Subgame Perfect Nash Eqm

You misunderstand subgame perfect equilibrium.

'No move' is not part of the strategy. If you have more branches you still have to specify in your equilibrium strategy those branches of the game that are not even reached.

E.g. Imagine someone can choose left middle righ, then the second person can choose high low. even if the first will never choose left and middle, you still have to specify what Player 2 would pick in both cases.

So in your example, since Player 2 gets 5 no matter what, we actually do have an equilibrium that is:

Player 1: R
Player 2: L

Then the payoffs are 10 and 5. 2 can't get better off by choosing R still gets 5. 1 Cant be better off by choosing L still gets 10. This may confuse you but the way to check for Nash equilibrium is a one off deviation by one player. You fix all other decisions, and see if one player can do better by deviating. So yes PLayer 2 deviating to R would make 1 worse off, but that would not be your strategy anymore.

This is a nash equilibrium because you have specified 2 take L. Another in this game is

Player 1: L
Player 2: L

or

Player 1: L
Player 2: R

both work too.

(edited 13 years ago)

Reply 2

OP

Thanks a lot for your help.

So, can I confirm:

Pure Nash eqm: (L,R) & (L,L) & (R,L)?

Subgame Perfect Nash eqm: (L,R) & (L,L)?

Thanks once again.

Original post by danny111

You misunderstand subgame perfect equilibrium.

'No move' is not part of the strategy. If you have more branches you still have to specify in your equilibrium strategy those branches of the game that are not even reached.

E.g. Imagine someone can choose left middle righ, then the second person can choose high low. even if the first will never choose left and middle, you still have to specify what Player 2 would pick in both cases.

So in your example, since Player 2 gets 5 no matter what, we actually do have an equilibrium that is:

Player 1: R
Player 2: L

Then the payoffs are 10 and 5. 2 can't get better off by choosing R still gets 5. 1 Cant be better off by choosing L still gets 10. This may confuse you but the way to check for Nash equilibrium is a one off deviation by one player. You fix all other decisions, and see if one player can do better by deviating. So yes PLayer 2 deviating to R would make 1 worse off, but that would not be your strategy anymore.

This is a nash equilibrium because you have specified 2 take L. Another in this game is

Player 1: L
Player 2: L

or

Player 1: L
Player 2: R

both work too.

Reply 3

Original post by BJP

Thanks a lot for your help.

So, can I confirm:

Pure Nash eqm: (L,R) & (L,L) & (R,L)?

Subgame Perfect Nash eqm: (L,R) & (L,L) & (R,L)?

Thanks once again.

Because if 1 knows that 2 takes L, then he knows he will have a payoff of 10. So by picking L or R he gets the same.

This is a pretty poor example, since the payoffs are so similar everywhere.

Reply 4

OP

Original post by danny111

Because if 1 knows that 2 takes L, then he knows he will have a payoff of 10. So by picking L or R he gets the same.

This is a pretty poor example, since the payoffs are so similar everywhere.

That's what I thought too! I think that's why it's catching me out to be honest.

So all three equilibriums mentioned are pure Nash and subgame perfect Nash?

With regards to the (R,L) one, I've read somewhere a section about credible promises/threats related to subgame perfect eqm. So I did wonder if II choosing L wouldn't be credible (and thus make it not a subgame perfect eqm) because II stands to gain nothing by choosing L over R.

(edited 13 years ago)

prisoner's dilemma ?

Original post by BJP

That's what I thought too! I think that's why it's catching me out to be honest.

So all three equilibriums mentioned are pure Nash and subgame perfect Nash?

With regards to the (R,L) one, I've read somewhere a section about credible promises/threats related to subgame perfect eqm. So I did wonder if II choosing L wouldn't be credible (and thus make it not a subgame perfect eqm) because II stands to gain nothing by choosing L over R.

A threat is credible if the player that is threatening actually would play the action that he threatens with. If he would lose out as well by the action he threatens to play then it is not credible.

Here the threat is not credible in the sense that why would he make a threat? His payoffs are the same everywhere. He has no incentive to change.

Reply 7

OP

To be a pure Nash eqm, no player should be able to improve their pay-off by unilaterally changing their strategy right?
For (R,R) I can obviously attain higher pay-off (10>5) but to do this, he/she must go back to the first decision node in the game...so can he still do that (and thus it's not a pure Nash eqm)? Or, because it's a previous node, can he now not go back and change strategy (and thus it is a Nash eqm, because nobody can improve by changing just their own strategy). I.e. should I look at possibilities of changing previous node decisions for pure Nash eqm...or is that just to be looked at for subgame perfect eqm?
Thanks if anybody knows.

Reply 8

Original post by BJP

To be a pure Nash eqm, no player should be able to improve their pay-off by unilaterally changing their strategy right?
For (R,R) I can obviously attain higher pay-off (10>5) but to do this, he/she must go back to the first decision node in the game...so can he still do that (and thus it's not a pure Nash eqm)? Or, because it's a previous node, can he now not go back and change strategy (and thus it is a Nash eqm, because nobody can improve by changing just their own strategy). I.e. should I look at possibilities of changing previous node decisions for pure Nash eqm...or is that just to be looked at for subgame perfect eqm?
Thanks if anybody knows.

It's really hard to explain without a pen and paper where I can draw a tree.

The unilateral deviation is like this:

Imagine there are a 10 actions. 5 from each player. Let the other player take a strategy and FIX that. If I have a strategy, it is a nash equilibrium (if the following holds for me and the other player taking my strategy as fixed) if i cannot do better by changing one of my 5 actions - but the other 4 stay the same.

So say I have actions

1
2
3
4
5

If I change one at a time, and i do no better then its NE. Of course it might be possible that given the 5 choices of the other player, that if I change say 1 & 5 that I might do better. But that is not the point of checking for NE, because changing 2 is not a unilateral deviation anymore. The reason why you only look at changing 1 action at a time comes from the definition of NE.

I will not go into that because there is no point and I also don't remember it. It is very difficult notation wise, but if you define everything correctly, you can show that a NE is equivalent to not being able to do better from a unilateral deviation.

Reply 9

Oh and the difference between NE and subgame perfect NE:

for NE you take the tree as whole. for subgame, every subgame (hence the name) of the tree, must be a NE.

Reply 10

OP

Original post by danny111

Oh and the difference between NE and subgame perfect NE:

for NE you take the tree as whole. for subgame, every subgame (hence the name) of the tree, must be a NE.

Thanks for your replies (and your patience - I realise I'm probably being unintentionally awkward).

So, in relation to my game:

1.) (L,R), (L,L) and (R,L) are NE because neither player can change strategy to improve their pay-offs. (R,R) is not NE because player I can go back to the first decision node of the game and change his decision to L, thus improving his pay-off?

2.) (L,R), (L,L) and (R,L) are subgame perfect equilibrium because each of the sub games is in equilibrium?

Thanks a lot again.

(edited 13 years ago)

Reply 11