Solutions to Chapter 19 Exercises

Solutions to Chapter 9 Exercises
SOLVED EXERCISES
S1. (A) YOUR NEIGHBOR HAS A SURE INCOME OF $100,000. IN ADDITION, UNDER
THE INSURANCE CONTRACT, HE WILL RECEIVE X WHEN YOU HAVE A GOOD YEAR AND PAY
YOU $60,000 WHEN YOU HAVE A BAD YEAR. THE LOWEST VALUE OF X SUCH THAT YOUR
NEIGHBOR PREFERS TO ENTER THE CONTRACT WILL BE THE X FOR WHICH HIS EXPECTED
UTILITY FOR ENTERING THE CONTRACT IS EQUAL TO HIS UTILITY FOR NOT ENTERING
THE CONTRACT:

0.6 * ?(100,000 + x) + 0.4 * ?40,000 = ?100,000

( ?(100,000 + x) = (?100,000 - 0.4 * ?40,000) / 0.6

( x = [(?100,000 - 0.4 * ?40,000) / 0.6]2 - 100,000 ? 55,009.8818

Rounding down to $55,009.88 would make your neighbor very slightly
prefer not entering the contract, so the minimum x that your neighbor will
agree to is $55,009.89.

(b) Here we are looking for the level of x where you are
indifferent between getting insurance (where you pay x in a good year and
receive 60,000 in a bad year) and not getting insurance. That is, we're
looking for the x for which your expected utility with the insurance is
equal to your expected utility without the insurance:

0.6 * ?(160,000 - x) + 0.4 * ?100,000 = 0.6 * ?160,000 + 0.4 *
?40,000

( 0.6 * ?(160,000 - x) + 0.4 * ?100,000 = 320

( ?(160,000 - x) = (320 - 0.4 * ?100,000) / 0.6

( x = 160,000 - [(320 - 0.4 * ?100,000) / 0.6]2 ? 55,984.1891

Rounding up and paying $55,984.19 would make you very slightly prefer
not having insurance. The highest x you would be willing to pay in a good
year and still (barely) prefer to have insurance is $55,984.18.



S2. (a) To achieve separation, t must be such that:

t2 / 160 > 10 and t2 / 320 < 10

( t2 > 1,600 and t2 < 3,200

( t > 40 and t < 56.57

The minimum wait time that achives separation is the smallest moment
longer than forty minutes.

(b) The expected benefit has changed for the college students, so t
must now satisfy:

t2 / 160 > 0.5*(10) + 0.5*(-5) = 2.5 and t2 / 320 < 10

( t2 > 400 and t2 < 3200

( t > 20 and t < 56.57

The minimum wait time that achives separation is now a shade more
than twenty minutes. The partial identification of college students reduces
the minimum wait time required to achieve separation. When the charity has
more information about the patrons-even if only partial information-this
allows it to distinguish between the two types by means of a less stringent
test.



S3. (a) Buyers expect a random Citrus to be an orange with probability
f = 0.6 and a lemon with probability 0.4. Risk-neutral buyers are then
willing to pay up to:

0.6 * $18,000 + 0.4 * $8,000 = $14,000

(b) The willingness to accept (WTA) of owners of oranges is
$12,500. Since the willingness to pay (WTP) of buyers is $14,000 for a
random Citrus, sellers will be willing to sell and buyers will be willing
to buy for any price in the range [$12,500, $14,500]. Thus, there will
indeed be a market for oranges.

(c) If f = 0.2, risk-neutral buyers will be willing to pay up to:

0.2 * $18,000 + 0.8 * $8,000 = $10,000

(d) There will not be a market for oranges when f = 0.2. The
probability of a random Citrus being an orange is so low that a buyer would
only be willing to pay up to $10,000, but owners of oranges require at
least $12,500. No oranges will be sold.

(e) The minimum f such that oranges are sold is the f that
satisfies:

f threshhold * $18,000 + (1 - f threshhold) * $8,000 = $12,500


( $10,000 * f threshhold = $4,500

< f threshhold = 0.45

(f) One way to look at it is to redo part (e) in the abstract:

f threshhold * WTPorange + (1 - f threshhold) * WTPlemon =
WTAorange

( (WTPorange - WTPlemon) * f threshhold = WTAorange - WTPlemon

< f threshhold = (WTAorange - WTPlemon) / (WTPorange - WTPlemon)

Increasing WTPorange increases the size of the denominator of the
fraction, so f threshhold decreases. Increasing WTPlemon increases the size
of the both the numerator and the denominator of the fraction by the same
amount, which also causes f threshhold to decrease (assuming that WTPorange
> WTAorange).

In an intuitive sense, the more highly buyers value oranges or lemons
(or both), the more willing they will be to take a greater gamble on a
random Citrus. That is, they'll be willing to pay a higher price for a
random Citrus for each possible value of f (compared to before the increase
in demand). A buyer's willingess to pay for a random Citrus will thus be
sufficiently high for owners of oranges to accept at a lower threshhold
value of f than before.



S4. A competent electrician's payoff after obtaining the signal
(certification) is ?100 - C; in the absence of the signal, this
electrician's payoff is ?25. The competent electrician will signal as long
as ?100 - C ( ?25, or as long as 10 - C ( 5, or as long as 5 ( C. The
incompetent electrician earns only ?100 - 2C after certification versus ?25
without certification. He signals only if ?100 - 2C ? ?25, or as long as 10
- 2C ? 5, or as long as 5/2 ? C. The range of values of C for which the
competent electrician signals but the incompetent electrician does not is 5
( C ( 2.5. If the program is restricted to an integer number of months, 5 (
C ( 3.



S5. (a) The game table in terms of z is:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|166z + 3 , - 34z + 5 |175z + 25 , 0 |
| |Honest |141z + 28 , - 34z + 5 |172z + 28 , - 5z + 5 |
| |(LH) | | |


(b) When z = 0 (that is, when Tudor is definitely high cost), the
game table, with best responses underlined, is:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|3 , 5 |25 , 0 |
| |Honest |28 , 5 |28 , 5 |
| |(LH) | | |


There are thus two pure-strategy Nash equilibria: (Honest,
Regardless) and (Honest, Conditional). Tudor will always signal that it is
high cost (which it is), and Fordor will always enter.

(c) When z = 1 (that is, when Tudor is definitely low cost), the
game table, with best responses underlined, is:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|169 , - 29 |200 , 0 |
| |Honest |169 , - 29 |200 , 0 |
| |(LH) | | |


There are thus two pure-strategy Nash equilibria: (Bluff,
Conditional) and (Honest, Conditional). Tudor will always signal that it is
low cost (which it is), and Fordor will never enter.
(d) When z is strictly between 0 and 1, the best responses are:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|166z + 3 , - 34z + 5 |175z + 25 , 0 |
| |Honest |141z + 28 , -34z + 5 |172z + 28 , - 5z + 5 |
| |(LH) | | |


If Fordor plays Regardless, Tudor's best response is Honest, because
141z + 28 > 166z + 3 ( 25 > 25z for all z such that 0 < z < 1.

If Fordor plays Conditional, Tudor's best response is Honest, because
172z + 28 > 175z + 25 ( 3 > 3z for all z such that 0 < z < 1.

If Tudor plays Bluff, Fordor's best response is Regardless if z (
5/34 and Conditional if z ( 5/34, - 34z + 5 ( 0 when z ( 5/34, and - 34z +
5 ( 0 when z ( 5/34.

If Tudor plays Honest, Fordor's best response is Conditional, because
- 34z + 5 < - 5z + 5 for all z such that 0 < z < 1.

When z is strictly between 0 and 1 there is thus a unique pure-
strategy Nash equilibrium: (Honest, Conditional). Since a low-cost Tudor
will always send a different signal from that of a high-cost Tudor, this is
a separating equilibrium.

S6. (a) The extensive form with a risk-averse Tudor (with square-root
utility) is:



[pic]

(b) The equivalent game table for the tree in part (a), with z =
0.4, is:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|13 ( 0.4 + (3 ( 0.6 = 6.2 ,|10(2 ( 0.4 + 5 ( 0.6 = 8.7 |
| | | |, |
| | |- 29 ( 0.4 +5 ( 0.6 = - 8.6|0 |
| |Honest |13 ( 0.4 + 2(7 ( 0.6 = 8.4 |10(2 ( 0.4 + 2(7 ( 0.6 = |
| |(LH) |, |8.8 , |
| | |- 29 ( 0.4 + 5 ( 0.6 = - |5 ( 0.6 = 3 |
| | |8.6 | |


The unique pure-strategy Nash equilibrium of the game is (Honest,
Conditional). Since the low-cost Tudor and the high-cost Tudor send
different signals in equilibrium, this game has a separating equilibrium.

(c) The equivalent game table for the tree in part (a), with z =
0.1, is:

| | |Fordor |
| | |Regardless (II) |Conditional (OI) |
|Tudor|Bluff (LL)|13 ( 0.1 + (3 ( 0.9 = 2.9 ,|10(2 ( 0.1 + 5 ( 0.9 = 5.9 |
| | | |, |
| | |- 29 ( 0.1 +5 ( 0.9 = 1.6 |0