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Nash equilibria and bifurcation point

Mixed-strategy Nash equilibriaNash equilibria and best response functionsGame theory: proof V is convex and compactTopology of the set of Nash equilibriaFinding mixed strategy Nash equilibria for a game with infinite strategiesNash Equilibria Vs. IESDSExamples of behavioural breakdowns in game theory?How to find Nash equilibria through KKT conditions (convex optimization)?Pure strategy Nash equilibriaHow are all kinds of equilibra in game theory connected to each other? (Please correct my summary)

1

$begingroup$

I have a non-cooperative game involving a number of players. I computed the Nash equilibria which depend on a parameter $alpha$. For $alpha<alpha_0$ there exist two Nash equilibria, e.g. $P$ and $Q$, for $alpha=alpha_0$ we have that $Pequiv Q$ while for $alpha>alpha_0$ we have only one of them, $P$.

Then, is $alpha_0$ a bifurcation point? I'm asking because I know that the notion of bifurcation point is usually associated to ODE (or to a system of ODEs) that depends on a parameter, while in my case I don't have ODE (but players with utility functions).

game-theory nash-equilibrium bifurcation

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asked Mar 21 at 12:59

MarkMark

3,46151947

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is a notion of a catastrophe when a change in critical points of smooth function or mapping depending on parameters is being discussed. However, I presume, in game theory there is very rarely a smooth objective function, it is more likely to be piecewise-smooth at best?
  $endgroup$
  – Evgeny
  Mar 22 at 10:08
  

  
  
  
  

add a comment | 

1

$begingroup$

I have a non-cooperative game involving a number of players. I computed the Nash equilibria which depend on a parameter $alpha$. For $alpha<alpha_0$ there exist two Nash equilibria, e.g. $P$ and $Q$, for $alpha=alpha_0$ we have that $Pequiv Q$ while for $alpha>alpha_0$ we have only one of them, $P$.

Then, is $alpha_0$ a bifurcation point? I'm asking because I know that the notion of bifurcation point is usually associated to ODE (or to a system of ODEs) that depends on a parameter, while in my case I don't have ODE (but players with utility functions).

game-theory nash-equilibrium bifurcation

share|cite|improve this question

asked Mar 21 at 12:59

MarkMark

3,46151947

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is a notion of a catastrophe when a change in critical points of smooth function or mapping depending on parameters is being discussed. However, I presume, in game theory there is very rarely a smooth objective function, it is more likely to be piecewise-smooth at best?
  $endgroup$
  – Evgeny
  Mar 22 at 10:08
  

  
  
  
  

add a comment | 

$begingroup$

I have a non-cooperative game involving a number of players. I computed the Nash equilibria which depend on a parameter $alpha$. For $alpha<alpha_0$ there exist two Nash equilibria, e.g. $P$ and $Q$, for $alpha=alpha_0$ we have that $Pequiv Q$ while for $alpha>alpha_0$ we have only one of them, $P$.

Then, is $alpha_0$ a bifurcation point? I'm asking because I know that the notion of bifurcation point is usually associated to ODE (or to a system of ODEs) that depends on a parameter, while in my case I don't have ODE (but players with utility functions).

game-theory nash-equilibrium bifurcation

share|cite|improve this question

asked Mar 21 at 12:59

MarkMark

3,46151947

$endgroup$

share|cite|improve this question

asked Mar 21 at 12:59

MarkMark

3,46151947

share|cite|improve this question

asked Mar 21 at 12:59

MarkMark

3,46151947

asked Mar 21 at 12:59

MarkMark

3,46151947

asked Mar 21 at 12:59

MarkMark

3,46151947