<p>Take any <span class="math-container">$a\in(0,3)$</span>.
Since <span class="math-container">$a&gt;0$</span>, X is a best response (BR) to x. Since <span class="math-container">$a&lt;3$</span>, Y is BR against y, and Z is BR against z.
Similarly, x is BR against X, z against Y, and y against Z.
Hence, your only pure strategy NE is (X,x). The latter is also true if <span class="math-container">$a=3$</span>.
If <span class="math-container">$a&gt;3$</span>, X and x are strictly dominant strategies such that (X,x) is the unique NE.</p>
<p>Now back to case <span class="math-container">$a\in(0,3)$</span>.
Suppose 1 randomizes with probabilities <span class="math-container">$p_X,p_Y,p_Z$</span>, and 2 randomizes with <span class="math-container">$q_x,q_y,q_z$</span>. If all those probabilities are positive, both must be indifferent over all three options. Let <span class="math-container">$\pi^i_a$</span> be the expected payoff of player <span class="math-container">$i$</span> playing action <span class="math-container">$a$</span>.</p>
<p><span class="math-container">$\pi^2_x=a, \pi^2_y=3p_Z, \pi^2_z=3p_Y$</span> and the <span class="math-container">$\pi^1$</span> look similar. You need <span class="math-container">$\pi^2_x=a = \pi^2_y=3p_Z = \pi^2_z=3p_Y$</span> if 2 shall randomize over all three. Hence, we need <span class="math-container">$p_Y=p_Z=\frac{a}{3}$</span>, which is a probability as <span class="math-container">$a&lt;3$</span>. But since <span class="math-container">$p_X+p_Y+p_Z=1$</span> we also need <span class="math-container">$a\leq \frac{3}{2}$</span>. From the same exercise for player 1, we get <span class="math-container">$q_y=q_z=\frac{a}{3}$</span> and also condition <span class="math-container">$a\leq \frac{3}{2}$</span>.</p>
<p>Hence, if <span class="math-container">$a&lt;\frac{3}{2}$</span>, we have a mixed strategy Nash equilibrium in which both players randomize over all three of their actions with probabilities <span class="math-container">$q_y=q_z=pY=p_Z=\frac{a}{3}$</span> and <span class="math-container">$q_z=p_Z=1-2\frac{a}{3}=\frac{3-2a}{3}$</span>. Such that both of them expect equilibrium payoff <span class="math-container">$a$</span>. Any unilateral deviation would also lead to payoff <span class="math-container">$a$</span>.</p>
<p>We also have an equilibrium in which both players only randomize over <span class="math-container">$Y,Z$</span> and <span class="math-container">$y,z$</span>, respectively, all with probability <span class="math-container">$\frac{1}{2}$</span>. In this case, equilibrium utility of <span class="math-container">$\frac{3}{2}$</span> cannot be improved upon by a deviation to <span class="math-container">$X$</span> or <span class="math-container">$x$</span>, respectively, as <span class="math-container">$a&lt;\frac{3}{2}$</span>, and the other two pure strategies yield the equilibrium utility by construction.</p>
<p>Both these equilibria would not exist if <span class="math-container">$a&gt;\frac{3}{2}$</span>, as a deviation to  <span class="math-container">$X$</span> or <span class="math-container">$x$</span>, respectively, would be better or the indifference condition could not be fulfilled.</p>
<p>For any <span class="math-container">$a&gt;0$</span>, there cannot be an NE in which 1 mixes between X and Y and 2 between x and y, because 2 can only be indifferent between x and y if 1 also plays Z, <span class="math-container">$p_Z=0 \Rightarrow \pi^2_x=a&gt;\pi^2_y=0$</span>.
Similarly, the other combinations (such as X and Z and x and z) would violate the indifference condition.</p>

CC BY-SA 4.0
2020-11-19T14:59:17.710Z
2020-11-19T14:59:17.710Z
Bayesian
<p><span class="math-container">$$\begin{aligned}
&amp;\begin{array}{|c|c|c|c|}
\hline \text{Player 1 / 2}&amp; x &amp; y &amp; z \\
\hline X &amp; a, a &amp; a, 0 &amp; a, 0 \\
\hline Y &amp; 0, a &amp; 3,0 &amp; 0,3 \\
\hline Z &amp; 0, a &amp; 0,3 &amp; 3,0 \\
\hline
\end{array}
\end{aligned}$$</span></p>
<p>Consider that <span class="math-container">$a \in (0, +\infty)$</span> I know that if <span class="math-container">$a=1$</span> then the NE is <span class="math-container">$(X,x)$</span> yet I don't know how to solve for the NE when <span class="math-container">$a \in (1,3)$</span> and <span class="math-container">$a \in(0,1)$</span> . My attempt was writting the conditions of indifference to find the mixed strategy NE but somehow I still don't arrive to a clear answer. Also I don't now if I should consider all of the possibilities of the support or if there is any shortcut.</p>

<game-theory>
Parameter in normal form NE
ceins
# Parameter in normal form NE

(Asked by: username_0 on 2020-11-06)

$$\begin{aligned}
&\begin{array}{|c|c|c|c|}
\hline \text{Player 1 / 2}& x & y & z \\
\hline X & a, a & a, 0 & a, 0 \\
\hline Y & 0, a & 3,0 & 0,3 \\
\hline Z & 0, a & 0,3 & 3,0 \\
\hline
\end{array}
\end{aligned}$$

Consider that $a \in (0, +\infty)$ I know that if $a=1$ then the NE is $(X,x)$ yet I don't know how to solve for the NE when $a \in (1,3)$ and $a \in(0,1)$ . My attempt was writting the conditions of indifference to find the mixed strategy NE but somehow I still don't arrive to a clear answer. Also I don't now if I should consider all of the possibilities of the support or if there is any shortcut.

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2020-11-19. Score: 4

Take any $a\in(0,3)$.
Since $a>0$, X is a best response (BR) to x. Since $a<3$, Y is BR against y, and Z is BR against z.
Similarly, x is BR against X, z against Y, and y against Z.
Hence, your only pure strategy NE is (X,x). The latter is also true if $a=3$.
If $a>3$, X and x are strictly dominant strategies such that (X,x) is the unique NE.

Now back to case $a\in(0,3)$.
Suppose 1 randomizes with probabilities $p\_X,p\_Y,p\_Z$, and 2 randomizes with $q\_x,q\_y,q\_z$. If all those probabilities are positive, both must be indifferent over all three options. Let $\pi^i\_a$ be the expected payoff of player $i$ playing action $a$.

$\pi^2\_x=a, \pi^2\_y=3p\_Z, \pi^2\_z=3p\_Y$ and the $\pi^1$ look similar. You need $\pi^2\_x=a = \pi^2\_y=3p\_Z = \pi^2\_z=3p\_Y$ if 2 shall randomize over all three. Hence, we need $p\_Y=p\_Z=\frac{a}{3}$, which is a probability as $a<3$. But since $p\_X+p\_Y+p\_Z=1$ we also need $a\leq \frac{3}{2}$. From the same exercise for player 1, we get $q\_y=q\_z=\frac{a}{3}$ and also condition $a\leq \frac{3}{2}$.

Hence, if $a<\frac{3}{2}$, we have a mixed strategy Nash equilibrium in which both players randomize over all three of their actions with probabilities $q\_y=q\_z=pY=p\_Z=\frac{a}{3}$ and $q\_z=p\_Z=1-2\frac{a}{3}=\frac{3-2a}{3}$. Such that both of them expect equilibrium payoff $a$. Any unilateral deviation would also lead to payoff $a$.

We also have an equilibrium in which both players only randomize over $Y,Z$ and $y,z$, respectively, all with probability $\frac{1}{2}$. In this case, equilibrium utility of $\frac{3}{2}$ cannot be improved upon by a deviation to $X$ or $x$, respectively, as $a<\frac{3}{2}$, and the other two pure strategies yield the equilibrium utility by construction.

Both these equilibria would not exist if $a>\frac{3}{2}$, as a deviation to  $X$ or $x$, respectively, would be better or the indifference condition could not be fulfilled.

For any $a>0$, there cannot be an NE in which 1 mixes between X and Y and 2 between x and y, because 2 can only be indifferent between x and y if 1 also plays Z, $p\_Z=0 \Rightarrow \pi^2\_x=a>\pi^2\_y=0$.
Similarly, the other combinations (such as X and Z and x and z) would violate the indifference condition.


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