8
$\begingroup$

I've been given a question which I'm struggling with:

Take the standard Prisoner's Dilemma game and consider it is played twice. (Players observe the outcome of the first game before playing the second). Consider beliefs in terms of which node player 2 is in their information set.

Find a weak perfect bayesian equilibrium (strategies and beliefs) where the strategies aren't a sub-game perfect equilibria.

So in the Prisoner's Dilemma:

(Defect, Defect) is unique nash and so is also the unique sub-game perfect equilibrium.

But how can we get a weak perfect bayesian equilibrium that doesn't involve Defect? Surely this is strictly dominant . . .

Is the question wrong?

It then goes on to ask for sequential equilibria (where we consider the sequence of mixed strategies).

Is this question wrong or am I misunderstanding these concepts?

$\endgroup$
  • $\begingroup$ That isn't answering the question but just providing a pedantic point . . . In fact the strategy has to consist of 5 elements. $\endgroup$ – Brian Apr 23 '15 at 21:53
  • $\begingroup$ Given your comment I now think that your problem lies elsewhere: If you choose a dominated strategy in a subgame that is off the equilibrium path (so one that does not actually occur) your payoff does not decrease. $\endgroup$ – Giskard Apr 23 '15 at 22:02
  • $\begingroup$ So I understand that beliefs off-the-equilibrium path may be arbitrary (and thus don't have to be according to Bayesian updating) but I am under the impression that sequential rationality must hold (i.e. given those beliefs, the individual must be playing their best strategy). So in response to your suggestion, wouldn't a dominated strategy violate sequential rationality? $\endgroup$ – Brian Apr 23 '15 at 22:13
  • 3
    $\begingroup$ @denesp: Weak PBE is "weak" not because it doesn't require sequential rationality off equilibrium path, but because it does not require beliefs to be consistent with Bayes rule off equilibrium path. While I agree that in the case of twice repeated prisoner's dilemma (PD) there is no WPBE with non-subgame perfect strategies, this conclusion doesn't hold in general. The reason is because defect is a strictly dominant strategy in PD, thus for any belief off equilibrium path (even if inconsistent with Bayes' rule), defect is still sequentially rational. $\endgroup$ – Herr K. Apr 24 '15 at 18:17
  • 1
    $\begingroup$ However, for games without a dominant strategy, we could manipulate off equilibrium beliefs in such a way to make non-subgame perfect strategies to be sequentially rational. If we strengthens the consistency requirement on beliefs (such as required in sequential equilibrium) by forcing Bayes rule to hold even off equilibrium, then we can rule out non-subgame perfect strategies. Thus we have the result that sequential equilibrium implies both WPBE and SPE. $\endgroup$ – Herr K. Apr 24 '15 at 18:20
2
$\begingroup$

Let the strategy of player 1 be represented by $(x1_1,xDD_1,xDC_1,xCD_1,xCC_1)$ where $x1$ is the first round action of player 1, $xDD_1$ is the action taken at the information set where both players have defected in the first round, $xDC_1$ is the action taken at the information set where player 1 has defected and player 2 has cooperated in round 1, etc.. Note that something like $(x1_1,x2_1)$ (with $x2_1$ being the action taken in round 2) is never a full specification of the strategy of player 1, since we need to specify behavior at each information set separately. Define the strategies of player 2 similarly. However, a perfect Bayesian equilibrium must also specify the beliefs of the player, $\mu_1,\mu_2$. This is an important part of the specification of an equilibrium. As we will see below, the question is geared towards understanding that a different equilibrium does not require the strategies to differ. A difference in beliefs is sufficient to count as a different equilibrium.

The perfect equilibrium is given by: $((D,D,D,D,D),\mu_1)$ for player 1 and $((D,D,D,D,D),\mu_2)$ for player 2, where $\mu_1$ and $\mu_2$ are consistent beliefs at all information sets.

As has been noted in the comments, since "defect" is a dominated strategy irrespective of beliefs, even in a weak perfect Bayesian equilibrium the strategy profiles must be $(D,D,D,D,D)$ for both players. However, the following is now also a weak perfect Bayesian Nash equilibrium: $((D,D,D,D,D),\mu_1')$ and $((D,D,D,D,D),\mu_2')$ with $\mu_1'$, $\mu_2'$ consistent on the equilibrium path.

Thus, the question is not wrong, it simply shows that two weak perfect Bayesian Nash equilibria can have identical strategies as long as they differ in beliefs off the equilibrium path.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.