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?
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.