University of Kansas, Fall 2005
Philosophy 666: Rational Choice Theory
Ben Eggleston—eggleston@ku.edu

Mid-term exam: answer key

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.

Questions:

1. (10 points) You have won a free vacation and can choose any one of these three options: Miami, Rome, or Tokyo. If you go to Miami, you will have a good time. If you go to Rome, you will have a wonderful time unless you are sick, in which case you will have a bad time. If you go to Tokyo, you will have a great time unless it is cold there, in which case you will have a bad time. Write a decision table for this situation.

2. (5 points) A transformation t(x) is an ordinal one just in case, for all utilities w and v, it is the case that w ≥ v if and only if t(w) ≥ t(v). Following is a series of lines someone might offer as a proof of the second half of this biconditional as it applies to the transformation t(x) = 15 – 4x. Are all of the lines truly justified (not just claimed to be justified)? If not, which line is the smallest-numbered line that is not really justified (despite an apparent justification being provided for it)?

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.)

3. (10 points) Either prove that t(x) = 5 is an ordinal transformation or give a counter-example showing that it’s not.

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):

• contraction condition: If you choose some item from a menu and this item remains available in a more restricted menu, then you also choose it from the restricted menu.
• expansion condition: If you choose some item in pairwise choices with every other item on the menu, then you choose it, though not necessarily alone, from the full menu.
• revelation condition: If you ever choose one item when a second item is available, then whenever you choose the second and the first is available, you also choose the first.

4. (5 points) Specify a set of no more than five choices violating the expansion condition (or, if there is no such set, say so).

From AB, choose A.
From AC, choose A.
But from ABC, choose B.

(This is just one of many possible correct answers.)

5. (5 points) Specify a set of no more than five choices violating the contraction condition but not the expansion condition (or, if there is no such set, say so).

From ABC, choose A.
But from AB, choose B.

(This is just one of many possible correct answers.)

6. (5 points) Specify a set of no more than five choices violating the revelation condition but not the contraction condition (or, if there is no such set, say so).

From ABC, choose A.
From ABD, choose B.

(This is just one of many possible correct answers.)

7. (5 points) Specify a set of no more than five choices violating the expansion condition but not the revelation condition (or, if there is no such set, say so).

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:

8. (5 points) Which act(s) would be selected by the maximin rule?

A₁ and A₃

(Explanation: the minima for the four acts are 5, 4, 5, and 3.)

9. (5 points) Which act(s) would be selected by the optimism-pessimism rule, assuming an optimism index of 0.3?

A₃

(Explanation: The a-indexes for the acts are as follows:
A₁: 0.3*7 + (1 – 0.3)*5 = 2.1 + 3.5 = 5.6
A₂: 0.3*8 + (1 – 0.3)*4 = 2.4 + 1.2 = 3.6
A₃: 0.3*9 + (1 – 0.3)*5 = 2.7 + 3.5 = 6.2
A₄: 0.3*7 + (1 – 0.3)*3 = 2.1 + 2.1 = 4.2)

10. (5 points) Which act(s) would be selected by the principle of insufficient reason?

A₃

(Explanation: The expected utilities of the acts, assuming the states to be equiprobable, are as follows:
A₁: (1/3)*5 + (1/3)*6 + (1/3)*7 = (1/3)*(5 + 6 + 7) = (1/3)*18 = 18/3
A₂: (1/3)*8 + (1/3)*4 + (1/3)*6 = (1/3)*(8 + 4 + 6) = (1/3)*18 = 18/3
A₃: (1/3)*6 + (1/3)*9 + (1/3)*5 = (1/3)*(6 + 9 + 5) = (1/3)*20 = 20/3
A₄: (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:

11. (10 points) Prove that you can figure out the optimism index of an agent who uses the optimism-pessimism rule by ascertaining the value of c that makes the agent indifferent between A₁ and A₂. That is, prove that if an agent uses the optimism-pessimism rule and is indifferent between A₁ and A₂, then the agent’s optimism index is equal to c.

12. (5 points) Give an example of an ordinal transformation that is not a linear transformation, by giving a short list of utilities and a corresponding list that is an ordinal transformation of the first list but is not a linear transformation of the first list.

(This is just one of many possible correct answers.)

13. (10 points) Prove that no linear transformation converts your first list (in problem 12) into your second one.

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.

14. (5 points) Suppose the probability that you will make money in the stock market this year is 3/5, and that the probability that you will read a free online stock-market newsletter is 1/4. Suppose, finally, that the probability that you will make money in the stock market this year, given that you read the newsletter, is 3/5. Is the event of your making money in the stock market this year independent of the event of your reading the newsletter? Why or why not?

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.)

15. (10 points) Suppose Mike is suspected, by his employer, of being a marijuana user. The employer believes that in the population at large, 4 percent of the people use marijuana. To test Mike for marijuana use, he administers a test that correctly detects marijuana use 95 percent of the time, and correctly detects nonuse 90 percent of the time. If Mike tests positive for marijuana use, how probable should the employer think it is that Mike uses marijuana? Use Bayes’s Theorem and show your work. (This problem is adapted from the Stanford Encyclopedia of Philosophy entry on Bayes’s Theorem.)

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) = 0.1

So 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