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I am preparing for an exam. I have found an old exam but I have no solutions for it, so I tried to solve it, but I dont know if I did it correctly and need therefore your help.

The problem looks as follows:

Ultimatum game, proposer (Player 1) and responder (Player 2) bargain about the distribution of 80 Euro. The proposer offers an amount$\ [0; 80]$ to the responder who may accept or reject this offer. If responder accepts, he gets$\ z_2$ and the proposer gets$\ z_1=80-z_2$. In case of rejection, both get nothing.

I am looking for the subgame perfect Nash equilibrium.

$\ U_i=z_i-0,4max(z_i-z_j,0)-0,8max(z_j-z_i,0) , i\neq j , \epsilon (1,2) $

First of all, I am confused since usually the assumption is $ \underline{\beta>\alpha}$. This, does not apply here in my opinion. Since the term $\ 0,4max(z_i-z_j,0)$ expresses envy, where the$\ 0,4$ is generally expressed as an $\alpha$, but given here as a number. On the other side, the term$\ 0,8max(z_j-z_i,0)$ expresses fariness, generally given by the preference paramter $\beta$.

Anyway, I tried to solve it via backwardinduction as follows:

2nd Stage:

i) Responder accepts if $\ z_2>z_1$ and it holds that

$\ U_2=z_2-0,4(z_2-z_1)>0$

$\ z_2>0,4(z_2-z_1)$

I think, this condition will never be satisfied, because the propeser will never offer more than half of the amount in question, the 80 Euro here in this example? For me that is the reason why the term $\ 0,4max(z_i-z_j,0)$ in the utility function just gets zero.

ii) Responder accepts if $\ z_1\geq z_2$

$\ U_2=z_2-0,8(z_1-z_2)>0$

$\ z_2>0,4(z_1-z_2)$ with $\ z_1=80-z_2$

$\ z_2>0,4(80-2z_2)$

$\ z_2>\frac{320}{13} $

I think, I dont need the 1. Stage anymore since I already calculated the optimal share of the 80?

That is: $ z_2*= \frac{320}{13}+\varepsilon $ , where $\varepsilon $ is the smallest possible amount to add.

$ z_1*= 80-\frac{320}{13}-\varepsilon $

Questions:

  1. Is my categorization regarding envy and fairness correct?

  2. Is it of any importance for calcutlations that $\beta>\alpha$?

  3. Is it correct that$\ z_2>z_1$ will never be fullfilled?

  4. Is my subgame Nash equilibrium correct?

  5. Did I miss something?

  6. Is there a more easy approach, if preferences towards envy and fairness are given?

I would be very greatful for some help on this problem.

Many thanks in advance.

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There are a couple of things here. First of all, envy is when your own payoff is smaller than the other player's payoff, which is the case when $z_i<z_j$. So the coefficients here are $0.8$ for envy and $0.4$ for compassion (envy looms larger).

Solving the game using backward induction, we need to first find the minimum value that player 2 is willing to accept. A purely self-regarding player will accept any positive offer from player 1. But with fairness preferences, player 2 will actually decline low positive offers as unfair. To find the threshold, we first note that player 2 gets a utility of $0$ if she declines the offer (in which case both receive 0 payoff). Suppose player 1 offers player 2 an amount $z_2$. Player 1 would keep the remainder, $80-z_2$. The utility is piece-wise linear, with a kink at 40 (equal split), like this:

Utility function

We need to find the value for $z_2$ where player 2's utility crosses the x-axis (red dot). Since this happens below 40, the relevant part of the utility function is the envy case. Thus, we solve $z_2-0.8(80-z_2-z_2)=0$, which turns out to be $z_2^0=\frac{320}{13}\approx 24.6$.

Knowing the acceptance threshold we can now turn to player 1. Player 1 has the same Fehr-Schmidt preferences as player 2. Since the utility function is strictly increasing in own payoff, player 1 wants to keep as much of the pie as possible. Anticipating player 2's acceptance threshold, player 1 will offer $z_2^0=\frac{320}{13}$, maybe adding a small $\varepsilon$ to resolve player 2's indifference.

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