<p>The term I often come across is 'inefficient Nash equilibrium' or 'Pareto-inefficient Nash equilibrium'. Here is an older well cited article discussing inefficient NEs: <a href="http://www.jstor.org/stable/3690047?seq=1#page_scan_tab_contents" rel="nofollow">http://www.jstor.org/stable/3690047?seq=1#page_scan_tab_contents</a> </p>

<p>Here is the frequency of the term 'inefficient Nash equilibrium' used in scientific papers:
<a href="https://scholar.google.com/scholar?q=%22Inefficient+Nash+Equilibrium%22&amp;btnG=&amp;hl=en&amp;as_sdt=0%2C5" rel="nofollow">https://scholar.google.com/scholar?q=%22Inefficient+Nash+Equilibrium%22&amp;btnG=&amp;hl=en&amp;as_sdt=0%2C5</a></p>

<p>'Inefficient outcome' is also frequently used to describe NEs that are not Pareto-optimal: <a href="https://scholar.google.com/scholar?q=%22Inefficient+outcome%22&amp;btnG=&amp;hl=en&amp;as_sdt=0%2C5" rel="nofollow">https://scholar.google.com/scholar?q=%22Inefficient+outcome%22&amp;btnG=&amp;hl=en&amp;as_sdt=0%2C5</a>
But it has to be put into game-theoretical context and specified that it is NE.</p>

CC BY-SA 3.0
2015-12-01T02:30:33.050Z
I think these terms do not quite fit. A single Nash equilibrium may be inefficient, but another Nash equilibrium may not be, in which case the two sets are not disjoint.
HRSE
CC BY-SA 3.0
2015-12-01T02:41:23.967Z
@HRSE I agree. It would make sense only for a game with either one NE (like prisoner's dilemma) or many NE's, all of which are inefficient. So maybe there is a room for a better answer if anyone came across anything more specific.
Arthur Tarasov
CC BY-SA 3.0
2016-06-08T04:26:41.493Z
I propose "inefficient game".  Thoughts?
capybaralet
CC BY-SA 3.0
2016-05-16T19:37:35.707Z
I propose that we begin referring to such games as exhibiting the "user2429920 property"
John Q. Noob
CC BY-SA 3.0
2015-12-01T00:39:08.743Z
2015-12-01T00:39:08.743Z
Arthur Tarasov
<p>Is there a term for a game whose pareto optimal solutions and nash equilibria are disjoint?</p>

<p>(e.g. prisoner's dilemma)</p>

<game-theory><nash-equilibrium>
Is there a term for a game whose pareto optimal solutions and nash equilibria are disjoint?
capybaralet
# Is there a term for a game whose pareto optimal solutions and nash equilibria are disjoint?

(Asked by: username_0 on 2015-11-30)

Is there a term for a game whose pareto optimal solutions and nash equilibria are disjoint?

(e.g. prisoner's dilemma)

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


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## 1 answers

### Answer by username_1 on 2015-12-01. Score: 1

The term I often come across is 'inefficient Nash equilibrium' or 'Pareto-inefficient Nash equilibrium'. Here is an older well cited article discussing inefficient NEs: <http://www.jstor.org/stable/3690047?seq=1#page_scan_tab_contents>

Here is the frequency of the term 'inefficient Nash equilibrium' used in scientific papers:
<https://scholar.google.com/scholar?q=%22Inefficient+Nash+Equilibrium%22&btnG=&hl=en&as_sdt=0%2C5>

'Inefficient outcome' is also frequently used to describe NEs that are not Pareto-optimal: <https://scholar.google.com/scholar?q=%22Inefficient+outcome%22&btnG=&hl=en&as_sdt=0%2C5>
But it has to be put into game-theoretical context and specified that it is NE.

> **Comments:**
> **username_2**: I think these terms do not quite fit. A single Nash equilibrium may be inefficient, but another Nash equilibrium may not be, in which case the two sets are not disjoint.  
> **username_1**: @username_2 I agree. It would make sense only for a game with either one NE (like prisoner's dilemma) or many NE's, all of which are inefficient. So maybe there is a room for a better answer if anyone came across anything more specific.  
> **username_3**: I propose that we begin referring to such games as exhibiting the "user2429920 property"  
> **username_0**: I propose "inefficient game".  Thoughts?  

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