7
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – joriki May 3 '16 at 23:26
  • $\begingroup$ I wrote as it is given and I checked I wrote it correctly so the number of distinct pure strategies Bayesian equlibria is three $\endgroup$ – Melina May 4 '16 at 7:47
  • $\begingroup$ Could you provide the source please? $\endgroup$ – joriki May 4 '16 at 9:00
  • $\begingroup$ @joriki I just can take a picture because it is printed and given in class. $\endgroup$ – Melina May 4 '16 at 9:12
  • 1
    $\begingroup$ @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. $\endgroup$ – Melina May 30 '16 at 11:51
2
+100
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.