Credible Threat in the Perfect Bayesian Equilibrium

Consider a simple game between P1 and P2 shown in Figure 1.

We consider a Perfect Bayesian Equilibrium. P1 uses a cutoff strategy based on his $$c_1$$ that has the Cumulative Distribution Function of $$F$$. If his $$c_1$$ is lower than a threshold $$c_1^*$$, which depends on P2's strategy, P1 takes the strategy $$x$$.

Similarly, P2 uses a cutoff strategy based on his $$c_2$$ that has the Cumulative Distribution Function of $$G$$. If his $$c_2$$ is lower than a threshold $$c_2^*$$, which depends on P1's strategy, P2 takes the strategy $$a$$.

We denote the optimal cutoff strategies as $$F(c_1^*)$$ and $$G(c_2^*)$$.

Suppose the game has slightly changed as shown in Figure 2.

Now P1 has one more turn. Suppose that, given the initial $$F(c_1^*)$$ and $$G(c_2^*)$$, we set P1's gain from the strategy j as always being equal to the strategy k. We assume that P1 takes the strategy j if P1 is indifferent between j and k.

In this case, does P2 still continue to use the same cutoff strategy $$G(c_2^*)$$? For P2, the node from the path $$k$$ is never reached while P2 cannot verify it. As long as P2 uses the same cutoff strategy $$G(c_2^*)$$, P1 will take the strategy $$j$$. I am wondering if P2's threat is credible.

In the new game, P2's gain remains unchanged. However, P2 may try to increase his gain by deviating from the optimal cutoff strategy $$G(c_2^*)$$ if P1 never use the strategy k. But, P1 will be aware of it and may use the strategy k again. If so, P2 will again use the same optimal cutoff strategy $$G(c_2^*)$$. I got stuck here.

I appreciate any help.

• What is the question? – Giskard May 16 '19 at 15:23
• Thank you for your comment. The question is whether P2 continues to use the same cutoff strategy in the new game. – Tom M. May 16 '19 at 15:26
• I don't understand what you mean by a cut-off strategy. That term usually implies that there is some random variable that is out of the control of the player. I feel that you instead mean to say that the players are playing a random strategy where strategy $x$ is played with probability $p_1=F(c_1^*)$ and $y$ with complementary probability. Similarly for player $2$. Are you considering this strategies to be properly mixed, i.e. $F(c_1^*)\neq\{0,1\}$. If that is the case, all information sets are riched with positive probability so every Bayesian equilibrium is Perfect. – Regio May 16 '19 at 19:33

I am going to assume that $$F$$ and $$G$$ are continuous distribution functions (I.e. players are playing properly mixed strategies). If this is not the case there is not enough information to answer your question. Given that assumption, player 1 and 2 must be indifferent among their actions in game 1 given the strategies of the other player.

In the second game, you are assuming that the node given by $$x,k$$ is never reached (since $$j$$ and $$k$$ give the same payoff and when indifferent, player 1 always choose $$j$$).

Under that behavioral assumption (which must be verified to be consistent with equilibrium), player 2 must believe she is at the node "$$y$$" with probability 1. Since Player 2's beliefs have changed with respect to game 1, the mixed strategy given by $$G$$ is, in general, no longer optimal (unless $$a$$ and $$b$$ give the same payoff regardless of the action of player 1).

In short, player 2 will, in general, not use the same strategy in the first and the second game; so threatening to use $$G$$ is not credible.

Btw all threats are credible in a PBE since strategies are best responses given some beliefs that are Bayesian.

• Thanks so much for the great explanation. May I understand that it is impossible to artificially change P1's gain for the strategy j so that P1 always use the strategy j? As you wrote, even if we artificially make P1 choose the strategy j, P2 will no longer use the same strategy G if she is aware that P1 no longer chooses k. I assumed that P1's gain of the strategy j equals P1's gain of the strategy k only when P2 uses the same strategy G. In the new game, how should I derive the new equilibrium? Sorry for my poor understanding. I came back to academia after working as an engineer for a while. – Tom M. May 16 '19 at 20:18
• No, if strategy $x,j$ was better than every other strategy, then in equilibrium the information set of player 2 is never reached, in that case, any beliefs can be assumed, in particular you can assume that player 2 has the same beliefs as in game 1, in which case strategy $G$ is a best response, and thus would be credible. – Regio May 16 '19 at 20:25
• Thank you so much again. In both old and new games, P1 always use the strategy y with some probability 1-F(C1), and thus the information set of player 2 is reached (the node y). As you wrote, P2 will change her strategy G given her new belief. But this means that the strategy j is not better than the strategy k. Then P1 will again start to use the strategy k...it seems that this goes back to the equilibrium in the old game. – Tom M. May 16 '19 at 20:51
• That will depend on the payoffs, but I can see how that may occur for some payoffs. One last comment, it may be the case that in order to find an equilibrium you must drop the assumption that whenever indifferent between j and k, P1 always chooses j. However very little can be said about that without the actual payoffs. – Regio May 17 '19 at 5:05