[infinitely many correct answers—see me if you have questions about this one]
[infinitely many correct answers—see me if you have questions about this one]
[infinitely many correct answers—see me if you have questions about this one]
- a P b
- c P d
- a P d
Do these preferences satisfy the completeness condition?
Answer: No. (You don’t know about preferences between a and c, or b and c, or b and d.)
Answer: Yes. (They satisfy the transitivity condition as long as they don’t violate it. And no violation of the transitivity condition can be derived from the given preferences.)
Answer: There are multiple acceptable ways to go here. The most elegant approach is to impute to Ralph preference such as f P z and z P m. Then, if it remains true that Ralph has no preference between f and m, he is violating not only the completeness condition (in the say way as before), but also the transitivity condition, because the preferences we’ve newly imputed to him would imply (if his preferences did satisfy the transitivity condition) that f P m.
Another acceptable approach is to impute to Ralph some preferences that don’t connect with f and m in the way just described, but that result in a cycle of preferences on their own. For example, you could impute the following preferences to Ralph, and Ralph's preferences would then violate the transitivity condition: a P b, b P c, c P a. And Ralph’s preferences would still violate the completeness condition, because of his having no preference between f and m.
# | claim | justification |
1 | a P b | given |
2 | b P c | given |
3 | c P d | given |
4 | d P a | given |
5 | a P d | 1–3, transitivity |
6 | contradiction | 4 and 5, completeness |
What is the largest number n that makes the following sentence true? “Lines 1 through n are all correct and correctly justified.” (If the proof is valid, give the number of the last line. If the proof has one or more mistakes, give the number of the line immediately preceding the one containing the first mistake.)
Answer: 4.
rain or no running water | running water and no rain | |
bring assistant | happy with your decision | mad that you wasted some of your payroll budget |
not bring assistant | tired at the end of the day | happy with your decision |
leaves in state A and roots in state P | leaves in state B and roots in state P | leaves in state C and roots in state P | roots in state Q | |
use proteins | iron | tin | tin | aluminum |
use radiation | gold | silver | bronze | platinum |
Use the choice situation represented by the following matrix for problems 10–14:
S_{1 }(1/2) | S_{2} | S_{3} | |
A_{1} | 5 | 19 | 18 |
A_{2} | 13 | 8 | 11 |
A_{3} | 18 | 7 | 13 |
A_{4} | 6 | 7 | 20 |
- Would any option be selected by the dominance principle? If so, which one?
No.
- Which option(s) would be selected by the maximin rule?
A_{2}.
- Which option(s) would be selected by the maximax rule?
A_{4}.
- Which option(s) would be selected by the optimism/pessimism rule with an optimism index of 1/4?
A_{3}.
- Which option(s) would be selected by the optimism/pessimism rule with an optimism index of 2/3?
A_{4}.
Yes, Yes, 0 and 1/10. (Any α of less than 1/6 will make the α-index of A_{2} less than the α-index of each of the other options.)
A_{2}, 11.
A_{3}, 14.
(Note: Problems 15–18 are very similar to problems for credit nos. 16–19 that I assigned in my Fall 2006 section of this course. Those problems, and their answers, are at the web site that I set up for that course, at http://web.ku.edu/~utile/courses/rct2. Problems 19-20 are involve some slightly more complicated math than is found in nos. 15–18, but not additional conceptual complexities.)
EU($400) > EU(30-percent chance at $1,500 and 70-percent chance at $0)
u($400) > (30/100)u($1,500) + (70/100)u($0)
100u($300) > 30u($1,500) + 70u($0)
100[u($0) + x] > 30[u($0) + x + y] + 70u($0)
100u($0) + 100x > 30u($0) + 30x + 30y + 70u($0)
100u($0) + 100x > 100u($0) + 30x + 30y
100x > 30x + 30y
70x > 30y
(7/3)x > y
y < (7/3)x
There are infinitely many correct answers. Here is one:
u($0) = 0
u($400) = 10
u($1,500) = 11
EU(60-percent chance at $1,000 and 40-percent chance at $0) > EU($700)
(60/100)u($1,000) + (40/100)u($0) > u($700)
60u($1,000) + 40u($0) > 100u($700)
60[u($0) + x + y] + 40u($0) > 100[u($0) + x]
60u($0) + 60x + 60y + 40u($0) > 100u($0) + 100x
100u($0) + 60x + 60y > 100u($0) + 100x
60x + 60y > 100x
60y > 40x
y > (2/3)x
There are infinitely many correct answers. Here is one:
u($0) = 0
u($700) = 1
u($1,000) = 11
EU(25-percent chance at $600 and 75-percent chance at $500) > EU(60-percent chance at $1,000 and 40-percent chance at $0)
(25/100)u($600) + (75/100)u($500) > (60/100)u($1,000) + (40/100)u($0)
25u($600) + 75u($500) > 60u($1,000) + 40u($0)
25[u($0) + x + y] + 75[u($0) + x] > 60[u($0) + x + y + z] + 40u($0)
25u($0) + 25x + 25y + 75u($0) + 75x > 60u($0) + 60x + 60y + 60z + 40u($0)
100u($0) + 100x + 25y > 100u($0) + 60x + 60y + 60z
40x > 35y + 60z
8x > 7y + 12z
There are infinitely many correct answers. Here is one:
u($0) = 0
u($500) = 10
u($600) = 11
u($1,000) = 12
Use dominance considerations to analyze the next two games (which you can assume to be zero-sum games). Circle all outcomes not ruled out by such considerations.
C_{1} | C_{2} | C_{3} | C_{4} | |
R_{1} | 8 | 5 | 3 | 3 |
R_{2} | 7 | 4 | 1 | 1 |
R_{3} | 6 | 2 | 7 | 6 |
Answer: Four outcomes are not ruled out:
- the 5 (R_{1}, C_{2})
- the second 3 (R_{1}, C_{4})
- the 2 (R_{3}, C_{2})
- the last 6 (R_{3}, C_{4})
C_{1} | C_{2} | C_{3} | C_{4} | |
R_{1} | 2 | 6 | 4 | 3 |
R_{2} | 7 | 8 | 7 | 8 |
R_{3} | 8 | 5 | 5 | 8 |
R_{4} | 4 | 6 | 6 | 9 |
Answer: One outcome is not ruled out: the second 7 (R_{2}, C_{3}).
Circle every equilibrium outcome in the next two games (which you can assume to be zero-sum games).
C_{1} | C_{2} | C_{3} | C_{4} | |
R_{1} | 3 | 5 | 4 | 3 |
R_{2} | 1 | 4 | 5 | 5 |
R_{3} | 3 | 3 | 4 | 8 |
Answer: the first 3 (R_{1}, C_{1}) and the third 3 (R_{3}, C_{1})
C_{1} | C_{2} | C_{3} | C_{4} | |
R_{1} | 7 | 1 | 2 | 9 |
R_{2} | 7 | 8 | 5 | 7 |
R_{3} | 4 | 5 | 9 | 7 |
R_{4} | 7 | 9 | 8 | 9 |
Answer: the last 7 (R_{4}, C_{1})
Answer:
# | claim | justification |
1 | Suppose the outcome of (R_{i}, C_{j}) is an equilibrium outcome but that C_{j} is a dominated column. | assumption for proving conditional |
2 | The outcome of (R_{i}, C_{j}) is an equilibrium outcome. | 1 |
3 | Neither player has an incentive to unilaterally deviate from the strategy pair (R_{i}, C_{j}). | 2, definition of equilibrium |
4 | The column player does not have an incentive to unilaterally deviate from the strategy pair (R_{i}, C_{j}). | 3 |
5 | v_{ij} is at least as small as any other value in row i. | 4 |
6 | C_{j} is a dominated column. | 1 |
7 | There is some column C_{k}, with k ≠ j, that dominates C_{j}. | 6 |
8 | v_{ik} is at least as small as v_{ij}. | 7 |
9 | v_{ik} is at least as small as any other value in row i. | 8, 5, transitivity of ≤ |
10 | The column player does not have an incentive to unilaterally deviate from the strategy pair (R_{i}, C_{k}). | 9 |
11 | The row player does not have an incentive to unilaterally deviate from the strategy pair (R_{i}, C_{k}). | shortcut |
12 | Neither player has an incentive to unilaterally deviate from the strategy pair (R_{i}, C_{k}). | 10, 11 |
13 | The outcome of (R_{i}, C_{k}) is an equilibrium outcome. | 11, definition of equilibrium |
14 | This game has at least two equilibrium outcomes—the outcome of (R_{i}, C_{j}) and the outcome of (R_{i}, C_{k}). | 2, 13 |
15 | v_{ij} is at least as small as v_{ik}. | 5 |
16 | The values of the outcome of (R_{i}, C_{j}) and the outcome of (R_{i}, C_{k}) are equal to each other. | 8, 15 |
17 | The game has at least two equilibrium outcomes whose values are equal to each other. | 14, 16 |
18 | If a game has an equilibrium outcome in a dominated column, then it has at least two equilibrium outcomes whose values are equal to each other | 1–17 |
C_{1} | C_{2} | |
R_{1} | 8 | 3 |
R_{2} | 2 | 6 |
Answer:
To verify that [(4/9 R_{1}, 5/9 R_{2}), (1/3 C_{1}, 2/3 C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his or her strategy.
First, consider the game from the row player’s point of view. If the column player is playing strategy (1/3 C_{1}, 2/3 C_{2}), then the expected utility of the game for the row player, with his strategy (p R_{1}, (1 – p) R_{2}), is (p)[(1/3)(8) + (2/3)(3)] + (1 – p)[(1/3)(2) + (2/3)(6)], or p(8/3 + 6/3) + (1 – p)(2/3 + 12/3), or p(14/3) + (1 – p)(14/3), or p(14/3) + 14/3 – p(14/3), or 14/3. Because this expected utility does not depend on p, in that the p drops out of the expression as we simplify it, the row player’s expected utility is unaffected by his own strategy, and thus he has no incentive to unilaterally deviate from whatever strategy he happens to be playing. This goes for (4/9 R_{1}, 5/9 R_{2}) as much as for any other strategy.
Second, consider the game from the column player’s point of view. If the row player is playing strategy (4/9 R_{1}, 5/9 R_{2}), then the expected utility of the game for the column player, with her strategy (q C_{1}, (1 – q) C_{2}), is (q)[(4/9)(8) + (5/9)(2)] + (1 – q)[(4/9)(3) + (5/9)(6)], or q(32/9 + 10/9) + (1 – q)(12/9 + 30/9), or q(42/9) + (1 – q)(42/9), or q(42/9) + 42/9 – q(42/9), or 42/9. Because this expected utility does not depend on q, in that the q drops out of the expression as we simplify it, the column player’s expected utility is unaffected by her own strategy, and thus she has no incentive to unilaterally deviate from whatever strategy she happens to be playing. This goes for (1/3 C_{1}, 2/3 C_{2}) as much as for any other strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(4/9 R_{1}, 5/9 R_{2}), (1/3 C_{1}, 2/3 C_{2})] is an equilibrium pair.
C_{1} | C_{2} | |
R_{1} | a | –b |
R_{2} | –b | a |
Answer:
To compute the value of the game, we find the expected utility of the playing of [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})], by computing the weighted average of the values of the four possible outcomes of the game: (½)(½)(a) + (½)(½)(–b) + (½)(½)(–b) + (½)(½)(a) = a/4 – b/4 – b/4 + a/4 = 2a/4 – 2b/4 = a/2 – b/2 = ½(a – b).
C_{1} | C_{2} | |
R_{1} | 8 | 3 |
R_{2} | 2 | 6 |
Answer:
EU_{column}(game)
= q[(1/3)(8) + (2/3)(2)] + (1 – q)[(1/3)(3) + (2/3)(6)]
= q(8/3 + 4/3) + (1 – q)(3/3 + 12/3)
= q(12/3) + (1 – q)(15/3)
= q(12/3) + 15/3 – q(15/3)
= q(–3/3) + 15/3
= –q + 5
Answer:
The column player wants to minimize the score she gets upon the playing of the game. To minimize –(1/2)q + 3/2, the column player would make q as large as possible, by setting q (which must be between 0 and 1) equal to 1.
C_{1} (q) | C_{2} (1 – q) | |
R_{1} (p) | a | –b |
R_{2} (1 – p) | –c | a |
Answer:
There are two fairly manageable techniques for this one. I’ll use one technique to derive p, and the other to derive q.
First, let's derive p by figuring out what value of p would make the column player’s expected utility for this game not depend on q:
EU_{Column}(game)
= q[(p)(a) + (1 – p)(–c)] + (1 – q)[(p)(–b) + (1 – p)(a)]
= q(ap – c + cp) + (1 – q)(–bp + a – ap)
= q(ap + cp – c) + (–ap – bp + a) – q(–ap – bp + a)
= q(ap + cp – c + ap + bp – a) + (–ap – bp + a)
= q(2ap + bp + cp – a – c) + (–ap – bp + a)
Now it’s time to figure out what value of p would make the coefficient of q equal to 0:
2ap + bp + cp – a – c = 0
p(2a + b + c) = a + c
p = (a + c)/(2a + b + c)Second, let’s derive q by figuring out what value of q would make the row player’s expected utility for R_{1} equal to his expected utility for R_{2}:
EU(R_{1}) = EU(R_{2})
qa + (1 – q)(–b) = q(–c) + (1 – q)(a)
qa – b + qb = –qc + a – qa
qa + qb + qc + qa = a + b
2qa + qb + qc = a + b
q(2a + b + c) = a + b
q = (a + b)/(2a + b + c)
C_{1} | C_{2} | |
R_{1} | 1 | 9 |
R_{2} | 6 | 5 |
For this game, the strategy pair [(1/9 R_{1}, 8/9 R_{2}), (5/9 C_{1}, 4/9 C_{2})] is in equilibrium, and the value of the game is 49/9. If this game is played a large number of times using the specified strategy pair, what proportion of the time would you expect the outcome to be 49/9?
Answer:
None—even though 49/9 is the value of the game, it is not a possible outcome of the game.
C_{1} (q) | C_{2} (1 – q) | |
R_{1} (p) | a | b |
R_{2} (1 – p) | c | d |
Show that solving for p by assuming EU(C_{1}) = EU(C_{2}) yields a formula for p that makes the coefficient of q (in the formula for the expected utility of the game from the column player’s point of view) equal to 0.
Answer:
The answer involves fractions that are hard to read unless the numerator can be written above the denominator, which does not seem to be an option in this format (i.e., I don’t know HTML code that allows for that), so here I’m just going to say that this answer is closely parallel to the demonstration (done in class) that solving for q by assuming EU(R_{1}) = EU(R_{2}) yields a formula for q that makes the coefficient of p (in the formula for the expected utility of the game from the row player’s point of view) equal to 0. Please see me if you need more explanation.
C_{1} (q) | C_{2} (1 – q) | |
R_{1} (p) | 13 | 16 |
R_{2} (1 – p) | 18 | 4 |
Answer:
p = 14/17, q = 12/17
C_{1} (q) | C_{2} (1 – q) | |
R_{1} (p) | 19 | 1 |
R_{2} (1 – p) | 5 | 9 |
Answer:
p = 4/22, q = 8/22
C_{1} | C_{2} | |
R_{1} | 5, 4 | 9, 6 |
R_{2} | 6, 8 | 3, 2 |
Answer:
(R_{1}, C_{2}) and (R_{2}, C_{1}) are the pure-strategy equilibrium pairs.
The game is a coordination game.
C_{1} | C_{2} | |
R_{1} | 6, 8 | 1, 1 |
R_{2} | 8, 3 | 3, 7 |
Answer:
(R_{2}, C_{2}) is the pure-strategy equilibrium pair.
The game is neither a coordination game nor a prisoner’s dilemma.
C_{1} | C_{2} | |
R_{1} | 2, 7 | 7, 5 |
R_{2} | 3, 2 | 9, 1 |
Answer:
(R_{2}, C_{1}) is the pure-strategy equilibrium pair.
This game is a prisoner’s dilemma.
profile 1 | society | profile 2 | society | |||||||
Othello | Desdemona | Cassio | Iago | Othello | Desdemona | Cassio | Iago | |||
c | b | d | a | b | b | b | c | a | c | |
a | a | b | d | a | a | a | d | c | d | |
d | d | a | c | d | c | d | a | d | b | |
b | c | c | b | c | d | c | b | b | a |
Answer:
b (look at alternatives a and d)
profile 1 | society | profile 2 | society | |||
Antony | Cleopatra | Antony | Cleopatra | |||
a | b | c | b | a | c | |
b | d | b | d | d | a | |
c | a | a | c | b | b | |
d | c | d | a | c | d |
Answer:
c (no violation of condition I is evident, but we cannot be sure that M doesn’t violate condition I elsewhere)
profile 1 | society | profile 2 | society | |||||
Hamlet | Ophelia | Horatio | Hamlet | Ophelia | Horatio | |||
c | b | a | b | [unreadable] | [unreadable] | a | ||
b | [unreadable] | [unreadable] | c | [unreadable] | [unreadable] | c | ||
a | [unreadable] | [unreadable] | a | b | a | b |
What social preference ordering for profile 2 would make it the case that, regardless of how the unreadable entries are filled in, it could not be concluded from these two profiles and their corresponding social orderings that the social welfare function used to generated those two social preference orderings violates condition I?
correction added October 31: ‘generated’ should be ‘generate’
Answer:
b; c; a (violations of condition I arise only when the social preference orderings for two profiles are different and certain other conditions are not satisfied—if the social preference orderings are the same, there is no violation of condition I)
Answer: a
profile 1 | society | profile 2 | society | |||
Antony | Cleopatra | Antony | Cleopatra | |||
a | b | a | b | a | b | |
b | d | b | d | d | d | |
c | a | c | c | b | c | |
d | c | d | a | c | a |
Answer: c
Answer:
There are several correct answers. Based on profile 1, the only possible violation of condition ND would come from Ophelia’s being a dictator. So, as long as the social ordering corresponding to profile 2 did not match Ophelia’s ordering there, condition ND would be assured of being satisfied. We know that Ophelia’s ordering there has a in third place, so any ordering with a in first or second place would be fine.
Answer: c
profile 1 | society | profile 2 | society | |||
Antony | Cleopatra | Antony | Cleopatra | |||
a | b | d | b | a | d | |
b | d | b | d | d | b | |
c | a | a | c | b | a | |
d | c | c | a | c | c |
Answer: c
Answer:
a; c; b (then the social orderings for profiles 1 and 2 would have each pair of alternatives ranked in each of their two possible orders)
special instructions: Type your answers. You can present these proofs in paragraphs rather than numbered lists of claims and accompanying justifications.
Answer:
Proposition 2 is true, and can be proved as follows. Consider the following profile:
A B C a c b b a c c b a For this profile, pairwise minority rule would say that in the social ordering, b P a and a P c and c P b. Those three preferences, being cyclical, do not admit of being represented by a preference ordering. Thus, pairwise minority rule does not specify a preference ordering for every profile. (This is essentially the Concorcet paradox, in the case of pairwise minority rule rather than pairwise majority rule.)
Answer:
Proposition 1 is true, and can be proved as follows. Suppose condition I is violated. Then there is some pair of alternatives a and b, and some pair of profiles—call them Profile 1 and Profile 2—such that every individual’s pairwise ranking of a and b is the same in Profile 2 as in Profile 1, but such that the social preference orderings corresponding to the two profiles contain different pairwise rankings of a and b. Since every individual’s pairwise ranking of a and b is unchanged from Profile 1 to Profile 2, but the social pairwise ranking of a and b does change from Profile 1 to Profile 2, there must not be any individual whose preference ordering is always matched by the social preference ordering. Thus, condition ND will be satisfied.
Answer:
Proposition 1 is true, and can be proved as follows. Suppose condition ND is violated. Then there is some individual—the dictator—whose preference ordering is always matched by the social preference ordering. By condition U, every possible profile is included in the domain. This means that, for every pair of alternatives a and b, there is a profile in which the dictator prefers a to b. Since this individual is a dictator, this profile’s corresponding social preference ordering ranks a above b. So, for every pair of alternatives a and b, there is a social preference ordering in which a is ranked above b, and thus condition CS is satisfied.
Answer:
Proposition 2 is true, and there are a couple of good ways to prove it. One is to point to a Paretian social welfare function that violates condition I, such as the Borda count. Another is to point to a Paretian social welfare function that violates condition PA, such as pairwise minority rule modified to respect unanimity and to deal with Condorcet-paradox cases. The one strategy that won’t work is to point to a Paretian social welfare function that violates condition CS, since there aren’t any (see handout 24, section 24.3.4).
Which of the following claims is true?
Answer: a
For problems 52–54, consider the following partially specified profile and corresponding social preference ordering, and let M (which may or may not satisfy condition I—don’t assume anything either way) be the social welfare function generating the indicated social preference ordering:
A | B | C | society |
p | r | r | |
r | p | p | |
q | q | q |
Answer:
No, there is no such preference ordering for person C. Any preference ordering for person C would make the resulting profile and corresponding social preference ordering (1) consistent with the claim that M makes the set consisting of persons A and B quasi-decisive for r over q, because with society already ranking r over q there is no way to counter that claim with this profile and corresponding social preference ordering. By similar reasoning, no preference ordering for person C would make the resulting profile and corresponding social preference ordering (2) not consistent with the claim that M makes the set consisting of persons A and B decisive for r over q. Therefore, there is no preference ordering for person C that has both characteristics.
Answer:
Yes, there is such a preference ordering for person C. Any preference ordering for person C in which p is ranked above r would make the resulting profile and corresponding social preference ordering (1) consistent with the claim that M makes the set consisting of person A quasi-decisive for p over r; for if C ranks p above r, then the profile and its corresponding social preference ordering are prevented from being a counter-example to that claim. And any preference ordering for person C would make the resulting profile and corresponding social preference ordering (2) not consistent with the claim that M makes the set consisting of person A decisive for p over r, because with r being ranked above p in the social preference ordering (and A ranking p above r), we know that A is not decisive for p over r. Therefore, any preference ordering for person C in which p is ranked above r will do.
Answer:
No, there is no such preference ordering. A single profile and its corresponding preference ordering can’t prove a universal statement such as a claim of quasi-decisiveness.
correction added November 19: The first ‘if’ should have been ‘is’.
Answer: c
Answer: a
George | everyone except George |
Answer:
George | everyone except George |
x | a |
a | b |
b | x |
or
George | everyone except George |
a | x |
b | a |
x | b |
special instructions: If you turn in your answers to my office by the end of Tuesday, November 27, or send them to me by e-mail by that time (in which case I will waive the usual penalty for submitting answers electronically), I will grade them and return them in class on Wednesday, November 28.
1 | 2 |
e | d |
c | a |
d | b |
b | c |
a | e |
Suppose conditions P and L are in force, and person 1 has control (in the condition-L sense) over a versus b, and person 2 has control (in the condition-L sense) over c versus d. Does this profile show that condition U cannot be satisfied? If so, exhibit a set of cyclical preferences that follow from applying conditions P and L, and indicate, for each preference in the set of cyclical preferences, which condition it follows from. If not, give a social preference ordering for this profile that is consistent with conditions P and L.
Answer:
No, it does not. The following social preference ordering for this profile is consistent with conditions P and L:
d
b
a
c
eAny social preference ordering consistent with the following constraints will work:
d P a (Pareto)
d P b (Pareto)
b P a (L—person 1)
d P c (L—person 2)
1 | 2 |
c | b |
a | d |
e | c |
b | e |
d | a |
Answer:
Yes, it does. Conditions P and L imply the following preferences, which are cyclical and therefore do not admit of being represented by a social preference ordering:
a P b (L—person 1)
b P d (P)
d P c (L—person 2)
c P a (P)
Answer:
Yes, there is. Here it is:
1 | 2 |
c | b |
a | c |
b | a |