# Grab the dollar Bayesian game

$$\epsilon_i$$ is distributed uniformly on $$[-\phi, \phi]$$ where $$\phi>1$$. My working so far is as follows: $$\mathbb{E}U_1[I]=1+\epsilon_1\\\mathbb{E}U[N]=0\\$$ Therefore, the optimal strategy for player 1 is for all $$\epsilon_1>-1$$ to play $$I$$ and for all $$\epsilon_2<-1$$ to play $$U$$. This strategy specification is the same for player 2. In terms of pinning-down a BNE, I am lost beyond this point. Thank you.

• Why do you think that $\mathbb{E}U_1[I]=1+\epsilon_1$? And what does the LHS even mean - there is no indication of what strategy player 2 plays...? – VARulle Jun 4 '20 at 8:48
• @VARulle Because if player 1 plays $I$ then player 2 will respond by playing N, and so player 1 gets $1+\epsilon_1$. – Charles Jun 4 '20 at 8:54
• Are you saying that player 1 moves first? – VARulle Jun 4 '20 at 8:56
• @VARulle No; it is a simultaneous move game. – Charles Jun 4 '20 at 8:57
• Then how can player 2 "respond by playing N", if he doesn't observe player 1's choice of I? – VARulle Jun 4 '20 at 9:00