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Let's say there are two players in a game. In the first period, P1 makes an offer to split 3 dollars with P2. P2 accepts or rejects the offer. If P2 accepts, they get the offer made by P1, and P1 gets what is left of the 3 dollars. If P2 rejects, then P2 makes an offer to P1 in period 2. P1 can then accept or reject the offer. If P1 accepts, then P1 gets the offer made by P2, and P2 gets the 3 dollars minus the offer they made to P1. If P1 rejects the offer, they both get 1 dollar. A discount factor of δ applies in the second period, i.e. one should consider 3δ dollars in the pot, and 1δ dollars as being the value of the outside option.

How do I find the Sub Game Perfect Equilibrium of this game?

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So, we have to solve this game by Backward induction. We know if $P_1$ rejects the offer at the end note, both will get $P_1=P_2=\delta $. So, to make $P_1$ better off and make the proposal accepted,$P_2$ will offer $P_1$ an amount of $(\delta +\epsilon ) $, where $\epsilon \rightarrow 0$.(I'm considering the case of continuous prices or you can think of a $\epsilon =0.001$ i.e. a very small amount which just makes the $P_1$ strictly better off in accepting the proposal. So, $P_2$ will get $3\delta -(\delta +\epsilon )$ or $3\delta -\delta +0.001=2\delta +0.001$. Now, $P_1$ knows that if $P_1$ offers any amount $< 2\delta +\epsilon $( strictly less than), then $P_2$ will reject the offer. So, to make $P_2$ strictly better off, he'll offer $P_2$ an amount of $(2\delta +\epsilon + \theta )$, where $\epsilon \to 0,\theta \to 0 $ i.e. infinitesimal changes such as $\epsilon =\theta =0.001$. So, $P_1$ will get an amount of $(3-2\delta -\epsilon -\theta )$ where, $\epsilon ,\theta > 0$

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  • $\begingroup$ @The Artist Have you got your answer? $\endgroup$ – Henam Oct 24 '18 at 13:55

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