# Pure-Strategy Bayesian Nash equilibrium with general common prior

I'm doing a problem set on the subject of Bayesian Nash equilibrium. I'm asked to find the pure-strategy BNE of the following. I've calculated to matrix shown below. My first concern is if I've calculated the expected payoff matrix correctly, and second how do I find all of the pure-strategy BNE when the common prior is not concrete.

Matrix looks correct. To final all pure strategy BNEs, you'll have to discuss cases based on the value of $$p$$.

For example, if $$p\in(0,1)$$, then $$FT$$ is player 2's unique best response to $$F$$. Thus, to have a BNE, you'd want $$F$$ to be player 1's best response to $$FT$$ as well, meaning that you'd require $$3p>1-p$$, or $$p>\frac14$$. Hence, $$(F,FT)$$ is a BNE if $$p\in(\frac14,1)$$.

You should be able to find other BNEs following a similar line of reasoning. (Hint: don't forget the edge cases where $$p=0$$ and $$p=1$$.)

• Ah so the idea would be to solve for $p$ in the columns $FT$ and $TF$ (the first one as you did in your comment above and the second one in a similar way). Then apply the same idea for the rows or am I wrong? – Justin Malik Jul 25 at 15:33
• @JustinMalik: Yes that's right. Just note that $TT$ and $FF$ are also best responses for boundary values of $p$. – Herr K. Jul 25 at 18:25
• Great, it is definitely much clearer now. Thank you so much. – Justin Malik Jul 25 at 18:58
• @Tomcat: The equivalence result is mentioned in Prop 8.E.1 of MWG (p.255), as well as in the textbook by Fudenberg and Tirole (1991, p.215). Regarding the game tree, I fail to see the issue you mentioned. In the tree I drew, player 1 does move at only one information set (the two decision nodes after Nature's move is connected by a dashed line representing player 1's information set). If you still think it's wrong, I would really appreciate that you draw a correct version of the tree so that I can learn from it. – Herr K. Aug 2 at 17:07
• I get the point - the strategies are still type dependent in the matrix. That is what I was overlooking. In this case the ex-ante payoff is a simple convex combination of the ex-interim payoffs (with positive weights). Hence the optimal solution should be same. Apologise the confusion. – Tomcat Aug 2 at 18:16

The best way to visualise what's going on would be to use Harsanyi's transformation. I'm not drawing the game tree here (but I think Tirole has it in his example).

Let's set up notations first. We will denote player 1's strategy by $$x=Pr(T)$$. We will call the decision of player 2 following the realisation of game $$i$$ by $$y_i=Pr(T)$$ - i.e. player 2, following realisation of game A, chooses T with probability $$y_A$$.

Simple calculation of the best responses give us the following expressions: $$y_A=\begin{cases} 0 & \text{ if } x<\frac{1}{4}\\ [0,1] & \text{ if } x=\frac{1}{4} \\ 1 & \text{ if } x>\frac{1}{4}\\ \end{cases}$$

$$y_B=\begin{cases} 0 & \text{ if } x>\frac{1}{4}\\ [0,1] & \text{ if } x=\frac{1}{4} \\ 1 & \text{ if } x<\frac{1}{4}\\ \end{cases}$$

$$x=\begin{cases} 0 & \text{ if } \rho[4y_A-3] +(1-\rho)[4y_B-3]< 0\\ [0,1] & \text{ if } \rho[4y_A-3] +(1-\rho)[4y_B-3]=0\\ 1 & \text{ if } \rho[4y_A-3] +(1-\rho)[4y_B-3]> 0\\ \end{cases}$$

The set of BNE of the above game is the tuple $$(x,y_A,y_B)\in[0,1]^3$$ that satisfies the above three equations. The solution is quite simple:

1. For any $$\rho\in[0,1]$$, $$x=\frac{1}{4},\;y_A=y_B=\frac{3}{4}$$ is an equilibrium.

2. For $$\rho\in[\frac{1}{4},1]$$, $$x=0,\;y_A=0,\;y_B=1$$ is an additonal equilibrium to the above.

3. For $$\rho\in[\frac{3}{4},1]$$, $$x=1,\;y_A=1,\;y_B=0$$ is an additonal equilibrium to the above two.

I hope there are no calculation errors, but the key idea should remain the same.