \frac{z-y}{x-y}\}$$
$$\cup \{((pA+(1-p)B), (ra+(1-r)b)b)|r=\frac{z-y}{x-y}, 0\leq p\leq 1\}$$

Where $(ra+(1-r)b)b$ specifies that player 2 randomizes in the first node and chooses b in the second node.


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