Pure-Strategy Bayesian Nash equilibrium with general common prior

Question

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.

Herr K. · Accepted Answer

Matrix looks correct. To final all pure strategy BNEs, you'll have to discuss cases based on the value of $p$.
For example, if $pin(0,1)$, 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 $pin(frac14,1)$.
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 $p=1$.)

user28372 · Answer

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:

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

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

Pure-Strategy Bayesian Nash equilibrium with general common prior

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