Page 1 of 3 Problem Set 1 Due in class on Tuesday Jul.docx
- 1. Page 1 of 3 Problem Set 1: Due in class on Tuesday July 7. Solution s to this homework will be posted right after class, hence no late submissions will be accepted. Test 1 on the content of this homework will be given on Tuesday July 14 at 9:00am sharp. Group solutions are welcomed and encouraged. (There is no limit on the group size.) Problem 1 (20p) Figure 1 Figure 2
- 2. Figure 3 Figure 4 Figure 5 Figure 6
- 3. Figure 7 Figure 8 Figure 9 Figure 10 Figures 1-10 above depict preference and indifference relations of ten different people on a set of three or four alternatives. (For each decision maker her set of alternatives constitutes
- 4. the entire domain of choice.) Alternatives are marked as dots and preferences are marked Page 2 of 3 with arrows: the direction of an arrow is the direction of the preference relation, no arrow between two dots means that an individual is indifferent between these two alternatives. In each of the ten figures preference relation either is or is not transitive and also, independently of the transitivity of the preference relation, the indifference relation is or is not transitive. In the following table put “true” or “false” in each of the 20
- 5. empty cells. Preference relation is transitive Indifference relation is transitive Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 Figure 6 Figure 7
- 6. Figure 8 Figure 9 Figure 10 Problem 2 (10p) Assume a voter has a strict preference over any two candidates in the set of Clinton, Obama and Edwards; in other words, it is not the case that he is indifferent between some two candidates in this set. In addition, assume that the voter is not rational. (1) Prove that a voter who prefers Obama over Clinton has to prefer Edwards over Obama and Clinton over Edwards. Present your reasoning.
- 7. (2) Prove that if all voters are not rational then either (i) all of them have identical preferences or (ii) the set of all voters is split into two subsets such that in each subset all voters have identical preferences. Problem 3 (8p) Let’s see now how likely would we observe a rational outcome if a decision maker acted randomly. Consider the case of three alternatives X, Y and Z. Assume that deciding between any two of these alternatives, e.g., X and Y, the decision maker rolls a die and dependin each with equal probability. (More specifically, assume that a decision maker takes any two alternatives, say X and Y,
- 8. and rolls a die; if one or two dots come up he decides that he is indifferent between X and Y, if three or four dots come up he decides that he prefers X over Y and if five or six dots come up he decides that he prefers Y over X. Then he does the same for X and Z and for Y and Z.) Page 3 of 3 ADVICE/HIT: Write out all possible outcomes. Each outcome should be depicted as a graph with three dots corresponding to X, Y and Z and arrows, or lack thereof, between any two dots. Now, each graph depicts a rational or a non-rational preference system. Which
- 9. graphs correspond to rational systems? What is the relative frequency of the rational ones in the set of all systems? Note that this frequency is the probability that individual’s preferences determined at random, as described above, will end up being rational? Problem 4 (3 extra credit points) Let’s go back to explaining the survey data in which 60% chose Edwards over Obama, 60% chose Edwards over Clinton, 60% chose Clinton over Obama and then when asked to cast a single vote for one of the three candidates, 45% chose Clinton, 30% Obama, and 25% Edwards.
- 10. (1) Prove that it is not possible that all voters are rational (have transitive preferences.) (2) Suppose now that the distribution of votes was not 45% chose Clinton, 30% Obama, and 25% Edwards but 38% Clinton, 32% Obama and 30% Edwards; pair-wise data remains unchanged. Prove that it is possible now that all voters are rational (have transitive preferences) and calculate the percentage of votes for each of the six different orderings of the three candidates. Page 1 of 3 Problem Set 1: Due in class on Tuesday July 7.