Problem 1 a Alice’s own preffered curfew time ignoring the presence Bob

Problem 1.
a. Alice’s own preffered curfew time, ignoring the presence Bob:

max(UA), (UA)’x = 6(6-x); 6(6-x)=0; x=6.

For Bob: (UB)’x = 2(2-x); x=2.

If the coop board sets x=3, and there are not payment between them, their utility will be:
UA (x=3,y=0) = 1000 – 0 – 27 = 973
UB (x=3,y=0) = 1500 + 0 – 4 = 1496

x=3, y=0 isn’t Pareto-efficient.
If it’s Pareto-efficient, MRSUb (3,0) = MRSUa. (3,0).
But

MRSUa = MUx/MUy = 6(6-x)/(-1) MRSUa. (3,0) = -18
MRSUb = MUx/MUy = 2(2-x)/1 MRSUb (3,0) = -2
-18≠ -2=> MRSUa ≠MRSUb
b.
Ua+Ub (x,y) = 2500 – 3 (6-x)2 – (2-x)2
(Ua+Ub)’x = 6(6-x) +2(2-x) = 0;
36 – 6x + 4 – 2x = 0
40 – 8x = 0
8x=40; x=5;

Ua+Ub (5,y) = 2500 – 3 – 9 = 2488;
Yes, the curfew time x=5 is Pareto-efficient. We see, that it doesn’t depend on payment y, we can’t determine payment y from this maximization.

c.
Ub(x=2)-Ub(x=5)=y;
1500+y-(2-2)2 – 1500 – y + (5-2)2 = y
y=9
Ub (5,9) = 1500+9 – 9 = 1500
Ua (5,9) = 1000-9 – 3 = 988
Yes, this curfew is Pareto-efficieny

d. Ua = Ub and Ua+Ub = 2488 (Pareto-efficient curfew time x=5),
So, Ua = Ub = 2488/2 = 1244 and
y = 1000 – Ua -3(6-xpareto-efficient)2 = 1000 – 1244 -3 = -247

e. If Ua – 1000 = Ub – 1500 and Ua+Ub = 2488 (Pareto-efficient curfew), then
Ua – Ub = 1000-1500 = -500
and
Ua+Ub = 2488
2*Ua = 1988 and 2*Ub = 2988
Ua = 994 and Ub = 1494

In this case y = 1000 – 994 -3(6-xpareto-efficient)2 = 1000 – 994 -3 = 3

f. Bob’s utility is maximum in the case c, and minimum in the case d.
Alice’s utility is maximum in the case d, and minimum in the case c.
I think the most ethical the case e, because Alice and Bob’s utilities change least of all with respect to utilities in their preferred curfew time.
(∆Ub=∆Ua= -6)

Problem 2
a.
Bruce’s utility maximizing level of fishing.

Ub = yb – (eb)2 = 10*eb – (eb + ea) – (eb)2 = – (eb)2 + 9eb – ea
(Ub)’ = -2eb + 9 => -2eb + 9 = 0 => ebmax = 9/2 (I notice, that it is function ebmax (ea) )
Ub (9/2) = -81/4 + 81/2 – ea = 81/4 – ea

Armanda’s utility maximizing level of fishing.
Similar calculations as well as Bruce:
eamax = 9/2, (I notice, that it is function eamax(eb) )
Ub (9/2) = 81/4 – eb

b.
The Nash equilibrium is a solution concept of a non-cooperative game involving two or more players, in which each player is assumed to know the equilibrium strategies of the other players, and no player has anything to gain by changing only their own strategy. If each player has chosen a strategy and no player can benefit by changing strategies while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitutes a Nash equilibrium. The reality of the Nash equilibrium of a game can be tested using experimental economics method.

A Nash equilibrium is a pair (ea*, eb*) such that ea* = eamax (eb*) and 
eb* = ebmax (ea*). Thus, a Nash equilibrium is a solution of the equations

ea* = 9/2
eb* = 9/2

We conclude that the game has a unique Nash equilibrium, in which each amount of time ea*= eb* = 9/2

c.
In real-world practice, such compensations have unintended consequences. They can lead to incentive distortions over time as agents anticipate such compensations and change their actions accordingly. Under certain idealized conditions, it can be shown that a system of free markets will lead to a Pareto efficient outcome. This is called the first welfare theorem. It was first demonstrated mathematically by economists Kenneth Arrow and Gerard Debreu

max (Ua+Ub)= max (ya – (ea)2 + yb – (eb)2) = max(8*(ea + eb) – (ea)2 – (ea)2)
Denote U= Ua+Ub
U’ea = 8 – 2ea; U’ea = 0; => 8 – 2ea = 0; => ea = 4.
U’eb = 8 – 2eb; U’eb = 0; => 8 – 2eb = 0; => eb = 4

U”ea,ea = -2
U”eb,eb = -2
U”ea,eb = U”eb,ea = 0

(-2; 0)
(0 ;-2)

det= 4>0 => ea = 4 and eb = 4 (Pareto-eficient level) maximize the sum of the two utilities
ya=yb= 10*4 – (4+4)=32 (quantity of fish every of them catches)
Ua= Ub=32 – 42 = 32 – 16=16 (utility level at the Pareto efficient fishing level)

d. If, for instance, Bruce starts to work more, his utility will be less, and Armanda will catch less fishes with less utility. Armanda can decide increase utility by increasing time. In this case, situation will be the same for Bruce.

So, it’s necessary to coordinate their time.

Problem 3
a.
Bruce’s utility maximizing level of fishing.

Ub = yb – (eb)2 = 10*eb – teb – (eb + ea) – (eb)2 = – (eb)2 + (9 – t)eb – ea
(Ub)’ = -2eb + 9 – t => -2eb + 9 = 0 => ebmax = (9 – t)/2

b.
A Nash equilibrium is a pair (ea*, eb*) such that ea* = eamax (eb*) and 
eb* = ebmax (ea*). Thus a Nash equilibrium is a solution of the equations

ea* = (9-t)/2
eb* = (9-t)/2

ya=yb= 10*eb – teb – (eb + ea) = (10-t)eb – (eb + ea) = (10 – t)(9-t)/2 – 9 + t = 0.5*t2 – 9.5*t + 45 – 9 + t = 0.5*t2 – 8.5*t + 36
Ua= Ub= (0.5*t2 – 8.5*t + 36)2 – (9-t)2/4

c.
Using solving problem 3

max (Ua+Ub)= max (ya – (ea)2 + yb – (eb)2) = max(8*(ea + eb) – (ea)2 – (ea)2)
Denote U= Ua+Ub
U’ea = 8 – t – 2ea; U’ea = 0; => 8 –t – 2ea = 0; => ea = 4 – t/2.
=> eb = 4 – t/2

eapareto = 4 – t/2 and ebpareto = 4 – t/2 (Pareto-efficient level) maximize the sum of the two utilities

Problem 4
The implemented game is a three-person (simultaneous move) ultimatum game with one proposer and two responders. The proposer proposes a split 10 pieces of candy between himself and the two responders. Both responders simultaneously decide whether to accept or reject the proposal. If both responders accept all players’ earnings are according to the proposal. If at least one responder rejects the proposer earns zero.

Experiments
Proposal
Decision,
responder 1 Decision, responder 2
For me
Responder 1 Responder 2

4 3 3 Yes
Yes
4 3 3 Yes
Yes
2 4 4 Yes
Yes
3 3 4 Yes
Yes
8 1 1 Yes
No
5 2 3 Yes
Yes
1 2 7 Yes
Yes
10 0 0 No
No
6 3 1 Yes
No
5 2 3 No
Yes

Important notice: in experiment members were not familiar with each other. And the value of candies were about the same.

When carried out between members of a shared social group people offer “fair” (i.e., 33:33:33) splits, and offers of less than 20% are often rejected.

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