# Coordination game

Consider the following game in normal form:

$$\begin{bmatrix} & x_B=0& x_B=1 \\ x_A=0& 1-\theta_A,1-\theta_B & 0,0 \\ x_A=1 & 0,0 & 1+ \theta_A,1+\theta_B \end{bmatrix}$$

where $\theta_A$ and $\theta_B$ are mutually independent random variables, both uniformly distributed on $[-T,T]$, where T>0. Namely cumulative distribution of $\theta_i$ is $F(\theta_i)=\frac{\theta_i + T}{2T}$ on $[-T,T]$. Parameter $\theta_A$ is player A's private information, while parameter $\theta_B$ is player B's private information.

• Formulate this game as a Bayesian game. Write down the definition of a Bayesian equilibrium for this game.

• If T<1, find three distinct pure strategies Bayesian equilibria of this game.

• If T>1, explain why only one out of the three equilibria identified in 2) is a Bayesian equilibrium.

• How would the answer to 3) change if both the realizations of $\theta_A$ and $\theta_B$ were common knowledge before the game is played?

From comments of joriki and my experience I could find only two pure strategies Bayesian equilibria. Is any possibility to find the third pure Bayesian strategy for T <1?

My two pure strategies Bayesian equilibria I found like this:

Since $\theta_A$ and $\theta_B$ are bounded by $[-T,T]$ (because they are uniformly distributed on that interval) and for whatever values of $\theta_A$ and $\theta_B$ and for $T<1$ pure strategies equilibria are: $(x_A=0,x_B=0),(x_A=1,x_B=1)$.

An idea to find three pure strategies Bayesian equilibria:

• Nature moves by 0.5p and 0.5(1-p) for two types let say t and t' .

• Then, player 1 will play with probability 1 for type t and 0 for type t' (because he is completely sure for private information that he has for type t and not for type t').

• Game continues as before for type t and also for t'. And from above we know that for type t we have two pure Bayesian strategies (T<1) and for type t' we would have payoff (0,0) and thus there would be three pure Bayesian strategies.

Is this correct?

• Something seems to be wrong. The game is invariant under the transformation $x_A\to1-x_A$, $x_B\to1-x_B$, $\theta_A\to-\theta_A$, $\theta_B\to-\theta_B$, so there should be an even number of pure strategy equilibria for any value of $T$. Am I missing something? May 3, 2016 at 23:26
• I wrote as it is given and I checked I wrote it correctly so the number of distinct pure strategies Bayesian equlibria is three May 4, 2016 at 7:47
• Could you provide the source please? May 4, 2016 at 9:00
• @joriki I just can take a picture because it is printed and given in class. May 4, 2016 at 9:12
• @joriki I agree with your argument since then I have been trying and I couldn't find three pure Bayesian strategies - only two of them. May 30, 2016 at 11:51

I will write the monotone strategies and playing according to these strategies will be an equilibrium. To understand the logic of this strategy note that as $\theta_i$ increases playing second strategy, strategy $x_i=1$, becomes better. Also as $\theta_i$ decreases the first strategy, strategy $x_i=0$, becomes better. Thus there should exist some number $p_i$ for player $i$ such that player $i$ chooses $x_i=0$ when $\theta_i<p_i$ and chooses $x_i=1$ when $\theta_i>p_i$.

Now consider expected payoff of playing $x_i=0$ with the strategy given above:

$$(1-\theta_i)\Pr(\theta_j<p_j)+0$$

Similarly, playing $x_i=1$ yields $$0+(1+\theta_i)\Pr(\theta_j>p_j).$$ To make player $i$ indifferent between $x_i=0$ and $x_i=1$ at the critical value $p_i$ these expected payoffs should be same when $\theta_i=p_i$ i.e., realized value of $\theta_i$ is $p_i$. So,

$$(1-p_i) F(p_j)=(1+p_i)(1-F(p_j))\implies 2F(p_j)=1+p_i$$ Since $F$ us uniform we have $p_i=p_j T$ but the situation is symmetric so $p_j=p_iT$. Combining these two we have $p_i=p_i T^2$. If $T<1$ or $T>1$ then $p_i=0$. So equilibrium strategy of player $i$ playing $x_i=0$ when $\theta_i<0$ and playing $x_i=1$ when $\theta_i>0$. Using these essentially pure strategies is the third equilibrium you are looking for.

Now if $T>1$ the two pure equilibria that you found will not be equilibrium anymore, but the one given above will still be equilibrium (Try to convince yourself).

I believe you can do first and the last part without any difficulty.