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

Problems for credit

(due at the beginning of class unless otherwise specified)

due Wednesday, August 24 (Resnik, section 1-2):

1. You have $10,000 to invest for one year. You could put it under your mattress (resulting, let us say, in $10,000 in one year), put it in a savings account with an interest rate of 3 percent (resulting in $10,300 in one year), or invest it in some junk bonds that, after one year, will have either appreciated by 20 percent or become worthless. Write a decision table for this situation.

2. In problem 1, do any of the options dominate any of the others? Do any of the options dominate all of the others? Are any dominated by all of the others?

Yes, the savings-account option dominates the mattress option. (Remember the criterion for dominance: an option A dominates an option B if, in every possible state, option A leads to at least as as good an outcome as option B, and there is at least one state in which option A leads to a better outcome than option B.) None of the options dominates all of the others, and no option is dominated by all of the others.

3. For lunch, you can have either an egg-salad sandwich or spaghetti. If you have the sandwich, you will be somewhat satisfied, but will get food poisoning if the egg salad is bad. If you have spaghetti, you will be very satisfied, but will get at least a little bit of the sauce on your shirt unless you are having a particularly lucky day. Write a decision table for this situation. (Remember that when you specify the states, you must make them mutually exclusive and exhaustive. Thus, you cannot just specify one state as 'the egg salad is bad' and another as 'you are having a particularly lucky day'. Rather, in specifying each of the possible states—hint: there are four of them—you will need to refer to both of these factors.)

	egg salad is not bad and you are not lucky	egg salad is not bad and you are lucky	egg salad is bad and you are not lucky	egg salad is bad and you are lucky
egg salad	somewhat satisfied	somewhat satisfied	somewhat satisfied, but food poisoning as well	somewhat satisfied, but food poisoning as well
spaghetti	very satisfied but some sauce on shirt	very satisfied	very satisfied but some sauce on shirt	very satisfied

4. In problem 3, why does the spaghetti option not dominate the egg-salad option?

The spaghetti option does not dominate the egg-salad option because there is at least one state—the first one in the table as drawn above—in which it is not clear that the spaghetti outcome is better than the egg-salad outcome. (Technically, this question should have been written, “Why are we not entitled to say that the spaghetti option dominates the egg-salad option?” And then the answer would be, “Because we are not entitled to say that ‘very satisfied but some sauce on shirt’ is at least as good as ‘somewhat satisfied’.)

due Friday, August 26 (Resnik, section 1-4):

5. You will probably need to go to San Francisco in a couple of months, but there is a chance you will not. You can either buy a ticket right now or wait until the last minute, when you will know for sure. Right now, you can buy a ticket for $300, but if you don’t end up taking the trip this money will be lost forever. If you wait until the last minute and know you have to go, you might be able to buy a please-take-this-empty-seat ticket for $100, but you might have to buy a we-know-you’re-desperate-and-we’re-your-last-option ticket for $1,000. Draw a decision tree for this situation.

Your tree should begin with a decision node and a branch labeled ‘buy now’ and a branch labeled ‘wait until last minute'. The 'buy now’ branch should have a chance node with a branch labeled 'take trip’ and a branch labeled ’don’t take trip’. The outcomes corresponding to these should be labeled ’take trip for $300’ and '$300 lost forever’. The ‘wait until last minute’ branch should have a chance node with three branches: one labeled ‘$100’, one labeled ’$1,000’, and one labeled ‘don’t take trip’. The outcomes corresponding to these should be labeled ‘take trip for $100’, ‘take trip for $1,000’, and ‘don’t take trip’.

6. Write a decision table for the situation described in problem 5.

due Wednesday, August 31 (Resnik, section 2-1):

7. Convert a scale of 32, 33, 34, 35 into a scale whose numbers are between 5 and 7, inclusive.

There are infinitely many correct solutions, but one is 5, 6, 6.5, 7; another is 5.32, 5.33, 5.34, 5.35.

8. Either prove that t(x) = 8x – 8 is an ordinal transformation or give a counter-example showing that it is not. (Assume that x can be any real number.)

proof that if w ≥ v, then t(w) ≥ t(v):

proof that if t(w) ≥ t(v), then w ≥ v:

Thus, we have a proof that w ≥ v if and only if t(w) ≥ t(v). This means that t(x) is an ordinal transformation.

9. Either prove that t(x) = x/3 + 4 is an ordinal transformation or give a counter-example showing that it is not. (Assume that x can be any real number.)

proof that if w ≥ v, then t(w) ≥ t(v):

proof that if t(w) ≥ t(v), then w ≥ v:

Thus, we have a proof that w ≥ v if and only if t(w) ≥ t(v). This means that t(x) is an ordinal transformation.

10. Either prove that 1 – x is an ordinal transformation or give a counter-example showing that it is not. (Assume that x can be any real number.)

counter-example: The scale {16, 15} gets transformed into {–15, –14}, and the numbers are no longer in order from largest to smallest.

due Wednesday, September 7 (Allingham, pp. 11–21):

11. Suppose you wanted to prove a statement of the form ‘A if and only if B’, where A and B are statements such as mathematical equations or mathematical inequalities or descriptions or preferences or some other kind of statement. What two statements could you prove in order to prove something like ‘A if and only if B’?

You could prove ‘if A then B’ and ‘if B then A’.

12. In providing a proof that a transformation satisfying condition c on p. 25 of Resnik also satisfies condition a on that page, we were somewhat hampered by the fact that condition c is written in terms of weak inequalities rather than strong inequalities. It is interesting to note, however, that if a transformation satisfies condition c, then it satisfies its strong-inequality variant. Prove this. That is, prove that if a transformation t satisfies condition c as stated, then it also satisfies this statement: w > v if and only if t(w) > t(v). (Each half of this proof should take you about ten lines, and will be similar in certain respects to the halves of the proof mentioned above (i.e., our proof that a transformation satisfying condition c on p. 25 of Resnik also satisfies condition a on that page. You should not, however, need to mention any function u—just stick to w, v, t(w), and t(v).)

proof that if w > v, then t(w) > t(v):

no.	claim	justification
1	w > v	assumption for proving conditional
2	w > v or w = v	1, ‘or’ introduction
3	w ≥ v	2, definition of ≥
4	t(w) ≥ t(v)	3, condition c
5	t(w) > t(v) or t(w) = t(v)	4, definition of ≥
6	t(w) = t(v)	5, assumption—exploring second possibility in line 5
7	t(w) ≥ t(v) and t(v) ≥ t(w)	6, definition of ≥
8	w ≥ v and v ≥ w	7, condition c
9	w = v	8, definition of ≥
10	not the case that w > v	9, definition of =
11	contradiction	1 and 10
12	line 6 is false	line 6 leads to the contradiction just mentioned
13	t(w) > t(v)	only remaining possibility in line 5

proof that if t(w) > t(v), then w > v:

due Friday, September 9 (Allingham, pp. 22–27):

13. Suppose you pick A from a menu of A, B, and C; and B from a menu of B and C. Have you violated the contraction condition? Why or why not?

No, because what you chose from the first menu (namely, A) was not available on the second (contracted) menu.

14. Suppose you pick C from a menu of A, B, C, and D; and B from a menu of B, C, and D. Have you violated the contraction condition? Why or why not?

Yes, because what you chose from the first menu (namely, C) was available on the second (contracted menu), and yet you chose something else from that menu.

15. Suppose you pick A from a menu of A and B; A from a menu of A and C; A from a menu of A and D; and B from a menu of A, B, C, and D. Have you violated the expansion condition? Why or why not?

Yes, because you chose one thing (namely, A) in a pair-wise choice with each of the items on the larger menu, and yet you chose something else from that menu.

16. Suppose you pick A from a menu of A and B; C from a menu of A and C; and B from a menu of A, B, C, and D. Have you violated the expansion condition? Why or why not?

No, because the item you chose from the larger menu is not something you chose in pair-wise choices with all the other things on that menu.

17. Can the choices in problem 13 be explained by a transitive preference relation? How about those in problem 14? (In each case, explain why or why not.)

13: yes; you could rank A highest, then B, then C and D.

14: no; your choices imply that C is better than everything else, but that B is better than C and D. The reversal between C and B violates transitivity.

18. Can the choices in problem 15 be explained by a transitive preference relation? How about those in problem 16? (In each case, explain why or why not.)

15: no; your choices imply that A is better than B, that A is better than C, that A is better than D, and that B is better than everything else. The reversal between A and B violates transitivity.

16: no; your choices imply that A is better than B, that C is better than A, and that B is better than everything else. The reversal between A and B violates transitivity.

due Monday, September 19 (Resnik, section 2-2):

19. Suppose you follow the maximin rule in a situation represented by the following decision table. What act or acts pass that test? Now suppose you follow the lexical maximin rule. What act or acts pass that test?

The maximin rule selects A₂ and A₃. The lexical maximin rule selects A₂.

20. Suppose you follow the maximin rule in a situation represented by the following decision table. What act or acts pass that test? Now suppose you follow the lexical maximin rule. What act or acts pass that test?

The maximin rule selects A₁ and A₃. The lexical maximin rule selects A₁.

due Wednesday, September 21 (Resnik, section 2-3):

21. Suppose you follow the minimax regret rule in a situation represented by the following decision table. What act or acts pass that test?

The minimax regret rule selects A₂.

22. Suppose you follow the minimax regret rule in a situation represented by the following decision table. What act or acts pass that test?

The minimax regret rule selects A₁ and A₃.

due Friday, September 23 (Resnik, section 2-3):

23. Prove that every positive linear transformation is an ordinal transformation.

To prove that every positive linear transformation is an ordinal one, we must prove that an arbitrary positive linear transformation—which we’ll represent as t(x) = ax + b, with a > 0—satisfies condition c stated on p. 25 of Resnik: namely, that for all w and v (which are untransformed utility values), w ≥ v if and only if t(w) ≥ t(v) (where t(w) and t(v) are the transformed utility values). This course, involves two proofs, one for each half of the biconditional. As mentioned above, we’ll let t(x) = ax + b, where a > 0.

Here’s a proof that if w ≥ v, then t(w) ≥ t(v):

no.	claim	justification
1	w ≥ v	assumption for proving conditional
2	aw ≥ av	1, multiply each side by a
3	aw + b ≥ av + b	2, add b to each side
4	t(w) ≥ t(v)	3, definition of t(x)

Here’s a proof that if t(w) ≥ t(v), then w ≥ v:

no.	claim	justification
1	t(w) ≥ t(v)	assumption for proving conditional
2	aw + b ≥ av + b	1, definition of t(x)
3	aw ≥ av	2, subtract b from each side
4	w ≥ v	3, divide each side by a

Together, these two proofs prove the biconditional that characterizes ordinal transformations.

24. Prove that not every ordinal transformation is a positive linear transformation. Do this by proving the stronger claim that not every ordinal transformation is a linear transformation, and then by appealing to the claim (which you may take as given) that if some transformation is not a linear transformation, then it is not a positive linear transformation.

There are a couple of different approaches that can be taken to the problem of showing that not every ordinal transformation is a linear transformation. One involves specifying a non-linear polynomial function that, although non-linear, nonetheless suffices to specify an ordinal transformation; another involves specifying enough of a transformation to ensure that it is non-linear, while allowing that it is ordinal.

A. Consider the transformation t(x) = x³. This is an ordinal transformation, but obviously non-linear. (To see that it is non-linear, consider how it transforms the numbers 1, 2, and 3 into 1, 8, and 27. The only linear transformation that could account for the transformations of the first two values (1 to 1, and 2 to 8) is u’ = 7u – 6. But this transformation does not account for the transformation of the third value (3 to 27), since 7*3 – 6 = 21 – 6 = 15, which is not equal to 27. Therefore, the transformation t(x) = x³ cannot be expressed as a linear transformation.)

B. Instead of coming up with an algebraic function that is non-linear, we can just assert that there exist ordinal transformations that transform 1 into 1, 2 into 8, and 3 into 27, and point out that, by the reasoning above (i.e., solving for a linear transformation fitting the first two values and not fitting the third value), no transformation that transforms 1, 2, and 3 in this way could possibly be a linear one. (Can we just assert, without proof, that there exist ordinal transformations that transform 1 into 1, 2 into 8, and 3 into 27? Sure—in fact, there exist infinitely many such transformations. Here is one of them:

1. For all x less than 1, let t(x) = x.
2. For x = 1, let t(x) = 1.
3. For all x between 1 and 2, let t(x) = x.
4. For x = 2, let t(x) = 8.
5. For all x between 2 and 3, let t(x) = x + 15.
6. For x = 3, let t(x) = 27.
7. For all x greater than 3, let t(x) = x + 100.

So, if you’re not comfortable just asserting that there are ordinal transformations that transform 1 into 1, 2 into 8, and 3 into 27, you can exhibit one, by specifying one in a piece-wise fashion, as above.)

So there are a couple of different approaches you can take to showing that not every ordinal transformation is a linear one. Then, having established that, you can point out that if some transformation is not a linear one, then it is certainly not a positive linear one. So not every ordinal transformation is a positive linear one.

due Monday, September 26 (Resnik, section 2-4):

25. Consider the following table.

Compute the a-indexes for the acts in this table. Assume that the agent’s optimism index is 0.6.

26. Consider the following table.

Prove that no ordinal transformation converts this table into the one specified in problem 25.

25. (revised) Consider the following table.

Compute the a-indexes for the acts in this table. Assume that the agent’s optimism index is 0.6.

The a-index for A₁ is 0.6*max + (1 – 0.6)*min = 0.6*max + 0.4*min = 0.6*8 + 0.4*6 = 4.8 + 2.4 = 7.2.

The a-index for A₂ is 0.6*max + (1 – 0.6)*min = 0.6*max + 0.4*min = 0.6*11 + 0.4*8 = 6.6 + 3.2 = 9.8.

The a-index for A₃ is 0.6*max + (1 – 0.6)*min = 0.6*max + 0.4*min = 0.6*10 + 0.4*8 = 6 + 3.2 = 9.2.

The a-index for A₄ is 0.6*max + (1 – 0.6)*min = 0.6*max + 0.4*min = 0.6*10 + 0.4*6 = 6 + 2.4 = 8.4.

26. (revised) Consider the following table.

Prove that no linear transformation converts this table into the one specified in problem 25.

Suppose there were a linear transformation that would convert this one into the one specified in problem 25. Then—looking at just the first two cells for A₁—it would have to convert 10 to 8 and 2 to 6. So we have 8=a*10 + b, and 6 = a*2 + b. The first equation implies b = 8 – 10a; substituting this into the second equation, we have 6 = 2a + 8 – 10a. Then we have –2 = –8a, or a = 1/4. Substituting this into b = 8 – 10a, we have b = 8 – 10*(1/4), or b = 8 – 2.5, or b = 5.5. So we have the transformation u’ = u/4 + 5.5.

Now we need to find a cell for which this transformation does not work. If we try the third cell for A₁, we see that if you substitute 6 for u, then you have 6/4 + 5.5, which is 1.5 + 5.5, which is 7. And that is what the third cell for A₁ is supposed to be. So, in an unfortunate coincidence, the transformation we found for the first two cells for A₁ also works for the third cell for A₁. We’ll have to keep looking in order to show that this transformation fails somewhere.

It won’t pay to use the first cell for A₂, since we know our transformation successfully converts 10 to 8—that is one of the conditions for which we solved in the first place. Let’s try the second cell for A₂. If you substitute 19 for u, then you have 19/4 + 5.5, which is 4.75 + 5.5, which is 10.25, which is not 11. So we have found a cell where the transformation we derived does not work, and thus we have proved that there is no transformation that works for every cell.

due Wednesday, September 28 (Resnik, section 2-5):

27. Resnik gives table 2-25 (on p. 35) as an example of a table with two acts between which the principle of insufficient reason—unlike many other principles—is not indifferent. What values does that principle assign to the acts, and which act does it select as the better one?

In table 2-25, the principle of insufficient reason assigns to act A₁ a value of 73/9, and it assigns to act A₂ a value of 28/9. So it selects A₁ as the better act.

28. Suppose you know the following information about a decision situation; that is, you know all of the utilities except for one (the one referred to by ‘x’). What must be true of x in order for the principle of insufficient reason to be indifferent between A₁ and A₂? (You should specify some equation or inequality that will be true when and only when the principle of insufficient reason is indifferent between A₁ and A₂.)

In order for the principle of insufficient reason to be indifferent between A₁ and A₂, the utilities associated with the possible outcomes of A₁ must have the same sum as the utilities associated with the possible outcomes of A₂. So it has to be the case that 10 + 4 + 2 = 8 + 5 + x, or x = 3.

29. Suppose you know that the utilities of the possible outcomes of act A₁ are 2, 4, 6, 8, and 10; and that the utilities of the possible outcomes of act A₂ are 3, 6, 9, 12, and 15. Without knowing which states these outcomes are associated with, can you say which act would be recommended by the principle of insufficient reason? Also, consider how the dominance principle might bear on these acts in light of the fact that you do not know which outcomes are associated with which states. Is it possible that A₁ dominates A₂? Is it possible that A₂ dominates A₁?

Yes, the principle of insufficient reason would recommend A₂, since the utilities associated with its outcomes have a greater sum than the utilities associated with the outcomes of A₁. It is not possible that A₁ dominates A₂, since the utilities of A₁ and A₂ cannot be arranged in any way that makes those of A₁ always greater than or equal to, and in some state greater than, those of A₂. But A₂ could dominate A₁, since the utilities of A₁ and A₂ can be arranged in a way (namely, ascending order) that makes those of A₂ always greater than or equal to, and in some state greater than, those of A₁. For example, just arranging the utilities in ascending order would suffice.

30. In answering question 2 of section 2-5 of the Resnik text, you proved that at least one ordinal transformation of a decision table can change the results of applying the principle of insufficient reason. It is significant that this is not true of any positive linear transformation—i.e., if one decision table is a positive linear transformation of another, then the principle of insufficient reason selects the same act(s) from each table. Your assignment is to prove this, though you can make certain assumptions in order to keep things simple: you can assume that there are just two acts, and just two states, and that the principle of insufficient reason selects just one act from the untransformed table (i.e., that the acts do not tie, when the principle of insufficient reason is applied). You assignment, then, is to prove that if decision table D₁ has two acts and two states, the principle of insufficient reason selects just one of the acts, and decision table D₂ is a positive linear transformation of D₁, then the principle of insufficient reason selects just that same act from D₂.

Here is a proof of the desired result:

due Friday, October 7 (Resnik, section 3-2):

31. Resnik writes on p. 51 that while axiom 3 “does not always apply, . . . there is no restriction on theorem 3.” In what circumstances does theorem 3 apply, while axiom 3 does not? What is the restriction on axiom 3 that is absent from theorem 3?

Theorem 3, but not axiom 3, applies when p and q are not mutually exclusive. The restriction on axiom 3 that is absent from theorem 3 is that p and q must be mutually exclusive.

32. Prove theorem 6. (You can prove it in just one direction and simply assert that the proof for the other direction is analogous.) You may, of course, draw on the proof of theorem 6 given in the text, but you must provide a proof meeting the standards of rigor to which we have adhered so far. That is, your proof must consist of a series of numbered statements and justifications, and each statement (whether it be an English sentence or a mathematical equation or inequality) must be explicitly justified in terms of preceding statements (or other legitimate bases for writing new statements, such as assuming something in order to prove a conditional statement).

To prove theorem 6, we’ll prove that if p is independent of q, then q is independent of p. An analogous proof could be constructed to establish the converse—that if q is independent of p, then p is independent of q. Together, these two conditionals would yield the biconditional of theorem 6.

no.	claim	justification
1	P(p) ≠ 0	given
2	P(q) ≠ 0	given
3	p is independent of q	assumption for proving conditional
4	P(p) = P(p/q)	3, definition of independence
5	P(q/p) = P(p & q)/P(p)	theorem 4
6	p & q and q & p are equivalent	nature of conjunction
7	P(p & q) = P(q & p)	6, theorem 2
8	P(q/p) = P(q & p)/P(p)	5, with substitution from 7
9	P(q & p) = P(q) * P(p/q)	axiom 4
10	P(q/p) = P(q) * P(p/q) / P(p)	8, with substitution from 9
11	P(q/p) = P(q) * P(p) / P(p)	10, with substitution from 4
12	P(q/p) = P(q)	11, simplified
13	q is independent of p	12, definition of independence

33. Both the inverse probability law and Bayes’s theorem can be used to compute conditional probabilities. Bayes’s theorem is more complicated, but allows you to compute P(p/q) without a piece of information that the inverse probability requires. What is that piece of information?

The extra piece of information required by the inverse probability law (relative to the information Bayes’s theorem requires) is P(q), where P(p/q) is the probability to be computed. That is, the the inverse probability law requires one to know the absolute probability of the conditioning event.

34. A new disease, called the Bucolic Plague, is sweeping the countryside. A test for the disease can tell whether you have it, but not flawlessly. If you have the disease, there is a 99-percent chance that the test will indicate that (and a 1-percent chance that it will give you a false negative result); and if you do not have the disease, there is a 97-percent chance that the test will indicate that (and a 3-percent chance that it will give you a false positive result). In the general population, 0.1 percent of the people have the disease. (So think of this as the prior probability that you have the disease.) You take the test, and your result is positive. Given this, what is the probability that you have the disease? Find this probability using Bayes’s theorem.

Let’s define a couple of variables as follows:

• d = have the disease
• p = positive test result

Now, deriving and applying Bayes’s theorem, we have the following: P(d/p)
= P(d & p) / P(p)
= P(d & p) / [P(d & p) + P(not d & p)]
= P(d) * P(p/d) / [P(d) * P(p/d) + P(not d) * P(p / not d)]
= .001 * .99 / [.001 * .99 + .999 * .03]
= .00099 / [.00099 + .02997]
= .00099 / .03096
= .03
(This problem is based on the Car Talk Puzzler of June 7, 2004.)

due Monday, October 10 (Resnik, section 3-2a):

35. Suppose you are 99-percent sure your hamster does not have telekinetic powers. Yet, you find that your VCR is programmed to tape several shows on Animal Planet, and you have no idea how this happened. You figure that the probability of this happening, on the assumption that your hamster is not psychic, is 3 percent. (That is, you think there is a 3-percent chance that you would do it and forget, or that a friend would sneak in and do it to gaslight you, or that it would happen in some other non-hamster way.) You also figure that the probability of this happening, on the assumption that your hamster is psychic, is 80 percent—it's just the sort of thing your media-savvy-but-couch-potato hamster would do. Given the mysterious programming of your VCR, what is the (posterior) probability that your hamster has telekinetic powers?

Let k be that your hamster has telekinetic powers, and let p be the programming of your VCR. Deriving and applying Bayes’s theorem, we have the following:
P(k/p)
= P(both) / P(p)
= P(k & p) / P(k & e or not k & p)
= P(k & p) / [P(k & p) + P(not k & p)]
= P(k) * P(p/k) / [P(k) * P(p/k) + P(not k) * P(p / not k)]
= 0.01 * 0.8 / [0.01 * 0.8 + 0.99 * 0.03]
= 0.008 / [0.008 + 0.0297]
= 0.008 / 0.0377
= 0.21

36. How high would your prior probability that your hamster does not have telekinetic powers have to be in order for the posterior probability (in the foregoing scenario) to be less than 10 percent?

Let’s represent P(not k) as x. Then, borrowing from work we just did, we need for solve the following inequality for x:
(1 – x) * 0.8 / [(1 – x) * 0.8 + x * 0.03] < 0.1
(0.8 – 0.8x) / [0.8 – 0.8x + 0.03x] < 0.1
All these decimals are a pain. Let’s multiply the top and bottom of the left side of the equality by 100:
(80 – 80x) / (80 – 80x + 3x) < 0.1
(80 – 80x) / (80 – 77x) < 0.1
Now notice that x is between 0 and 1, inclusive, since all probabilities are. So 77x is between 0 and 77, and the denominator on the left hand side is positive. So we can multiply both sides of the inequality by that value without affecting the direction of the inequality:
80 – 80x < (0.1) * (80 – 77x)
Let’s multiply both sides by 10:
800 – 800x < 80 – 77x
–723x < –720
Dividing both sides by –723 (a negative number requiring a reversal of the inequality sign), we have
x > –720/–723
x > 720/723
x > 0.995851
(So you’d have to be pretty darn sure your hamster didn’t have telekinetic powers in order for you to remain at least 90-percent sure of that in the face of the countervailing evidence described in the problem.)

due Friday, October 21 (Resnik, section 3-2b):

37. Suppose Clay is at a yard sale and sees an earthenware bowl for sale for a price of $5. Clay has an associate who is a crockery collector who will buy it from him for $50 if she deems it to be a valuable antique, which he regards as having a probability of 10 percent, and will buy it from him for $1 if she deems it to be junk, which he regards as having a probability of 90 percent. What is the expected monetary value of the act of Clay’s buying the bowl and selling it to his associate? (Be sure to include, in your calculations, the $5 Clay has to pay for the bowl as well as the possible prices he may receive upon selling it. You can do this by filling in the cells in the decision table with things like ‘–$5 + $50 = $45’, etc.)

EMV of buying bowl and selling it to his associate:

EMV(A₁) = (1/10)($45) + (9/10)(–$4) = $4.50 – $3.60 = $0.90

38. In the situation just described, Clay notices that another shopper has a cell phone that he could use to take a picture of the bowl, send it to his associate, and find out from her, with certainty, whether she regards it as valuable or as junk. Clay’s fellow shopper is going to charge him a little bit of money for the use of the phone. How much should Clay be willing to pay (assuming he is concerned to maximize expected monetary value)? In other words, what is the phone-usage price that would make Clay’s paying that price, using the phone, and getting definite information have the same expected monetary value as Clay’s just buying the bowl without the additional information he could get by using the phone?

value of phone usage, assuming completely reliable information:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of buying and selling regardless of what the information source says, its EMV is just the EMV of buying and selling, which we know (from problem 37) is $0.90.

EMV(A₂): Since A₂ has an outcome of $0 in every state, its EMV is obviously $0.

EMV(A₃) = (1/10)($45) + (0)($___) + (0)($___) + (9/10)($0)
= $4.50 + $0 + $0 + $0
= $4.50

EMV(A₄) = (1/10)($0) + (0)($___) + (0)($___) + (9/10)(–$4)
= $0 + $0 + $0 – $3.60
= –$3.60

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₃, which has an EMV of $4.50. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $0.90. That’s a difference of $3.60; you should be willing to pay that much to use the phone.

39. In the situation just described, Clay contemplates paying for the use of the phone, but believes that, whatever his associate says on the phone, there is a 20-percent chance that she’ll change her mind when she sees the bowl up close. Under these conditions, how much should Clay be willing to pay for the use of the phone?

value of phone usage, assuming 20-percent possibility of error of either kind:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of buying and selling regardless of what the information source says, its EMV is just the EMV of buying and selling, which we know (from problem 37) is $0.90.

EMV(A₂): Since A₂ has an outcome of $0 in every state, its EMV is obviously $0.

EMV(A₃) = (8/100)($45) + (2/100)($0) + (18/100)(–$4) + (72/100)($0)
= $3.60 + $0 – $0.72 + $0
= $2.88

EMV(A₄) = (8/100)($0) + (2/100)($45) + (18/100)($0) + (72/100)(–$4)
= $0 + $0.90 + $0 – $2.88
= –$1.98

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₃, which has an EMV of $2.88. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $0.90. That’s a difference of $1.98; you should be willing to pay that much to use the phone.

40. In the situation just described, Clay contemplates paying for the use of the phone. He believes that if his associate says she will pay $1 for the bowl, then she definitely will pay exactly that amount, but that if she says she will pay $50, that is not certain—she might, in fact, pay only $1, despite having said she’ll pay $50. In other words, if the bowl is junk, it is not certain that she’ll say so on the phone—she might say it’s valuable, and then later it will turn out to be junk. Let p be the probability that she’ll say it’s junk (on the phone), given that it is junk. How much Clay should be willing to pay for the use of the phone? (Your answer should be a mathematical expression involving p.)

value of phone usage, assuming probability p of correcting identifying junk as junk:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of buying and selling regardless of what the information source says, its EMV is just the EMV of buying and selling, which we know (from problem 37) is $0.90.

EMV(A₂): Since A₂ has an outcome of $0 in every state, its EMV is obviously $0.

EMV(A₃) = (1/10)($45) + (0)($___) + ((9 – 9p)/10)(–$4) + (9p/10)($0)
= $4.50 + $0 + (9/10)(–$4) – (9p/10)(–$4) + $0
= $4.50 – $3.60 + $36p/10
= $3.60p + $0.90

EMV(A₄) = (1/10)($0) + (0)($___) + ((9 – 9p)/10)($0) + (9p/10)(–$4)
= $0 + $0 + $0 – $36p/10
= –$3.60p

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₃, which has an EMV of $3.60p + $0.90. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $0.90. That’s a difference of $3.60p; you should be willing to pay that much to use the phone. (Notice that if p = 1, then we the situation described in problem 38, and our formula yields the same answer as the one we found there—namely, $3.60.)

due Monday, October 31 (Resnik, section 3-3c):

41. Provide a Dutch Book argument establishing Theorem 2 directly (i.e., without appeal to any other theorem or axiom).

We’ll need the following table:

		p		q
p	q	for	against	for	against
T	T	1 – a	–(1 – a)	1 – b	–(1 –b)
T	F	1 – a	–(1 – a)	–b	b
F	T	–a	a	1 – b	–(1 – b)
F	F	–a	a	–b	b

1. proof that if p and q are equivalent, then P(p) < P(q) leads to a Dutch Book:

no.	claim	justification
1	p and q are equivalent.	assumption for proving conditional
2	P(p) < P(q)	given
3	P(p) = a	definition of having a betting quotient of a
4	P(q) = b	definition of having a betting quotient of b
5	a < b	2, with substitution from 3 and 4
6	b – a > 0	5, subtract a from each side
7	Let strategy A1 be the strategy of betting for p and against q.	just defining strategy A1
8	Let the stakes on p be 1, and let the stakes on q be 1 as well.	specifying the stakes
9	Rows 1 and 4 of the foregoing table describe all the possibilities.	1 (equivalent statements have the same truth value).
10	In row 1, strategy A1 pays (1 – a) + –(1 – b) = 1 – a – 1 + b = b – a, which is positive.	reading table, with information from 6
11	In row 4, strategy A1 pays (–a) + (b) = – a + b = b – a, which is positive.	reading table, with information from 6
12	In the rows describing all the genuine possibilities, strategy A1 has a positive result.	8–10
13	A Dutch Book can be made against anyone who holds that P(p) < P(q).	12

2. The proof that if p and q are equivalent, then P(p) > P(q) leads to a Dutch Book is analogous.

42. Provide a Dutch Book argument establishing Theorem 3 directly (i.e., without appeal to any other theorem of axiom).

43. Suppose Larry believes that the probability that Microsoft will release a new version of Windows in 2006 is 0.5; that the probability that Google’s shares will double in 2006, given that Microsoft releases a new version of windows in 2006, is 0.4; and that the probability that both of these events will occur is 0.3. Larry is taking bets on these propositions, with stakes of $40 for the first proposition and $100 for each of the remaining two propositions. Answer the following two questions about this situation. First, what betting strategy can one use against Larry, so that one is certain to make money, regardless of what truth values the propositions end up having? Second, how much money is one guaranteed to make? Show your work.

44. Hoyt goes to the race track and notices that the odds on the horses are all of the form n to 1: 8 to 1, 10 to 1, 5 to 1, etc. Hoyt computes the implied probabilities of winning for the respective horses (1/9, 1/11, 1/6, etc.) and notices that they add up to a number greater than 1. Realizing that the race track is in violation of the axioms of the probability calculus, Hoyt immediately bets on every horse, thinking he has a Dutch Book against the race track. Hoyt ends up losing money at the track. Answer the following two questions about this situation. What is wrong with Hoyt’s assumption that betting on every horse is a good way to exploit violations of the calculus? Why would a smart race-track owner, who takes bets on horses but does not bet on any of them himself, make sure that he always gives odds whose implied probabilities sum to more than 1?

due Wednesday, November 2 (Resnik, section 4-1):

45. Suppose you prefer Nike to Adidas, Adidas to Converse, and Converse to Reebok. You prefer Nike to Adidas half as strongly as you prefer Adidas to Converse, which you prefer one-and-a-half times as strongly as you prefer Converse to Reebok. What is an interval scale that can be used to represent your preferences?

Here is one scale that works: 13, 10, 4, 0. Any positive linear transformation of this would work equally well.

46. What is the positive linear transformation converting the first scale mentioned near the bottom of p. 82 (i.e., 1, 2, 7, 10) to the second one (i.e., 15, 25, 75, 105)?

u’ = 10u + 5

47. Consider the scale 0, 1, 3, 7 and the scale 1, 5, 13, 29.
    1. As ordinal scales, are they equivalent?

Yes.

2. As interval scales, are they equivalent?

Yes. (Use u’ = 4u + 1.)

3. As ratio scales, are they equivalent?

No.

48. Consider the scale 1, 3, 6, 7 and the scale 3, 9, 18, 21.
    1. As ordinal scales, are they equivalent?

Yes.

2. As interval scales, are they equivalent?

Yes. (Use u’ = 3u.)

3. As ratio scales, are they equivalent?

Yes. (Use u’ = 3u.)

49. Suppose you have $2,000 to invest. One investment opportunity offers an 80-percent chance of a $200 gain, with a 20-percent chance of a $100 gain. Another investment opportunity offers a 30-percent chance of what was described to you as “hitting the big time,” and a 70-percent chance of leaving you with nothing. What must “hitting the big time” mean, in terms of a monetary gain over your initial investment of $2,000, in order for the expected monetary value of the second investment opportunity to equal that of the first?

The first investment has an EMV of (0.8)($2,200) + (0.2)($2,100) = $1,760 + $420 = $2,180.

The second investment has an EMV of (0.3)($2,000 + x) + (0.7)($0) = $600 + (0.3)(x).

In order for the second EMV to equal the first, we need
$600 + (0.3)(x) = $2,180
(0.3)(x) = $1,580
x = $1,580/0.3
x = $5,267

50. Suppose you have $2,000 to invest. One investment opportunity offers a sure gain of $200, while another will either triple your money or reduce your $2,000 to a mere $300. Let p be the probability that the second investment opportunity will triple your money. What inequality must p satisfy in order for the expected monetary value of the second investment opportunity to exceed that of the first?

The first investment has an EMV of $2,200.

The second investment has an EMV of (p)($6,000) + (1 – p)($300).

In order for the second EMV to exceed the first, we need
(p)($6,000) + (1 – p)($300) > $2,200
$6,000p + $300 – $300p > $2,200
$5,700p > $1,900
p > $1,900/$5,700
p > 19/57
p > 1/3

due Monday, November 28 (Resnik, section 4-3):

51. What chance at B does each of the following lotteries give?
    1. L(1/3, L(4/5, B, W), L(1/4, X, B))

chance at B = (1/3)(4/5) + (1 – 1/3)(1 – 1/4)
= (1/3)(4/5) + (2/3)(3/4)
= 4/5 + 6/12
= 4/15 + 1/2
= 8/30 + 15/30
= 23/30

2. L(3/4, L(1/3, X, W), L(2/5, B, W))

chance at B = (1 – 3/4)(2/5)
= (1/4)(2/5)
= 2/20
= 1/10

52. Consider the lottery L(a, (2/3, X, B), (3/4, B, W)), where 0 ≤ a ≤ 1.
    1. What inequality must a satisfy in order for the lottery just specified to yield at least a 50-percent chance at B?

chance at B = (a)(1 – 2/3) + (1 – a)(3/4)
= (a)(1/3) + 3/4 – (a)(3/4)
= (1/3)(a) – (3/4)(a) + 3/4
= (4/12)(a) – (9/12)(a) + 3/4
= (–5/12)(a) + 3/4

(–5/12)(a) + 3/4 > 1/2
–5a + 9 > 6
–5a > –3
a < 3/5

2. What are the smallest and largest chances at B that the lottery just specified can yield?

From 52a, we have chance at B = (–5/12)(a) + 3/4

So, when a = 0, chance at B = 3/4.

And when a = 1, chance at B = (–5/12)(1) + 3/4
= –5/12 + 9/12
= 4/12
= 1/3.

So the smallest possible chance at B is 1/3, and the largest is 3/4.

53. Confining yourself to the basic prizes A, B, and C, along with any lotteries containing no other basic prizes than A, B, and C,
    1. give an example of a set of preferences violating the better-prizes condition.

A P B
L(1/2, B, C) P L(1/2, A, C)

2. give an example of a set of preferences violating the better-chances condition.

A P B
L(1/3, A, B) P L(2/3, A, B)

3. give an example of a set of preferences violating the reduction-of-compound-lotteries condition.

L(1/2, L(1/2, A, B), L(1/2, A, B)) P L(1/2, A, B)

54. Using only the rationality conditions of the expected-utility theorem, prove the following statement: for any lotteries x, y, z, and w, and any number a (with 0 ≤ a ≤ 1), if xPy and zPw, then L(a, x, z) P L(a, y, w).

55. Consider the following statement: for any lotteries x and y and any numbers a and b (with 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1), if a > b and L(a, x, y) P L(b, x, y), then xPy.

What follows is a proof of this statement, with each line being justifiable using nothing more than the rationality conditions of the expected-utility theorem and the following statement, which we’ll call the indifferent-prizes condition: for any lotteries x, y, and z and any number a (with 0 ≤ a ≤ 1), xIy if and only if L(a, z, x) I L(a, z, y) and L(a, x, z) I L(a, y, z). You will notice, as you read this proof, that some of the justifications are incomplete, and some are missing altogether. Your task is to complete the incomplete justifications and to provide the missing ones, as specified in the questions that follow the proof.

no.	claim	justification
1	a > b and L(a, x, y) P L(b, x, y)	assumption for proving conditional
2	a > b	1, via conjunction elimination
3	L(a, x, y) P L(b, x, y)	1, via conjunction elimination
4	xPy or yPx or xIy	[missing justification]
5	yPx	considering second possibility
6	–a < –b	2, multiply each side by –1
7	1 – a < 1 – b	6, add 1 to each side
8	1 – b > 1 – a	7
9	L(1 – b, y, x) P L(1 – a, y, x)	[missing justification]
10	L(1 – (1 – b), x, y) P L(1 – (1 – a), x, y)	9, with each lottery rewritten
11	L(1 – 1 + b, x, y) P (1 – 1 + a, x, y)	10, simplified
12	L(b, x, y) P L(a, x, y)	11, simplified
13	It is not true that L(a, x, y) P L(b, x, y).	12, via [missing phrase]
14	Line 5 leads to a contradiction.	[missing justification]
15	Line 5 must be false.	14
16	It is not true that yPx.	15
17	xIy	considering third possibility
18	L(a, x, y) I L(a, x, x)	17, via [missing phrase]
19	L(a, x, x) I x	proved in problem 8 on p. 96 of Resnik
20	xIy	17, reiterated
21	y I L(b, y, y)	proved in problem 6 on p. 96 of Resnik
22	L(b, y, y) I L(b, x, y)	21, via [missing phrase]
23	L(a, x, y) I L(b, x, y)	18–22, via repeated applications of [missing phrase]
24	It is not true that L(a, x, y) P L(b, x, y).	[missing justification]
25	Line 17 leads to a contradiction.	[missing justification]
26	Line 17 must be false.	25
27	It is not true that xIy.	26
28	xPy	[missing justification]

1. What is the justification of line 4? [The answers to these nine questions are provided below, in bold.]
2. What is the justification of line 9?
3. What is the missing phrase in the justification of line 13?
4. What two line numbers should be cited as the justification of line 14?
5. What is the missing phrase in the justifications of lines 18 and 22? (It’s the same phrase for each one.)
6. What is the missing phrase in the justification of line 23?
7. What is the justification of line 24?
8. What two line numbers should be cited as the justification of line 25?
9. What three line numbers should be cited as the justification of line 28?

The parts that were missing in the original statement of this problem are now provided below, in bold.

56. Using only the rationality conditions of the expected-utility theorem, the assumption that B is at least as good as any other lottery, and the statement proved in problem 55, prove the following statement: there exist no numbers a and b (with 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1), with a > b, and no lottery x, such that L(a, x, B) P L(b, x, B).

57. Give the positive linear transformation that transforms a 1-to-4 scale to a 65-to-95 scale.

To find this positive linear transformation, we solve the following equations: 65 = 1a + b and 95 = 4a + b. The first yields b = 65 – a; substituting this into the second equation, we have 95 = 4a + (65 – a), or 30 = 3a, or a = 10. Substituting this into b = 65 – a, we have b = 65 – 10, or b = 55. So, our positive linear transformation is u’ = 10u + 55.

58. If we measure an agent's preferences on a von Neumann-Morgenstern utility scale, does it make sense to say that the agent prefers a given prize to a second prize twice as much as he or she prefers that second prize to some third prize?

Yes: a von Neumann-Morgenstern utility scale is an interval utility scale, and one of the essential features of an interval utility scale is that it accurately represents not only whether one lottery (such as a prize) is preferred to a second one, and whether that second one is preferred to some third one, but also the ratio of (1) the magnitude of the interval between the first one and and the second one and (2) the magnitude of the interval between the second one and and the third one.

To see that such measurements are not affected by positive linear transformations (which any interval utility scale is susceptible to), suppose that I prefer one prize to some second prize twice as much as I prefer that second prize to some third price. Then the following utility scale would work: 3, 1, 0 (since the difference between 3 and 1 is twice the difference between 1 and 0). If these utilities are transformed by some positive linear transformation, the resulting numbers would be 3a + b, 1a + b, and 0a + b. Now notice that the difference between the first two numbers is (3a + b) – (1a + b) = 3a + b – 1a – b = 2a, and the difference between the second and third numbers is (1a + b) – (0a + b) = 1a + b – 0a – b = a. So the first interval will continue to be twice as large as the second, regardless of what a and b turn out to be; and thus no positive linear transformation can interfere with interval utility scales’ consistent representations of the ratios of the magnitudes of preferences between lotteries.

due Friday, December 2:

Resnik, section 4-3a:

59. Suppose George is an expected-utility maximizer who prefers more money to less, but who is risk seeking—that is, he would rather have any lottery whose EMV is some particular amount of money rather than simply have that same amount of money for certain. Suppose also that we are representing George’s preferences with a utility function, and that we begin by assigning the utilities 10 and 19 to the prizes $200 and $350. Suppose, finally, that we want to assign a utility to the prize $300. What is the range within which the utility we assign to $300 must fall, given our suppositions? That is, what inequality of the form x < u($300) < y follows from our suppositions? (Hint: begin by ascertaining the precise utility that we must assign to the lottery L(1/3, $200, $350).)

The precise utility that we must assign to the lottery is the weighted average of the utilities of its prizes. So,
u(L(1/3, $200, $350))
= (1/3)*u($200) + (1 – 1/3)*u($350)
= (1/3)*u($200) + (2/3)*u($350)
= (1/3)*10 + (2/3)*19
= 10/3 + 38/3
= 48/3
= 16

Hopefully, this lottery will have an EMV of $300, or else finding its utility will have been useless. Does it? Well,
EMV(L(1/3, $200, $350))
= (1/3)($200) + (1 – 1/3)($350)
= (1/3)($200) + (2/3)($350)
= $200/3 + $700/3
= $900/3
= $300

This helps, because we know that George’s being risk seeking means that he prefers any lottery with an EMV of $300 to just $300 itself. Since we just assigned a utility of 16 to a lottery with an EMV of $300, then u($300) < 16.

We also know that George prefers more money to less. Since $300 > $200, it follows that u($300) > u($200); that is, u($300) > 10. So, the utility we assign to $300 must fall in the range between 10 and 16.

60. Suppose William is an expected-utility maximizer who prefers more money to less, but who is risk averse. Suppose also that we are representing William’s preferences with a utility function, and that we begin by assigning the utilities 120 and 200 to the prizes $800 and $1,000. Suppose, finally, that we want to assign a utility to the prize $850. What is the range within which the utility we assign to $850 must fall? (Hint: begin by constructing a lottery with prizes of $800 and $1,000, and whose EMV is $850, and then ascertain the precise utility that we must assign to that lottery.)

We want to find a lottery of the form L(a, $800, $1,000) that has an EMV of $800. So, we need to solve the following equation for a: a*$800 + (1 – a)*$1,000 = $850. So:
$800a + $1,000 – $1,000a = $850
–$200a = –$150
a = $150/$200
a = 3/4

So, the lottery L(3/4, $800, $1,000) is what we need to work with. Its utility is the weighted average of the utilities of its prizes:

u(L(3/4, $800, $1,000))
= (3/4)*u($800) + (1 – 3/4)*u($1,000)
= (3/4)*u($800) + (1/4)*u($1,000)
= (3/4)*120 + (1/4)*200
= 360/4 + 200/4
= 560/4
= 140

Now we know that William’s being risk averse means that he prefers $850 to any lottery with an EMV of $850. Since we just assigned a utility of 140 to a lottery with an EMV of $850, then u($850) > 140.

We also know that William prefers more money to less. Since $850 < $1,000, it follows that u($850) < u($1,000); that is, u($850) < 200. So, the utility we assign to $850 must fall in the range between 140 and 200.

Resnik, section 4-4:

61. Consider the Allais paradox as represented on pp. 103–104, and suppose Norbert is an expected-utility maximizer who prefers more money to less and prefers a to b (in situation A). Suppose also that we want to assign utilities to the prizes $0, $1,000,000, and $5,000,000, so that we can represent Norbert’s preferences with a utility function. If we start by assigning the utility of 0 to the prize $0 and the utility of 1 to the prize $1,000,000, what is the inequality of the form u($5,000,000) > x that the the utility we assign to $5,000,000 must satisfy? (Hint: remember that if Norbert is an expected-utility maximizer, as we are assuming, then Norbert values a lottery in accordance with the weighted average of the utilities of its prizes, and this enables you to represent the utility of b as a weighted averaged of the utilities assigned to the prizes $0, $1,000,000, and $5,000,000 (a three-part weighted average, unlike the two-part weighted averages we’ve been dealing with so far—but, as before, the weights add up to 1.00). Also remember that since Norbert prefers a to b, the utility of b must be less than that of a.)

Given that Norbert prefers a to b, we have
u(a) > u(b)
u($1,000,000) > u(0.1 chance at $5,000,000 and 0.89 chance at $1,000,000 and 0.01 chance at $0)
u($1,000,000) > 0.1u($5,000,000) + 0.89u($1,000,000) + 0.01u($0)
100u($1,000,000) > 10u($5,000,000) + 89u($1,000,000) + 1u($0)
100*1                  > 10u($5,000,000) + 89*1                  + 1*0
100                     > 10u($5,000,000) + 89
11                       > 10u($5,000,000)
11/10                  > u($5,000,000)
u($5,000,000) < 1.1

62. Stay with the Allais paradox, but forget about Norbert’s preference for a to b (in situation A), and focus now on Norbert’s (supposed) preference for c to d (in situation B). If (as before), we assign utilities of 0 and 1 to $0 and $1,000,000, what is the inequality of the form u($5,000,000) > y that the utility we assign to $5,000,000 must satisfy? (Hint: since we know that preferring a to b and c to d is not consistent with being an expected-utility maximizer—that’s the whole point of the Allais paradox, after all—we should expect those preferences to lead to incompatible inequalities that u($5,000,000) is supposed to satisfy. That is, preferring c to d should require u($5,000,000) to be greater than some number that preferring a to b requires u($5,000,000) to be less than. Then we will have a proof—simpler than the one in the text!—that anyone with the Allais preferences is not an expected-utility maximizer.)

Given that Norbert prefers c to d, we have
u(c) > u(d)
u(0.1 chance at $5,000,000 and 0.9 chance at $0) >
   u(0.11 chance at $1,000,000 and 0.89 chance at $0)
0.1u($5,000,000) + 0.9u($0) > 0.11u($1,000,000) + 0.89u($0)
10u($5,000,000) + 90u($0) > 11u($1,000,000) + 89u($0)
10u($5,000,000) + 90*0     > 11*1                      + 89*0
10u($5,000,000)               > 11
u($5,000,000)                  > 11/10
u($5,000,000) > 1.1