(due at the beginning of class unless otherwise specified)
junk bonds appreciate  junk bonds become worthless  
put under mattress  $10,000  $10,000 
put in savings account  $10,300  $10,300 
invest in junk bonds  $12,000  $0 
Yes, the savingsaccount 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.
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
The spaghetti option does not dominate the eggsalad 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 eggsalad outcome. (Technically, this question should have been written, “Why are we not entitled to say that the spaghetti option dominates the eggsalad 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’.)
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’.
take trip; $100 lastminute tickets available  take trip; only $1,000 lastminute tickets available  don’t take trip  
buy now  take trip for $300  take trip for $300  $300 lost forever 
wait until last minute  take trip for $100  take trip for $1,000  don’t take trip 
There are infinitely many correct solutions, but one is 5, 6, 6.5, 7; another is 5.32, 5.33, 5.34, 5.35.
proof that if w ≥ v, then t(w) ≥ t(v):
no.  claim  justification 
1  w ≥ v  assumption for proving conditional 
2  8w ≥ 8v  1, multiply each side by 8 
3  8w – 8 ≥ 8v – 8  2, subtract 8 from each side 
4  t(w) ≥ t(v)  3, definition of t(x) 
proof that if t(w) ≥ t(v), then w ≥ v:
no.  claim  justification 
1  t(w) ≥ t(v)  assumption for proving conditional 
2  8w – 8 ≥ 8v – 8  1, definition of t(x) 
3  8w ≥ 8v  2, add 8 to each side 
4  w ≥ v  3, divide each side by 8 
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.
proof that if w ≥ v, then t(w) ≥ t(v):
no.  claim  justification 
1  w ≥ v  assumption for proving conditional 
2  w/3 ≥ v/3  1, divide each side by 3 
3  w/3 + 4 ≥ v/3 + 4  2, add 4 to each side 
4  t(w) ≥ t(v)  3, definition of t(x) 
proof that if t(w) ≥ t(v), then w ≥ v:
no.  claim  justification 
1  t(w) ≥ t(v)  assumption for proving conditional 
2  w/3 + 4 ≥ v/3 + 4  1, definition of t(x) 
3  w/3 ≥ v/3  2, subtract 4 from each side 
4  w ≥ v  3, multiply each side by 3 
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.
counterexample: The scale {16, 15} gets transformed into {–15, –14}, and the numbers are no longer in order from largest to smallest.
You could prove ‘if A then B’ and ‘if B then A’.
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:
no.  claim  justification 
1  t(w) > t(v)  assumption for proving conditional 
2  t(w) > t(v) or t(w) = t(v)  1, ‘or’ introduction 
3  t(w) ≥ t(v)  2, definition of ≥ 
4  w ≥ v  3, condition c 
5  w > v or w = v  4, definition of ≥ 
6  w = v  5, assumption—exploring second possibility in line 5 
7  w ≥ v and v ≥ w  6, definition of ≥ 
8  t(w) ≥ t(v) and t(v) ≥ t(w)  7, condition c 
9  t(w) = t(v)  8, definition of ≥ 
10  not the case that t(w) > t(v)  9, definition of = 
11  contradiction  1 and 10 
12  line 6 is false  line 6 leads to the contradiction just mentioned 
13  w > v  only remaining possibility in line 5 
No, because what you chose from the first menu (namely, A) was not available on the second (contracted) menu.
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.
Yes, because you chose one thing (namely, A) in a pairwise choice with each of the items on the larger menu, and yet you chose something else from that menu.
No, because the item you chose from the larger menu is not something you chose in pairwise choices with all the other things on that menu.
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.
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.
S_{1}  S_{2}  S_{3}  
A_{1}  8  6  7 
A_{2}  8  11  8 
A_{3}  9  8  10 
A_{4}  10  9  6 
A_{5}  7  10  6 
The maximin rule selects A_{2} and A_{3}. The lexical maximin rule selects A_{2}.
S_{1}  S_{2}  S_{3}  
A_{1}  9  11  9 
A_{2}  11  6  9 
A_{3}  9  10  11 
A_{4}  7  7  8 
A_{5}  6  10  8 
The maximin rule selects A_{1} and A_{3}. The lexical maximin rule selects A_{1}.
S_{1}  S_{2}  S_{3}  
A_{1}  8  6  7 
A_{2}  8  11  8 
A_{3}  9  8  10 
A_{4}  10  9  6 
A_{5}  7  10  6 
The minimax regret rule selects A_{2}.
S_{1}  S_{2}  S_{3}  
A_{1}  9  11  9 
A_{2}  11  6  9 
A_{3}  9  10  11 
A_{4}  7  7  8 
A_{5}  6  10  8 
The minimax regret rule selects A_{1} and A_{3}.
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.
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 nonlinear polynomial function that, although nonlinear, nonetheless suffices to specify an ordinal transformation; another involves specifying enough of a transformation to ensure that it is nonlinear, while allowing that it is ordinal.
A. Consider the transformation t(x) = x^{3}. This is an ordinal transformation, but obviously nonlinear. (To see that it is nonlinear, 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) = x^{3} cannot be expressed as a linear transformation.)
B. Instead of coming up with an algebraic function that is nonlinear, 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:
 For all x less than 1, let t(x) = x.
 For x = 1, let t(x) = 1.
 For all x between 1 and 2, let t(x) = x.
 For x = 2, let t(x) = 8.
 For all x between 2 and 3, let t(x) = x + 15.
 For x = 3, let t(x) = 27.
 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 piecewise 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.
S_{1}  S_{2}  S_{3}  
A_{1}  8  6  7 
A_{2}  8  11  8 
A_{3}  9  8  10 
A_{4}  10  9  6 
A_{5}  7  10  6 
Compute the aindexes for the acts in this table. Assume that the agent’s optimism index is 0.6.
S_{1}  S_{2}  S_{3}  
A_{1}  13  19  13 
A_{2}  19  2  13 
A_{3}  13  17  19 
A_{4}  6  6  10 
Prove that no ordinal transformation converts this table into the one specified in problem 25.
S_{1}  S_{2}  S_{3}  
A_{1}  8  6  7 
A_{2}  8  11  8 
A_{3}  9  8  10 
A_{4}  10  9  6 
Compute the aindexes for the acts in this table. Assume that the agent’s optimism index is 0.6.
The aindex for A_{1} 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 aindex for A_{2} 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 aindex for A_{3} 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 aindex for A_{4} 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.
S_{1}  S_{2}  S_{3}  
A_{1}  10  2  6 
A_{2}  10  19  10 
A_{3}  13  10  17 
A_{4}  17  13  2 
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 A_{1}—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 A_{1}, 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 A_{1} is supposed to be. So, in an unfortunate coincidence, the transformation we found for the first two cells for A_{1} also works for the third cell for A_{1}. 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 A_{2}, 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 A_{2}. 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.
In table 225, the principle of insufficient reason assigns to act A_{1} a value of 73/9, and it assigns to act A_{2} a value of 28/9. So it selects A_{1} as the better act.
S_{1}  S_{2}  S_{3}  
A_{1}  10  4  2 
A_{2}  8  5  x 
In order for the principle of insufficient reason to be indifferent between A_{1} and A_{2}, the utilities associated with the possible outcomes of A_{1} must have the same sum as the utilities associated with the possible outcomes of A_{2}. So it has to be the case that 10 + 4 + 2 = 8 + 5 + x, or x = 3.
Yes, the principle of insufficient reason would recommend A_{2}, since the utilities associated with its outcomes have a greater sum than the utilities associated with the outcomes of A_{1}. It is not possible that A_{1} dominates A_{2}, since the utilities of A_{1} and A_{2} cannot be arranged in any way that makes those of A_{1} always greater than or equal to, and in some state greater than, those of A_{2}. But A_{2} could dominate A_{1}, since the utilities of A_{1} and A_{2} can be arranged in a way (namely, ascending order) that makes those of A_{2} always greater than or equal to, and in some state greater than, those of A_{1}. For example, just arranging the utilities in ascending order would suffice.
Here is a proof of the desired result:
no.  claim  justification 
1 
Let A_{1} and A_{2} be the acts, let u_{1,1} and u_{1,2} be the utilities associated with A_{1}, and let u_{2,1} and u_{2,2} be the utilities associated with A_{2}. 
definition of some variables 
2  u_{1,1} + u_{1,2} ≥ u_{2,1} + u_{2,2}  1, assumption that A_{1} dominates A_{2}. 
3  a(u_{1,1} + u_{1,2}) ≥ a(u_{2,1} + u_{2,2}), where a > 0  2, multiply each side by a positive number 
4  au_{1,1} + au_{1,2} ≥ au_{2,1} + au_{2,2}, where a > 0  3, simplify 
5  au_{1,1} + b + au_{1,2} + b ≥ au_{2,1} + b + au_{2,2} + b, where a > 0  4, add b to each side 
6 
The expected utility of A_{1}, when its utilities are transformed with any positive linear transformation, is greater than that of A_{2}. 
5, definition of expected utility 
7 
The expected utility of A_{1}, in D_{2}, is greater than that of A_{2}. 
6, definition of D_{2} 
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.
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
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.
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.)
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
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 90percent sure of that in the face of the countervailing evidence described in the problem.)
EMV of buying bowl and selling it to his associate:
valuable (1/10) 
junk (9/10) 

A_{1}: buy and sell to associate  –$5 + $50 = $45  –$5 + $1 = –$4 
A_{2}: just skip the whole thing  $0  $0 
EMV(A_{1}) = (1/10)($45) + (9/10)(–$4) = $4.50 – $3.60 = $0.90
value of phone usage, assuming completely reliable information:
valuable and “valuable” (1/10)*(1) = 1/10 
valuable and “junk” (1/10)*(0) = 0 
junk and “valuable” (9/10)*(0) = 0 
junk and “junk” (9/10)*(1) = 9/10 

A_{1}: buy and sell to associate  –$5 + $50 = $45  –$5 + $50 = $45  –$5 + $1 = –$4  –$5 + $1 = –$4 
A_{2}: just skip the whole thing  $0  $0  $0  $0 
A_{3}: buy and sell if “valuable” and skip it if “junk”  –$5 + $50 = $45  $0  –$5 + $1 = –$4  $0 
A_{4}: skip it if “valuable” and buy and sell if “junk”  $0  –$5 + $50 = $45  $0  –$5 + $1 = –$4 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} has an outcome of $0 in every state, its EMV is obviously $0.
EMV(A_{3}) = (1/10)($45) + (0)($___) + (0)($___) + (9/10)($0)
= $4.50 + $0 + $0 + $0
= $4.50EMV(A_{4}) = (1/10)($0) + (0)($___) + (0)($___) + (9/10)(–$4)
= $0 + $0 + $0 – $3.60
= –$3.60The informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{3}, which has an EMV of $4.50. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, 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.
value of phone usage, assuming 20percent possibility of error of either kind:
valuable and “valuable” (1/10)*(8/10) = 8/100 
valuable and “junk” (1/10)*(2/10) = 2/100 
junk and “valuable” (9/10)*(2/10) = 18/100 
junk and “junk” (9/10)*(8/10) = 72/100 

A_{1}: buy and sell to associate  –$5 + $50 = $45  –$5 + $50 = $45  –$5 + $1 = –$4  –$5 + $1 = –$4 
A_{2}: just skip the whole thing  $0  $0  $0  $0 
A_{3}: buy and sell if “valuable” and skip it if “junk”  –$5 + $50 = $45  $0  –$5 + $1 = –$4  $0 
A_{4}: skip it if “valuable” and buy and sell if “junk”  $0  –$5 + $50 = $45  $0  –$5 + $1 = –$4 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} has an outcome of $0 in every state, its EMV is obviously $0.
EMV(A_{3}) = (8/100)($45) + (2/100)($0) + (18/100)(–$4) + (72/100)($0)
= $3.60 + $0 – $0.72 + $0
= $2.88EMV(A_{4}) = (8/100)($0) + (2/100)($45) + (18/100)($0) + (72/100)(–$4)
= $0 + $0.90 + $0 – $2.88
= –$1.98The informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{3}, which has an EMV of $2.88. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, 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.
value of phone usage, assuming probability p of correcting identifying junk as junk:
valuable and “valuable” (1/10)*(1) = 1/10 
valuable and “junk” (1/10)*(0) = 0 
junk and “valuable” (9/10)*(1 – p) = (9 – 9p)/10 
junk and “junk” (9/10)*(p) = 9p/10 

A_{1}: buy and sell to associate  –$5 + $50 = $45  –$5 + $50 = $45  –$5 + $1 = –$4  –$5 + $1 = –$4 
A_{2}: just skip the whole thing  $0  $0  $0  $0 
A_{3}: buy and sell if “valuable” and skip it if “junk”  –$5 + $50 = $45  $0  –$5 + $1 = –$4  $0 
A_{4}: skip it if “valuable” and buy and sell if “junk”  $0  –$5 + $50 = $45  $0  –$5 + $1 = –$4 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} has an outcome of $0 in every state, its EMV is obviously $0.
EMV(A_{3}) = (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 + $0.90EMV(A_{4}) = (1/10)($0) + (0)($___) + ((9 – 9p)/10)($0) + (9p/10)(–$4)
= $0 + $0 + $0 – $36p/10
= –$3.60pThe informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{3}, which has an EMV of $3.60p + $0.90. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, 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.)
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
 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 A_{1} be the strategy of betting for p and against q.
just defining strategy A_{1} 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 A_{1} 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 A_{1} 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 A_{1} has a positive result.
8–10 13 A Dutch Book can be made against anyone who holds that P(p) < P(q).
12
 The proof that if p and q are equivalent, then P(p) > P(q) leads to a Dutch Book is analogous.
We’ll need the following table:
p  q  p or q  p & q  
p  q  for  against  for  against  for  against  for  against 
T  T 
1 – a 
–(1 – a) 
1 – b  –(1 –b)  1 – c  –(1 – c)  1 – d  –(1 – d) 
T  F 
1 – a 
–(1 – a) 
–b  b  1 – c  –(1 – c)  –d  d 
F  T  –a  a  1 – b  –(1 – b)  1 – c  –(1 – c)  –d  d 
F  F 
–a 
a  –b  b  –c  c  –d  d 
We want to prove that c ≠ a + b – d leads to a Dutch book.
 proof that c < a + b – d leads to a Dutch Book:
no.  claim  justification 
1  c < a + b – d  given 
2 
0 < a + b – c – d 
1, subtract c from each side 
3 
a + b – c – d > 0 
2, reversed 
4 
Let strategy A_{1} be the strategy of betting against p, against q, for p or q, and for p & q. 
just defining strategy A_{1} 
5 
Let the stakes on each proposition be 1. 
specifying the stakes 
6 
Rows 1–4 of the foregoing table describe all the possibilities. 
2 truth values for each of 2 statements, 2^{2} = 4 
7 
In row 1, strategy A_{1} pays –(1 – a) +
–(1 – b) + (1 – c) + (1 – d) 
reading table, with information from 3 
8 
In row 2, strategy A_{1} pays –(1 – a) +
b + (1 – c) + (–d) 
reading table, with information from 3 
9 
In row 3, strategy A_{1} pays a + –(1 –
b) + (1 – c) + (–d) 
reading table, with information from 3 
10 
In row 4, strategy A_{1} pays a + b +
(–c) + (–d) 
reading table, with information from 3 
11 
In the rows describing all the genuine possibilities, strategy A_{1} has a positive result. 
6–10 
12 
A Dutch Book can be made against anyone who holds that c < a + b – d. 
11 
 proof that c > a + b – d leads to a Dutch Book:
no.  claim  justification 
1  c > a + b – d  given 
2 
d + c – b – a > 0 
1, add d to each side and subtract a and b from each side 
3 
Let strategy A_{1} be the strategy of betting for p, for q, against p or q, and against p & q. 
just defining strategy A_{1} 
4 
Let the stakes on each proposition be 1. 
specifying the stakes 
5 
Rows 1–4 of the foregoing table describe all the possibilities. 
2 truth values for each of 2 statements, 2^{2} = 4 
6 
In row 1, strategy A_{1} pays (1 – a) +
(1 – b) + –(1 – c) + –(1 – d) 
reading table, with information from 2 
7 
In row 2, strategy A_{1} pays (1 – a) +
(–b) + –(1 – c) + d 
reading table, with information from 2 
8 
In row 3, strategy A_{1} pays (–a) + (1
– b) + –(1 – c) + d 
reading table, with information from 6 
9 
In row 4, strategy A_{1} pays (–a) +
(–b) + c + d 
reading table, with information from 6 
10 
In the rows describing all the genuine possibilities, strategy A_{1} has a positive result. 
5–9 
11 
A Dutch Book can be made against anyone who holds that c > a + b – d. 
10 
The situation can be represented as follows:
new OS doubling given new OS both new OS and doubling p q for against for against for against T T $20
–$20
$60 –$60 $70 –$70 T F $20
–$20
–$40 $40 –$30 $30 F T –$20 $20 0 0 –$30 $30 F F –$20
$20 0 0 –$30 $30 Then, one can use the following betting strategy to make money off of Larry: bet for the new OS, for the doubling given the new OS, and against the conjunction of the new OS and the doubling. This has the following payoff for rows 1–4:
 $20 + $60 + –$70 = $80 – $70 = $10
 $20 + –$40 + $30 = $50 – $40 = $10
 –$20 + $0 + $30 = $30 – $20 = $10
 –$20 + $0 + $30 = $30 – $20 = $10
So one is guaranteed to make $10 off of Larry with this betting strategy.
What is wrong with Hoyt’s assumption is that not just any set of bets against someone in violation of the calculus will end up constituting a Dutch Book. The wouldbe Dutch Bookie (in this case, Hoyt) must take care to decide which side of each possible bet he wants to be on, as well as what the stakes of the various bets are going to be. Hoyt appears to have ignored this first set of decisions, by betting for every horse rather than using some other set of bets, such as betting against every horse or betting for some horses and against some horses. (Hoyt would have had a Dutch book against the track if had simply bet against every horse, assuming this had been allowed at the track.)
A smart racetrack owner, who takes bets on horses but does not bet on any of them himself, makes sure that he always gives odds whose implied probabilities sum to more than 1 so that if his customers bet on all the horses equally, they are guaranteed to lose money to him—collectively, if not individually. For example, suppose there are four horses, and the odds on each one are 2 to 1. Then the owner is saying that he believes each of the four horses has a 1/(2 + 1) = 1/3 probability of winning. These probabilities add up to 4/3, which is obviously more than 1. But to see how this is profitable rather than a mistake that is liable to cost the owner money, suppose a single customer bets on all the horses—and assume, for simplicity, that tickets cost $1 each. Then the customer buys four tickets, and spends $4. When one of the hoses wins, the customer wins the $3 associated with the winning horse. But the customer’s net winnings, then, are –$4 + $3 = –$1. So the customer loses $1. From the owner’s point of view, it does not matter whether these four tickets are bought by one customer or many; the point is that the owner is guaranteed to make money off of every “complete set” of tickets sold. Of course they typically will not be sold in sets; they will be sold one by one. But they may add up to many “complete sets,” with some leftover tickets. Those leftover tickets may cost the owner money in some races, and add to his profit in others. In the long run, they will have little effect, leaving profits that are roughly proportional to the number of “complete sets” of tickets sold.
Here is one scale that works: 13, 10, 4, 0. Any positive linear transformation of this would work equally well.
u’ = 10u + 5
Yes.
 As interval scales, are they equivalent?
Yes. (Use u’ = 4u + 1.)
 As ratio scales, are they equivalent?
No.
Yes.
 As interval scales, are they equivalent?
Yes. (Use u’ = 3u.)
 As ratio scales, are they equivalent?
Yes. (Use u’ = 3u.)
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
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
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
 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
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
 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.
A P B
L(1/2, B, C) P L(1/2, A, C)
 give an example of a set of preferences violating the betterchances condition.
A P B
L(1/3, A, B) P L(2/3, A, B)
 give an example of a set of preferences violating the reductionofcompoundlotteries condition.
L(1/2, L(1/2, A, B), L(1/2, A, B)) P L(1/2, A, B)
no.  claim  justification 
1  xPy and zPw  assumption for proving conditional 
2  xPy  1, via conjunction elimination 
3  zPw  1, via conjunction elimination 
4  L(a, x, z) P L(a, y, z)  2, via betterprizes condition 
5 
L(a, y, z) P L(a, y, w) 
3, via betterprizes condition 
6 
L(a, x, z) P L(a, y, w) 
4 and 5, via ordering condition, part O5 
What follows is a proof of this statement, with each line being justifiable using nothing more than the rationality conditions of the expectedutility theorem and the following statement, which we’ll call the indifferentprizes 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]
 What is the justification of line 4? [The answers to these nine questions are provided below, in bold.]
 What is the justification of line 9?
 What is the missing phrase in the justification of line 13?
 What two line numbers should be cited as the justification of line 14?
 What is the missing phrase in the justifications of lines 18 and 22? (It’s the same phrase for each one.)
 What is the missing phrase in the justification of line 23?
 What is the justification of line 24?
 What two line numbers should be cited as the justification of line 25?
 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.
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  ordering condition, part O4 
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) 
5 and 8, via betterchances condition 
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 ordering condition, part O1 
14 
Line 5 leads to a contradiction. 
3 and 13 
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 indifferentprizes condition 
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 indifferentprizes condition 
23 
L(a, x, y) I L(b, x, y) 
18–22, via repeated applications of ordering condition, part O8 
24 
It is not true that L(a, x, y) P L(b, x, y). 
23, via ordering condition, part O3 
25 
Line 17 leads to a contradiction. 
3 and 24 
26 
Line 17 must be false. 
25 
27 
It is not true that xIy. 
26 
28 
xPy 
4, 16, and 27 
no.  claim  justification 
1  Suppose there were numbers a and b (with 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1), with a > b, and a lottery x, such that L(a, x, B) P L(b, x, B).  assumption for indirect proof 
2  x P B  1, via statement proved in problem 55 
3  It is impossible that x P B.  It is assumed that B is at least as good as any other lottery. 
4 
Line 1 leads to an impossibility. 
3 
5 
It is not true that there exist numbers a and b (with 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1), with a > b, and a lottery x, such that L(a, x, B) P L(b, x, B). 
4 
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.
Yes: a von NeumannMorgenstern 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.
Resnik, section 43a:
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
= 16Hopefully, 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
= $300This 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.
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 = 3/4So, 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
= 140Now 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 44:
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
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