Problem Set #9
Suggested Answers to Problems 8.3, 8.5, 8.7
8.3 An individual purchases a dozen eggs and must take them home. Although making
trips home is costless, there is a 50% chance that all of the eggs carried on any one trip
will be broken during the trip. The individual considers two strategies:
Strategy 1: Take all 12 eggs in one trip.
Strategy 2: Take two trips with 6 in each trip
a. List the possible outcomes of each strategy and the probabilities of these outcomes.
Show that on the average, 6 eggs will remain unbroken after the trip home under either
strategy.
Outcomes
Probability
Strategy 1:
12 eggs (unbroken) .5
0 eggs
.5
Strategy 2:
12 eggs
6 eggs
0 eggs
.25
.50
.25
Expected values
Strategy 1:
12(.5) +
0(.5)
Strategy 2:
12(.25) + 6(.5) + 0(.25)
=
=
6
6
b. Develop a graph to show the utility obtainable under each strategy. Which strategy
will be preferable?
U(X)
3
E(U), two trips
E(U), all or none
3
6
9
12
Strategy 1: (.5(12-6)2 + .5(0-6) 2).5 = 6
Strategy 2: (.25(12-6)2+.25(0-6) 2).5 = 32 = 4.24
Observe, of course, that strategy
2 has three possible outcomes.
We might illustrate the gain in
utility by considering the
Utilities of the expected value of
each strategy, as shown to the
lift. Alternatively, we can
calculate a measure of
variability, and show that
variability unambiguously falls
with the second strategy. Fro
example, calculate standard
deviations.
c. Could utility be improved further by taking more than two trips? How would this
possibility be affected if additional trips were costly?
Yes. Taking further trips would decrease further the variability of outcome. However,
observe that extra trip increases result in reduced utility increments. For example,
consider taking 4 trips. Then
Strategy 3:
12 eggs
9 eggs
6 eggs
3 eggs
0 eggs
1/16
4/16
6/16
4/16
1/16
Expected value
Strategy 3:
12(1/16) + 9(4/16) + 6(6/16)+ 4(4/16)+ 0(1/16)
=
6
Standard Deviation:
Strategy 3: ((1/16)(12-6)2 + (4/16(9-6) 2+ (4/16(3-6) 2 + (1/16(0-6) 2 ).5 = (144/16).5
=12/4=3. Notice, doubling the number of trips reduced the standard deviation by 1.24
=(from 4.24 to 3). Reducing risk through extra trips yields diminishing returns.
8.5 Ms. Fogg is planning an around-the-world trip on which she plans to spend $10,000.
The utility from the trip is a function of how much she actually spends on it (Y), given by
U(Y) =
ln(Y)
a. If there is a 25 percent probability that Ms. Fogg will lose $1000 of her cash on the
trip, what is the trip’s expected utility?
E[U(Y)]
=
.25 ln(9,000) +.75 ln(10,000) = 9.184
b. Suppose that Ms. Fogg can buy insurance against losing the $1000 (say, by purchasing
traveler’s checks) at an “actuarially fair” premium of $250. Show that her expected
utility is higher if she purchases this insurance than if she faces the chance of losing the
$1000 without insurance.
With Insurance U(9,750) =
uninsured (9.184)
9.185. Note that this exceeds the Expected utility of going
c. What is the maximum amount that Ms. Fogg would be willing to pay to insure her
$1000?
Solve ln(X) = 9.184, or . e9.184 = 9,740. She will pay at maximum $259.96 ($260).
8. A farmer believes there is a 50-50 chance that the next growing season will be
abnormally rainy. His expected utility function has the form
expected utility = (1/2) ln YNR + (1/2) ln YR
where YNR and YR represent the farmer’s income in the states of “normal rain” and
“rainy,” respectively.
a. Suppose the farmer must choose between two crops that promise the following income
prospects:
Crop
Wheat
Corn
YNR
$28,000
$19,000
YR
$10,000
$15,000
Which of the crops will he plant?
expected utility (Wheat)
expected utility (Corn)
= (1/2) ln (28,000) + (1/2) ln (10,000)
= (1/2) ln (19,000) + (1/2) ln (15,000)
= 9.725
= 9.734
He receives higher expected utility from Corn.
b. Suppose the farmer can plant half his field with each crop. Would he choose to do so?
Explain your result.
With a mix, the farmer will get 14,000 + 9,500 = 23,500 in a normal year, and 5,000 _
7,500 = 12,500 in a wet year. Thus,
Expected Utility (Mix) = (1/2) ln (23,500) + (1/2) ln (12,500)
=
9.74912
He would choose to diversify. Diversification allows him to enjoy some of the upside of
Wheat in a dry year without bearing all the costs of Wheat if the year is rainy.
c. What mix of what and corn would provide maximum expected utility to this farmer?
Let  denote the percentage of the acreage planted in wheat (with the remainder planted
in corn). Then
Max
E(U) =(1/2) ln (28,000 +(1-)19,000) + (1/2) ln (10,000 +(1-)15,000)

Max E(U)

=(1/2) ln (19,000+ 9,000) + (1/2) ln (15,000 -5,000)
Taking the derivative,
.5(9000)/(19,000+ 9000) -.5(5000)/(15,000- 5000) = 0
Solving: 9(15-5)
3-
2
=
5(19 + 9)
=
19/9 + 
=
8/9

=
4/9
44 Wheat and 56% Corn
d. Would wheat crop insurance, available to farmers who grow only wheat, which costs
$4000 and pays off $8000 in the event of a rainy growing season, cause this farmer to
change what he plants?
Let’s assume that the farmer plants the optimal mix (it’s okay if you selected another mix,
but I’ll use that for reference here) Under the optimal mix
Crop
Wheat
Corn
Optimal mix
YNR
.44 ($28,000)
.56($19,000)
$22,960
YR
.44($10,000)
.56($15,000)
$12,800
Eopt mix(U)
=(1/2) ln (22,960) + (1/2) ln (12,800) = 9.74935
Now, with the insurance, and planting wheat
expected utility (Wheat w/ ins) = (1/2) ln (24,000) + (1/2) ln (14,000)
= 9.81
9.81>9.75 Yes, the insurance price is low enough to make specializing in wheat optimal.