gabarito_lista1_2006.2

Prévia do material em texto

Respostas para os exercícios da 1a lista.
1. Von-Neumann Morgenstern utility
Recall from notes and lecture that a vNM utility function is (a) additively
separable across states, i.e. it can be written as u(c1, c2) =
P2
s=1 vs(cs) =
π1v1(c1) + π2v2(c2), where πs = probability of state s and
P2
s=1 πs = 1;
and (b) utility is independent of the state that occurred, i.e. that v1(c) =
v2(c). This question asks which of the preference relations represented by
the utility functions below can be re-written in this form, preserving the
preference structure.
Risk aversion is when the consumer prefers the sure payoff to the expected
value of a random payoff. In vNM shorthand, this amounts to v(E(x)) >P2
s=1 πsvs(cs). Recall that we’ve shown all strictly concave utility func-
tions to be risk averse, and that a differentiable function v(x) is strictly
concave iff v00(x) < 0.
i) u(c1,c2) = (c1 − 1300(c1)2) + (c2 −
1
300(c2)
2). This function is clearly
additively separable, with v(c) = c − 1300c2 regardless of the state
realized. This utility function is just an affine transformation [a +
bv(·)] of a utility function of the proper expected utility form. For
example we could transform u˜ = 12v1(c)+
1
2v2(c) with b = 2 to achieve
u. Thus, this function u(c1,c2) represents the same preferences as a
function u˜ of the vNM expected utility form. Note also that v00(c)
= −1150 < 0. This individual is risk-averse.
ii) u(c1,c2) = c71 · c22. Here we can take a monotonic transformation of u
(retaining the preference structure) to obtain u¯ = 7 log c1 + 2 log c2.
The function u˜(c1,c2) = 79 log c1 +
2
9 log c2, which is clearly a vNM
utility function, is an affine transformation with b = 9 of u¯. Thus,
with v(c) = log c, independent of the state realized, we can say that
u(c1,c2) does represent vNM preferences. Our function is concave, as
v00(c) = −1c2 < 0. Again, our individual is risk averse.
v) u(c1,c2) = c21 + c
2
2. Again, we have additive separability with v(c) = c
2
independent of s. This function is simply a transformation of u˜ =
1
2c
2
1 +
1
2c
2
2 [with b = 2]. These are vNM preferences. This time
v00(c) = 2 ≮ 0, so this utility function is not concave. (You should
be able to see that v(E(x) <
P2
s=1 πsvs(cs) and that this individual
is risk-seeking.)
vi) u(c1,c2) = c1 + log(c2). These are NOT vNM preferences. There
is no monotone transformation to be made such that v(c) shall be
independent of s.
iii) u(c1,c2) =
√
c1 · c22. We can rewrite these preferences as u¯(c1,c2) =
1
2 log c1 + 2 log c2. This function is simply a transformation of u˜ =
1
5 log c1 +
4
5 log c2 [with b = 2.5]. These are vNM preferences. As in
(ii), this individual is risk averse.
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iv) u(c1,c2) = e(c1·
√
c2) Rewrite u as uˇ(c1,c2) = c1 ·√c2 which represents
the same preferences as u¯(c1,c2) = log c1 + 12 log c2. This function is
simply a transformation of u˜ = 23 log c1 +
1
3 log c2 [with b = 1.5]. As
before, this is a risk averse, vNM preference set.
2. The Risk Premium
a) Recall that the risk premium is the R that solves: v(I +E(x)−R) =
E(v(I + x)).
Our risk premium solves v(I + 0−R) = 12(v(I + a)) +
1
2(v(I − a)).
Or − exp[−AI + IR] = 12 (− exp[−AI −Aa]) +
1
2(− exp[−AI +Aa])
=⇒ − exp[−AI] · exp[AR] = − exp[−AI] · 12(exp[−Aa] + exp[Aa])
AR = ln 12 + ln(exp[−Aa] + exp[Aa])
=⇒ R = 1A ln
1
2(exp[−Aa] + exp[Aa])
b) ∂R∂A = −
1
A2 ln
1
2(exp[−Aa] + exp[Aa]) +
1
A
−a exp[−Aa]+a exp[Aa]
exp[−Aa]+exp[Aa] ;
∂R
∂a =
1
A
−A exp[−Aa]+A exp[Aa]
exp[−Aa]+exp[Aa] ;
∂R
∂a = 0. For A > 0 and a > 0,
∂R
∂a is positive,
so the risk premium is increasing in a. It is hard to generally sign ∂R∂A ,
so depending on a and A the risk premium may be either increasing
or decreasing in A. Finally, R does not depend on I, i.e. as income
increases, the risk premium remains unchanged.
c) Recall that with log utility we had R = I−(I2−a2) 12 . Note that ∂R∂a =
a(I2 − a2)− 12 . Thus R was increasing in a. But with log utility, we
could show that R was decreasing in I. These are features of the two
utility functions we utilized in these problems. Log utility exhibits
diminishing absolute risk aversion and constant relative risk aversion,
reflecting an agent that is concerned about proportional gains or
losses of wealth. For the exponential utility function [−exp(−Ax)]
the willingness to pay to avoid risk is independent of wealth. Which
do you think is a better description of human behavior?
3. Insurance
a) In each state realized, the agent gets utility
√
c from consumption
c. This utility function is concave, as v00(c) = −14 c
−3
2 < 0. The agent
is thus risk-averse.
b) Our agent must solve maxC≥0 u(c1, c2, c3) = 110
q
w −D + 910C +
8
10
q
w − 110C, but it is easier to see the meaning of the results if
we leave our problem in generalized parameter form.
i) For v(c) =
√
c, our FOC is: ∂u∂C =
1
10
9
10 · 12(w − D + 910C∗)
−1
2 −
8
10
1
10 · 12(w − 110C∗)
−1
2 = 0
ii) Solving out the FOC gives us C∗ = 170657W +
640
657D. Thus for
W > 110D , C
∗ > D. Our agent over insures!
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For intuition on why this is, we can look at the solution to the
problem in parameter form. If we maximize π2(w − D + (1 −
q)C)
1
2 +π3(w− qC) w.r.t. C, we find C∗ = π
2
3q
2
π23q
2(1−q)+π22(1−q)2q
·
D + π
2
2(1−q)2−π23q2
π23q
2(1−q)+π22(1−q)2q
· W . The condition that makes C∗ =
D is π2π3 =
q
1−q . In our problem we have
π2
π3
= .1.8 <
.1
.9 =
q
1−q . These two ratios compare the probability of loss to the
probability of no-loss from the perspective of the agent and the
insurer. The shifting of the problem due to the outcome in which
insurance is irrelevant renormalizes the agent’s probabilities, so
the ’actuarially fair’ insurance is no longer = P (Experiencing the
Loss ) · (Amount of the Loss). Our agent is over insuring because
as she factors in considering the probability of her death the price
of insurance becomes more than fair.
4. Equilibrio em Estratégias Puras
a) Não. Nenhum dos dois jogadores tem estratégias dominadas
b) (m,M)
c) Por inspeção. Note a completa simetria do jogo
5. Equilíbrio em Estratégias Mistas
É instrutivo ver quais são as melhores repostas em estrégias puras. Clara-
mente, a melhor resposta de 1 para a estratégia h é H, e para a estratégia
t é T.Mas a melhor resposta de 2 para H é t e para T é h. Logo não há
equilíbrio em estratégias puras.
Vejamos em mistas agora. Note que a estratégia M é fracamente dominada
(tanto por H como por T), e estritamente dominada por qualquer mistura
de H e T. para o jogador 1. Ou seja, não há nenhum ponto, para o jogador
1, em jogar esta estratégia com probabilidade positiva. Sabemos então que
ele jogará uma mistura de somente H e T.
Seja πt a probabilidade com que o jogador 2 joge t. Buscamos um πt tal
que o jogador 1 esteja indiferente entre H e T. Dada a simetria, este πt
é claramente 12 . Seja agora πT a probabilidade com que o jogador 1 joge
T . Buscamos um πT tal que o jogador 2 esteja indiferente entre h e t.
Resolvendo, temo πT = 25 . O (único) equilíbrio em estratégias mistas é:
1 joga H com probabilidade 35 , M com probabiliddade 0, e T com proba-
bilidade 25 . 1 joga h com probabilidade
1
2 e t com probabilidade
1
2 .
6. Extensive form games
a) OMITTED
b) The equilibrium payoffs are (7, 5). This results from the equilibrium
strategies A : {Introduce,Not Sue} ; B : {Adopt}.
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c) There is a Nash equilibrium of the game with strategiesA : {Introduce, Sue}
; B : {Exit}. This can be seen by examining the normal form (also
called the strategic form) of the game. When we examine the matrix
and look for mutual best responses,
A\B Adopt Exit
Not, Sue (0, 0) (0, 0)
Not, Not Sue (0, 0) (0, 0)
Intro, Sue (1,−1) (22, 0)Intro, Not Sue (7, 5) (22, 0)
we can see that there is a Nash equilibrium where Company A takes
strategy {Introduce, Sue} and B takes the strategy{Exit}.As we saw
from the game tree in part (a) this is not subgame perfect, since
Company A would never choose to sue once Company B has entered,
and so this Nash equilibrium violates the outcome of that subgame.
One lesson here is that incorporating the timing of choices made
(analogously, modelling the extensive form) can allow us to determine
which of the Nash equilibria will be selected.
2) Repeated games
a) Game A has only one Nash equilibrium: A : {U} ; B : {l}.
Game B has four pure strategy equilibria: {U, l} ; {D, l} ; {M,m} ; {D, r}.
[There are also mixed Nash equilibria, such as player 1 playing any
mixture of U and D against player 2 playing l, but these are unnec-
essary for this problem.]
b) Nash equilibrium strategies are Player 1:{DA; always DB}; Player
2:{rA; if Player 1 plays DA then rB, else lB}.
You should realize that player 1 has a dominant strategy in Game
A, and thus player 2 will have to offer an inducement to get her to
play the dominated strategy. This is accomplished by offering a
strategy that offers player 1 a greater payoff over the two periods.
The existence of multiple Nash equilibria in the second game make a
carrot-stick strategy by player 2 possible. Since our theory predicts
any of the Nash equilibria is possible in the second stage game, player
2 devises a strategy to pick between them based upon the actions of
player 1 in Game A. If player 1 coordinates for non-Nash play in
Game A, then player 2 will reward her with the best of the Nash
equilibria in Game B.
7. Jogos Repetidos e Dinâmicos
a) No. The only equilibrium in the second game is confess, confess, no matter
what happens in at the first stage. Therefore at the first stage the only
equilibirum is confess, confess.
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b) Consider both players playing the following strategy:
⎧
⎨
⎩
Play c at the 1st stage
Play c at the 2nd stage, if c,c at 1st stage. Othersiwe play n thereafter
Play c at the 3rd stage, if c,c at 2nd stage. Otherwise play n
Let’s show now that β = 1 will sustain a positive payoff in a SPE. Given
that one player is following this strategy, the other at the 1st period can
deviate and have an immediate gain of 5, but will loose 6 at 2nd stage
and 9 at the 3rd stage. At the 2nd period, if he deviates, he also has an
immediate gain of 5, but looses 9 in the 3rd period. Therefore she has no
incentive to deviate from the prescribed strategy, neither at the 1st period
nor at the second. Off the equilibirum path it is SPE because it describes
playing Nash thereafter. Therefore, with β = 1, in this equilibrium payoffs
are 20 for both players.
Trivially, if β = 0, the future does not matter. They do not cooperate in
any equilibrium., and payoffs cannot be positive.
c) Consider both players playing the following strategy:½
Play c at t = 1
Play c at t = τ , if c,c at τ − 1. Otherwise play n forever
Let’s show now that β = .99 will sustain a positive payoff in a SPE. Given
that one player is following this strategy, the other at any period can
deviate and have an immediate gain of 5, but will loose 6 forever. With
β = .99, the loss in the next period (.99 ∗ 6 = 5.96) is already larger
than 5. Therefore she has no incentive to deviate from the prescribed
strategy. Off the equilibirum path it is SPE because it describes playing
Nash thereafter.Trivially if β = 0, players don’t value the future and
therefore no positive payoff can be sustained in SPE.
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