<p>Consider a two player simultaneous game where each player chooses a non-negative value, $x_i$, $i=1,2$. The payoff to each player is $π_i(x_1,x_2)=2x_i+2x_1x_2-x_i^2.$</p>

<p>a. What are the pure strategy Nash equilibria of this game?</p>

<p>b. Suppose that the set of strategies is limited to $[0,M],M&gt;0$. What are the pure strategy Nash equilibria of the game now?</p>

<p>For a) I got the answer that there are no PSNE since there's no intersection between the best response function of both players. Please correct me if that's wrong! </p>

<p>For b) I got very confused because I'm unsure how bounding the values of $x$ will make the conclusion any different from a). I realize that the payoff function provided doesn't necessarily mean that the larger $x_i$ is, the larger the payoff will be.</p>

<p>Below is my proof:</p>

<p>Consider $π_i(x_1,x_2)=2x_i+2x_1x_2-x_i^2.$. We first assume $x_1&gt; x_2$. In order to find out if a larger number leads to a higher payoff, i.e. </p>

<p>if $2x_1+2x_1x_2-x_1^2&gt;2x_2+2x_1x_2-x_2^2$</p>

<p>$2x_1-x_1^2&gt;2x_2-x_2^2$</p>

<p>$x_1(2-x_1)&gt; x_ 2(2-x_2)$</p>

<p>Following from $x_1&gt;x_2$,</p>

<p>$2-x_1&lt;2-x_2$</p>

<p>$x_1(2-x_1)&gt; x_ 2(2-x_2)$ for $0&lt;x_1\leq 1 $ and $0&lt;x_2&lt; 1$</p>

<p>and</p>

<p>$x_2(2-x_2)&gt; x_ 1(2-x_1)$ for $x_1&gt;1 $ and $x_2\ge 1$, where $x_2&lt;x_1$</p>

<p>Therefore, in order to maximize the payoff of each player and find the intersection of their best response such that they will have no incentive to deviate, with the condition that $[0,M]$, we find that:</p>

<p>As long as $0&lt;M\leq 1$, the PSNE will be {$P1: M, P2: M$}. When $M&gt;1$, the PSNE will be {$P1: 1, P2: 1$}.</p>

<p>I'm quite unsure if I'm right though, and would love to get some input!</p>

CC BY-SA 3.0
2017-11-05T13:56:13.600Z
So if $M = 2$ then $x_1 = x_2 = 1$ is a Nash-equilibrium? The constraints are not active here, meaning this is also a Nash-equilibrium of the original game. But that game had none!
Giskard
CC BY-SA 3.0
2017-11-05T14:41:28.493Z
@denesp oh that makes sense! But why did my working indicate that a higher $x_i$ value doesn't necessarily indicate a higher payoff? Wouldn't that mean that both players aren't maximizing their payoff if the only nash equilibrium is at $M,M$ regardless of the range?
iamtrying
CC BY-SA 3.0
2017-11-05T14:46:26.183Z
Comparing the two players' payoffs just does not mean anything. Let's play a game. Whoever says the bigger number wins a dollar from the other. If we  both say the same number, it is a draw. Now if we both say five, it is a draw. We both get a payoff of 0. Neither payoff is larger than the other. Yet both of us saying five is obviously not an equilibrium.
Giskard
CC BY-SA 3.0
2017-11-05T14:47:15.307Z
I recommend reading about constrained optimization. That seems to be your problem here, not game theory.
Giskard
CC BY-SA 3.0
2017-11-05T17:14:02.737Z
@denesp I just worked it out through brute force by drawing a table, and I think I see where you're coming from. Will look up constrained optimization to get a better understanding! Thanks so much for your help! (:
iamtrying
CC BY-SA 3.0
2017-11-08T20:52:12.607Z
@denesp: Write an answer?
Herr K.
CC BY-SA 3.0
2017-11-08T21:16:18.160Z
@HerrK. I would rather vote to close. The problem is homework level and very specific.
Giskard
<game-theory>
Game Theory PSNE of a two player simultaneous game
iamtrying
# Game Theory PSNE of a two player simultaneous game

(Asked by: username_0 on 2017-11-05)

Consider a two player simultaneous game where each player chooses a non-negative value, $x\_i$, $i=1,2$. The payoff to each player is $π\_i(x\_1,x\_2)=2x\_i+2x\_1x\_2-x\_i^2.$

a. What are the pure strategy Nash equilibria of this game?

b. Suppose that the set of strategies is limited to $[0,M],M>0$. What are the pure strategy Nash equilibria of the game now?

For a) I got the answer that there are no PSNE since there's no intersection between the best response function of both players. Please correct me if that's wrong!

For b) I got very confused because I'm unsure how bounding the values of $x$ will make the conclusion any different from a). I realize that the payoff function provided doesn't necessarily mean that the larger $x\_i$ is, the larger the payoff will be.

Below is my proof:

Consider $π\_i(x\_1,x\_2)=2x\_i+2x\_1x\_2-x\_i^2.$. We first assume $x\_1> x\_2$. In order to find out if a larger number leads to a higher payoff, i.e.

if $2x\_1+2x\_1x\_2-x\_1^2>2x\_2+2x\_1x\_2-x\_2^2$

$2x\_1-x\_1^2>2x\_2-x\_2^2$

$x\_1(2-x\_1)> x\_ 2(2-x\_2)$

Following from $x\_1>x\_2$,

$2-x\_1<2-x\_2$

$x\_1(2-x\_1)> x\_ 2(2-x\_2)$ for $0

and

$x\_2(2-x\_2)> x\_ 1(2-x\_1)$ for $x\_1>1 $ and $x\_2\ge 1$, where $x\_2

Therefore, in order to maximize the payoff of each player and find the intersection of their best response such that they will have no incentive to deviate, with the condition that $[0,M]$, we find that:

As long as $01$, the PSNE will be {$P1: 1, P2: 1$}.

I'm quite unsure if I'm right though, and would love to get some input!

**Tags:** game-theory
> **Comments:**
> **username_1**: So if $M = 2$ then $x_1 = x_2 = 1$ is a Nash-equilibrium? The constraints are not active here, meaning this is also a Nash-equilibrium of the original game. But that game had none!  
> **username_0**: @denesp oh that makes sense! But why did my working indicate that a higher $x_i$ value doesn't necessarily indicate a higher payoff? Wouldn't that mean that both players aren't maximizing their payoff if the only nash equilibrium is at $M,M$ regardless of the range?  
> **username_1**: Comparing the two players' payoffs just does not mean anything. Let's play a game. Whoever says the bigger number wins a dollar from the other. If we  both say the same number, it is a draw. Now if we both say five, it is a draw. We both get a payoff of 0. Neither payoff is larger than the other. Yet both of us saying five is obviously not an equilibrium.  
> **username_1**: I recommend reading about constrained optimization. That seems to be your problem here, not game theory.  
> **username_0**: @denesp I just worked it out through brute force by drawing a table, and I think I see where you're coming from. Will look up constrained optimization to get a better understanding! Thanks so much for your help! (:  
> **username_2**: @denesp: Write an answer?  
> **username_1**: @HerrK. I would rather vote to close. The problem is homework level and very specific.  

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