We have to answer this question and I think I have done part (a) right but get stuck at part (b). Since $-0.5 \le \varepsilon_i \le 0.5 \ \forall i$, I seem to get a solution of the NE being TR, which does not seem right. Please could you help?
John Harsayni showed that a mixed strategy equilibrium of a perfect information game can be thought of as an approximation to an equilibrium (i.e. Bayesian NE) of a game where each player has a slight amount of incomplete information about the exact preferences of the other players. Consider the following Bayesian game, with payoffs as given below:
$\begin{array}{|c|c|c|} \hline & L & R \\ \hline T & 1 + \varepsilon_1, \varepsilon_2 & \varepsilon_1, 1 \\ \hline B & 0, 2 + \varepsilon_2 & 2, 0 \\ \hline \end{array}$
Nature chooses $\varepsilon_1$ and $\varepsilon_2$ independently, and both of these are uniformly distributed on the interval $[-k, k]$ where $k\lt 0.5$: Player $i$ is informed of the realization of $\varepsilon_i$; but not of $\varepsilon_j$ for all $j\neq i$: Players then choose actions simultaneously.
$(a)$ Solve for an Nash equilibrium of this game when $k = 0$:
$(b)$ Solve for a Bayesian NE when $k \gt 0$; $k \lt 0.5$: What is the probability assigned by player $i$ to the event that his opponent plays his first action in this BNE?
$(c)$ What do the probabilities in $(b)$ converge to when $k\to 0$?
Interpret your results.
I have attempted (a) as when $k=0$, there is no range for $\varepsilon_1$ or $\varepsilon_2$ to be on, so they must both be $0$ too, and then one can find mixed strategies for the game. But I am unsure how to introduce probabilities or to draw the tree for the Bayesian game in part (b). Please can you help?
(a) If $k = 0$ , then $\varepsilon_1 = \varepsilon_2 = 0$, and both players know this. Using the normal method of finding mixed strategy which makes each player indifferent between playing either of their actions, we have player 1 playing $T$ with probability $\dfrac 2 3$ and $B$ with $\dfrac 1 3$, and player 2 playing $L$ with probability $\dfrac 2 3$ and $R$ with $\dfrac 1 3$.
(b) If $0 < k < 0.5$, it is still not possible for a PSNE to exist, so you will have to maximize each player's utility to find the BNE.
Computing it will give us player 1 playing $T$ with probability $\dfrac {2 + \varepsilon_2} 3$ and player 2 playing $L$ with probability $\dfrac {2 - \varepsilon_1} 3$. Note the magnitudes on the $\varepsilon$-terms differ between the players.
(c) As $k$ approaches $0$, $\varepsilon$ approaches $0$ as well, hence the mixed strategy NE in (a) can be used to approximate the BNE when $k$ is small.