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Econ 109 – Game Theory – Winter 2023

Problem Set 3 Solutions

1. Consider a variant of the ultimatum game we studied in class in which players have fairness considerations. me timing of the game is as usual. First, player 1 proposes the split (J00 - x, x) of a hundred dollars to player 2, where x e [0, J00]. Player 2 observes the split and decides whether to accept (in which case they receive money according to the proposed split) or reject (in which case they both get 0 dollars).

Now suppose that each player’s utility from an outcome equals the amount ofmoney she gets minus the ”unfairness disutility” which is proportional to the squared di住erence in the monetary outcomes. mat is, if an o住er of x is accepted by P2, then  the unfairness disutility” of player i equals βi(x - (J00 - x))S , where βi is a parameters of the game indicating how strongly

player i cares about fairness. Note that if an o住er is rejected, they both get t0 and so the disutility term (and, hence, the anal utility as well) equals zero. Note also that the case we considered in class corresponds to βJ = βS = 0.

(a)  [1pt] Represent this game in extensive form.

Answer: See agure below.

(b)  [1pt] Let βJ = , βS = 0. Which ofers will player 2 defnitely accept? reject? Describe all sequentially rational strategies

of player 2.

Answer: mis part is exactly as in the classic ultimatum game considered in class. me utility from accepting o住er x is x, and the utility from rejecting o住er x is 0. mus, we have the following x-subgame best response mapping

This means that any o住er x 5 0 will deanitely be accepted and there is no o住er that will deanitely be rejected. mere is one indi住erence point 0 and so we get two sequentially rational strategies depending on how the tie is broken:

(c)  [1pt] Let βJ  = , βS  = 0. For each sequentially rational strategy of player 2 you identifed in (c), either describe which proposal maximizes player 1’s continuation value or explain why it does not exist.

Answer: For sS(*), we get the following c口J(*)(x) = J00 - x - (J00 - Sx)S . Seting the derivative to zero, we obtain the maximum of this function on [0, J00] at 寸8.\2.

For sS(**), we get the following

This function also atains the maximum at 寸8.\2 as can be seen from the graph J00 - x - (J00 - Sx)S (below).

(d)  [0.5pt] Let β1 = , β2 = 0. Describe all SPNE of the game.

Answer: Given the analysis in parts b-c, we obtain two SPNE: (48.75, s2(∗)(.)), (48.75, s2(∗∗)(.)). (In the “acceptance-region notation”, they are writen as (48.75, [0, 100]), (48.75, (0, 100])) Note that we get two SPNE because P2 can either accept or reject an o住-equilibrium-path (hypothetical) o住er x = 0. mis is in contrast to the analysis of the classic game from class, where the o住er x = 0 was on the equilibrium path and P2 accepting it was crucial for the existence of optimal o住er of P1. In both SPNE, P1 o住ers 48.75 to P2 and it is accepted.

(e)  [1pt] Let β1 = 0, β2 =  . Which ofers will player 2 defnitely accept? reject? Describe all sequentially rational strategies

of player 2.

Answer: me utility from accepting o住er x is x − (100 − 2x)2, and the utility from rejecting o住er x is 0. Solving the equation x (100 − 2x)2 = 0, we get two points of indi住erence x = 40 and x = 62.5. mus, we have the following x-subgame best response mapping

This means that any o住er strictly between 40 and 62.5 will deanitely be accepted and any o住er strictly below 40 or strictly above 62.5 will deanitely be rejected. We get four sequentially rational strategies depending on how the tow ties are broken:

(In the“acceptance-region notation”, they are writen as A(] = (40, 62.5], A[] = [40, 62.5],A() = (40, 62.5), A[) = [40, 62.5).)

(f)  [1pt] Let β1 = 0, β2 =  . For each sequentially rational strategy of player 2 you identifed in (f), either describe which

proposal maximizes player 1’s continuation value or explain why it does not exist.

Answer: For each sequentially rational strategy, we get the corresponding continuation value of P1:

Note that only two of these functions, c口 and c口1([]) atain a maximum, both at 40. At the same time, c口 and c口 – don’t, for a similar reason as in the classic game we considered in class: P1 wants to set x as small as possible but still in the acceptance region. If the acceptance region has a open le尤 end, then the maximum does not exist.

(g)  [0.5pt] Let β1 = 0, β2 =  . Describe all SPNE of the game.

Answer: Given the analysis in parts f-h, we obtain two SPNE: (40, s), (40, s2([])(.)). (In the“acceptance-region notation”, they are writen as (40, [40, 62.5)), (40, [40, 62.5])) Note that we get two SPNE because P2 can either accept or reject an o-equilibrium-path (hypothetical) o住er x = 62.5. In both SPNE, P1 o住ers 40 to P2 and it is accepted.

2. Two producers of the same good repeatedly compete in prices for T periods. mey have the same discount factor 8. In the stage game, each arm chooses between three price levels given below in each part of the problem. me Bertrand stage game proats are given by:

In each of the following parts you must explain your positive or negative answer. If your answer says that some path can be implemented in an SPNE, then you must list the strategies.

Part I: Suppose the set of available prices is restricted to {2, 4, 10}.

(a)  [0.5pt] Suppose T = 2, and 8 = 1. Can the path ((4, 4), (4, 4)) be implemented in a SPNE?

Hint: Start by writing a 3 × 3 table representing the normal form of the stage game and fnd all stage NE.

Answer:  Let’s start by anding the normal form and NE of the stage game.

Yes, by Result 2: the sequence (4, 4), (4, 4) can be implemented in a SPNE with constant strategies since it is a sequence of stage NE.

(b)  [0.5pt] Suppose T = 2, and 8 = 1. Can the path ((10, 10), (10, 10)) be implemented in a SPNE?

Answer: No, by Result 3: (10, 10) cannot be played in the second period as it is not a stage NE.

(c)  [0.5pt] Suppose T = 2, and 8 = 1. Can the path ((10, 10), (4, 4)) be implemented in a SPNE?

Answer: Yes, with a pair of trigger strategies where any deviation from (10, 10) in period 1 is punished with (2, 2) in period 2.

(d)  [0.5pt] Suppose T = ∞, and 8 ∈ (0, 1). For which 8 can the path ((2, 2), (2, 2), …) be implemented in a SPNE using any strategies?

Answer: Yes, by Result 2: it can be implemented with constant strategies because it is a sequence of stage NE. Note that

in this case grim trigger strategies coincide with constant strategies si(t)(…) = 2 because players would‘punish’deviations

with the same stage NE.

(e)  [0.5pt] Suppose T = w, and 6 e (0, J). For which 6 can the path ((J0, J0), (J0, J0), …) be implemented in a SPNE using grim trigger strategies?

Answer: Consider grim trigger strategies that punish any deviation from this path with perpetual (S, S). Deviation in

period 1 is not proatable if and only if  = JQ + 6  . merefore, this path can be implemented for 6 = J\S.

Part II: Suppose the set of available prices is restricted to {?, 寸, J0}.

(f)  [0.5pt] Suppose T = S, and 6 = J. Can the path ((寸, 寸), (寸, 寸)) be implemented in a SPNE?

Answer:  Let’s start by anding the normal form and NE of the stage game.

me sequence (寸, 寸), (寸, 寸) cannot be implemented in a SPNE since (寸, 寸) cannot be played in the second stage as it is not a stage NE.

(g)  [0.5pt] Suppose T = S, and 6 = J. Can the path ((J0, J0), (J0, J0)) be implemented in a SPNE?

Answer: me sequence (J0, J0), (J0, J0) cannot be implemented in a SPNE since (寸, 寸) cannot be played in the second stage as it is not a stage NE.

(h)  [0.5pt] Suppose T = S, and 6 = J. Can the path ((J0, J0), (寸, 寸)) be implemented in a SPNE?

Answer: me sequence (J0, J0), (寸, 寸) cannot be implemented in a SPNE since (寸, 寸) cannot be played in the second stage as it is not a stage NE.

(i)  [0.5pt] Suppose T = w, and 6 e (0, J). For which 6 can the path ((J0, J0), (J0, J0), …) be implemented in a SPNE grim trigger strategies?

Answer: For the sequence (J0, J0), (J0, J0), … is implementable if and only ifplayer 1 does not have a proatable deviation

in period 1 under grim trigger strategies:  = JQ + 6  J-(寸.)6(2) . merefore, this path can be implemented for 6 =   0.\

(j)  [0.5pt] Suppose T = w, and 6 e (0, J). For which 6 can the path ((寸, 寸), (寸, 寸), …) be implemented in a SPNE using grim trigger strategies?

Answer: For the sequence (寸, 寸), (寸, 寸), … is implementable if and only if player 1 does not have a proatable deviation

in period 1 under grim trigger strategies:  = o + 6  J-(寸.)6(2) . merefore, this path can be implemented for 6 = J2   0.SS