<p>The profile <span class="math-container">$(D,L)$</span> is Pareto optimal at all values of <span class="math-container">$x,y$</span> that satisfy <span class="math-container">$(x&lt;-1)\vee (-1&lt;x&lt;1 \wedge y&lt;-1) \vee (x=-1 \wedge y\leq -1)$</span></p>

CC BY-SA 4.0
2024-09-26T09:59:12.287Z
2024-09-26T09:59:12.287Z
Amit
<blockquote>
<p>For Pareto Optimality, x &amp; y should have values such that any improvement in one players  performance should create an adverse effect on the others.</p>
</blockquote>
<p>For (D,L) to be Pareto Optimal,</p>
<p>(U,L) -&gt; Since Player 2 has 5 rather than -1, that player is better off. So, the only way (U,L) is not an improvement is that if x &lt; 1.</p>
<p>(D,R) -&gt; Since Player 2 has 0 rather than -1, that player is better off. So, the only way (D,R) is not an improvement is that if x &lt; 1.</p>
<p>(U,R) -&gt; Since from the above 2 points x &lt; 1, so x+2 &lt; 3. That creates 2 cases :</p>
<ol>
<li>If 1 <span class="math-container">$\leq$</span> (x+2) &lt; 3 or -1 <span class="math-container">$\leq$</span> x &lt; 1, that makes player 1 better off. So y &lt; -1.</li>
<li>If (x+2) &lt; 1 or x &lt; -1, that makes player 1 worse off so -1 <span class="math-container">$\leq$</span> y.</li>
</ol>
<p><strong>Therefore,</strong></p>
<p><strong>1. if 1 <span class="math-container">$\leq$</span> (x+2) &lt; 3 or -1 <span class="math-container">$\leq$</span> x &lt; 1 then y &lt; -1</strong></p>
<p><strong>2. (x+2) &lt; 1 if or x &lt; -1, then y <span class="math-container">$\geq$</span> -1</strong></p>

CC BY-SA 4.0
2024-08-26T20:00:50.163Z
What do you mean "*where A and B are player 1 and player 2 resp.*"?
Herr K.
CC BY-SA 4.0
2024-08-27T05:47:53.970Z
It is simply $(x < -1) \ \vee \ (-1 < x < 1 \  \wedge \ y < -1) \ \vee \ (x=-1 \ \wedge \ y \leq -1) $.
Amit
CC BY-SA 4.0
2024-09-26T09:39:49.623Z
@Amit You should post your comment as answer.
Herr K.
CC BY-SA 4.0
2024-08-26T17:05:40.410Z
2024-08-27T05:06:57.140Z
2024-08-27T05:06:57.140Z
Souvik Ganguly
<p>For what values <span class="math-container">$x$</span>, <span class="math-container">$y$</span> the profile <span class="math-container">$(D,L)$</span> is Pareto optimal?</p>

<p><span class="math-container">\begin{array}{c|ccc}
&amp; L  &amp; R \\ \hline
U &amp; x,5 &amp; x+2,y  \\
D&amp; 1,-1 &amp; x,0 \\
\end{array}</span></p>

<p>Is correct <span class="math-container">$x&lt;1$</span> ?</p>

<p>I suppose this because <span class="math-container">$D$</span> should not be weakly dominated.</p>

<p>And <span class="math-container">$y&lt;5$</span> for the same argument.</p>

<p>Is these values enough to answer that <span class="math-container">$(D,L)$</span> is Pareto optimal?</p>

<p>Thanks!</p>

CC BY-SA 4.0
2019-05-20T14:09:56.067Z
$(x,y) \in \{(a,b)  \mid a<-5, ~ a + b \leq - 2\}$
clueless
CC BY-SA 4.0
2019-05-20T20:43:56.387Z
clueless.  Could you explain me a little? Thank you.
Franciscolli
<microeconomics><game-theory><nash-equilibrium>
Optimal pareto in two-person game
Franciscolli
# Optimal pareto in two-person game

(Asked by: username_0 on 2019-05-20)

For what values $x$, $y$ the profile $(D,L)$ is Pareto optimal?

\begin{array}{c|ccc}
& L & R \\ \hline
U & x,5 & x+2,y \\
D& 1,-1 & x,0 \\
\end{array}

Is correct $x<1$ ?

I suppose this because $D$ should not be weakly dominated.

And $y<5$ for the same argument.

Is these values enough to answer that $(D,L)$ is Pareto optimal?

Thanks!

**Tags:** microeconomics, game-theory, nash-equilibrium
> **Comments:**
> **username_1**: $(x,y) \in \{(a,b)  \mid a<-5, ~ a + b \leq - 2\}$  
> **username_0**: clueless.  Could you explain me a little? Thank you.  

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## 2 answers

### Answer by username_2 on 2024-09-26. Score: 1

The profile $(D,L)$ is Pareto optimal at all values of $x,y$ that satisfy $(x<-1)\vee (-1


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### Answer by username_3 on 2024-08-26. Score: 0

> For Pareto Optimality, x & y should have values such that any improvement in one players performance should create an adverse effect on the others.

For (D,L) to be Pareto Optimal,

(U,L) -> Since Player 2 has 5 rather than -1, that player is better off. So, the only way (U,L) is not an improvement is that if x < 1.

(D,R) -> Since Player 2 has 0 rather than -1, that player is better off. So, the only way (D,R) is not an improvement is that if x < 1.

(U,R) -> Since from the above 2 points x < 1, so x+2 < 3. That creates 2 cases :

1. If 1 $\leq$ (x+2) < 3 or -1 $\leq$ x < 1, that makes player 1 better off. So y < -1.
2. If (x+2) < 1 or x < -1, that makes player 1 worse off so -1 $\leq$ y.

**Therefore,**

**1. if 1 $\leq$ (x+2) < 3 or -1 $\leq$ x < 1 then y < -1**

**2. (x+2) < 1 if or x < -1, then y $\geq$ -1**

> **Comments:**
> **username_4**: What do you mean "*where A and B are player 1 and player 2 resp.*"?  
> **username_2**: It is simply $(x < -1) \ \vee \ (-1 < x < 1 \  \wedge \ y < -1) \ \vee \ (x=-1 \ \wedge \ y \leq -1) $.  
> **username_4**: @Amit You should post your comment as answer.  

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