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$\epsilon_i$ 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. enter image description here

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    $\begingroup$ 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...? $\endgroup$ – VARulle Jun 4 '20 at 8:48
  • $\begingroup$ @VARulle Because if player 1 plays $I$ then player 2 will respond by playing N, and so player 1 gets $1+\epsilon_1$. $\endgroup$ – Charles Jun 4 '20 at 8:54
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    $\begingroup$ Are you saying that player 1 moves first? $\endgroup$ – VARulle Jun 4 '20 at 8:56
  • $\begingroup$ @VARulle No; it is a simultaneous move game. $\endgroup$ – Charles Jun 4 '20 at 8:57
  • $\begingroup$ Then how can player 2 "respond by playing N", if he doesn't observe player 1's choice of I? $\endgroup$ – VARulle Jun 4 '20 at 9:00

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