Following are the correct answers. After some answers, explanations are required. Stating these explanations was not necessary for getting full credit. They are added here only to provide additional information about the questions and answers.
you are sick; it is cold in Tokyo | you are sick; it is not cold in Tokyo | you are not sick; it is cold in Tokyo | you are not sick; it is not cold in Tokyo | |
Miami | good | good | good | good |
Rome | bad | bad | wonderful | wonderful |
Tokyo | bad | great | bad | great |
no. | claim | justification |
1 | t(w) ≥ t(v) | assumption for proving conditional |
2 | 15 – 4w ≥ 15 – 4v | 1, definition of t(x) |
3 | –4w ≥ –4v | 2, subtract 15 from each side |
4 | w ≥ v | 3, divide each side by –4 |
No, not all the lines are truly justified. Line 4 is not really justified.
(Explanation: When you divide both sides of an inequality by a negative number, you have to reverse the direction of the inequality sign. So what follows from line 3 is not line 4, but v ≥ w. You cannot derive line 4, as stated above, from the preceding lines.)
Here’s a counter-example. Suppose v = 1 and w = 0. Then t(v) = 5 and t(w) = 5.
(Explanation: If t(x) were an ordinal transformation, then, for all w and v, we would have w ≥ v if and only if t(w) ≥ t(v). But here, it is false that w ≥ v, even though it is true that t(w) ≥ t(v).) (This is just one of many possible correct answers.)
Questions 4–7 refer to conditions defined as follows (in the glossary in Allingham’s Choice Theory):
From AB, choose A.
From AC, choose A.
But from ABC, choose B.(This is just one of many possible correct answers.)
From ABC, choose A.
But from AB, choose B.(This is just one of many possible correct answers.)
From ABC, choose A.
From ABD, choose B.(This is just one of many possible correct answers.)
There is no such set.
(Explanation: There is no such set because any violation of the expansion condition is also a violation of the revelation condition.)
Questions 8–10 refer to the following decision table:
S_{1} | S_{2} | S_{3} | |
A_{1} | 5 | 6 | 7 |
A_{2} | 8 | 4 | 6 |
A_{3} | 6 | 9 | 5 |
A_{4} | 3 | 5 | 7 |
A_{1} and A_{3}
(Explanation: the minima for the four acts are 5, 4, 5, and 3.)
A_{3}
(Explanation: The a-indexes for the acts are as follows:
A_{1}: 0.3*7 + (1 – 0.3)*5 = 2.1 + 3.5 = 5.6
A_{2}: 0.3*8 + (1 – 0.3)*4 = 2.4 + 1.2 = 3.6
A_{3}: 0.3*9 + (1 – 0.3)*5 = 2.7 + 3.5 = 6.2
A_{4}: 0.3*7 + (1 – 0.3)*3 = 2.1 + 2.1 = 4.2)
A_{3}
(Explanation: The expected utilities of the acts, assuming the states to be equiprobable, are as follows:
A_{1}: (1/3)*5 + (1/3)*6 + (1/3)*7 = (1/3)*(5 + 6 + 7) = (1/3)*18 = 18/3
A_{2}: (1/3)*8 + (1/3)*4 + (1/3)*6 = (1/3)*(8 + 4 + 6) = (1/3)*18 = 18/3
A_{3}: (1/3)*6 + (1/3)*9 + (1/3)*5 = (1/3)*(6 + 9 + 5) = (1/3)*20 = 20/3
A_{4}: (1/3)*3 + (1/3)*5 + (1/3)*7 = (1/3)*(3 + 5 + 7) = (1/3)*15 = 15/3)
Question 11 refers to the following decision table:
S_{1} | S_{2} | |
A_{1} | 0 | 1 |
A_{2} | c | c |
no. | claim | justification |
1 | The agent uses the optimism-pessimism rule. | assumption for proving conditional |
2 | The agent is indifferent between A_{1} and A_{2}. | assumption for proving conditional |
3 | The a-index of A_{1} = the a-index of A_{2}. | 1 and 2 |
4 | The a-index of A_{1} = a*1 + (1 – a)*0 | definition of a-index, with substitution from table given above |
5 | The a-index of A_{2} = a*c + (1 – a)*c | definition of a-index, with substitution from table given above |
6 | a*1 + (1 – a)*0 = a*c + (1 – a)*c | 3, with substitution from 4 and 5 |
7 | a + 0 = ac + c – ac | 6, simplified |
8 | a = c | 7, simplified |
x | t(x) |
2 | 3 |
1 | 1 |
0 | 0 |
(This is just one of many possible correct answers.)
Suppose there were a linear transformation that converted 0 to 0 and 1 to 1. Then we would have the following equations: 0 = a*0 + b and 1 = a*1 + b. The first equation yields 0 = 0 + b, or b = 0. Substituting this into the second equation, we have 1 = a + 0, or a = 1. So the only linear transformation that converts 0 to 0 and 1 to 1 is the transformation t(x) = 1*x + 0, or t(x) = x. But clearly this transformation fails to convert 2 to 3. So, no linear transformation converts 2, 1, and 0 to 3, 1, and 0.
Yes, the event of your making money in the stock market this year is independent of the event of your reading the newsletter, since (1) the unconditional probability of your making money in the stock market is equal to (2) the conditional probability of that event, given that you read the newsletter—each probability is 3/5.
(Explanation: The basic idea is that since P(p/q) = P(p), we can conclude that p is independent of q. Note that you are also given P(q)—the probability that you will read the newsletter. But this is irrelevant—even if that information had been left out, the reasoning leading to the answer would have been exactly the same.)
Let u = Mike uses marijuana, and let p = Mike tests positive. Then we have
P(u) = 0.04
P(not u) = 0.96
P(p/u) = 0.95
P(p / not u) = 0.1So P(u/p)
= P(both) / P(p)
= P(u & p) / P(p)
= P(u & p) / P(u & p or not u & p)
= P(u & p) / [P(u & p) + P(not u & p)]
= P(u)*P(p/u) / [P(u)*P(p/u) + P(not u)*P(p / not u)]
= 0.04*0.95 / [0.04*0.95 + 0.96*0.1]
= 0.038 / [0.038 + 0.096]
= 0.038 / 0.134
= 38/134
= 19/67