Econ 171 Problem Set 5 Solutions Spring 2022
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Econ 171
Problem Set 5 Solutions
Spring 2022
1. Imagine a lottery that offers the following three prizes: a bicycle (b), a 1987 Chevy Malibu (c) and a Michael Moore dvd (d). For the purpose of this question assume that everyone prefers b to c and prefers c to d. Consider the following four lotteries.
W: (b, 0.2; c, 0.8)
X: (b, 0.3; c, 0.4; d, 0.3)
Y: (b, 0.6; c, 0. 1; d, 0.3)
Z: (b, 0.5; c, 0.5)
a. What preferences would lead to an Allais style paradox?
There are two (sensible) ways to get an Allais style paradox.
W X and Y Z (We’ll use this option for the remaining parts.) X W and Z Y
b. Illustrate your results to (a) in our triangle diagram.
0 P(d) 1
The indifference curve that goes through W must be flatter than the line connecting W and X. The indifference curve that goes through Y must be steeper than the line connecting Y and Z. Since the two connecting lines have the same slope, the indifference curves cannot be parallel lines.
c. Algebraically show how the preferences in (a) violate expected utility theory.
W X
EU W EU X
0.2u b 0.8u c 0.3u b 0.4u c 0.3u d
0.4u c 0. 1ub 0.3ud ineq 1
Y Z
EU Y EU Z
0.6u b 0.1u c 0.3u d 0.5u b 0.5u c
0.1u b 0.3u d 0.4u c
0.4u c 0. 1ub 0.3ud ineq 2
(ineq 1) and (ineq 2) cannot both hold.
d. Suppose 10% of people have the preferences you indicate in (a). What does this say about expected utility theory?
10% of the population has preferences that are not consistent with expected utility theory. The remaining 90% may or may not have preferences that are consistent with expected utility theory. They may have violations over different lotteries.
e. Could Kahneman & Tversky’s intuition explain the preferences in (a)? Explain.
We can rewrite (ineq 1) and (ineq 2):
0.2u b 0.8u c 0.3u b 0.4u c 0.3u d 0.8 0.4 u c 0. 1u b 0.3u d
ineq 1
0.6u b 0.1u c 0.3u d 0.5u b 0.5u c
0.1u b 0.3u d 0.5 0.1 u c
0.5 0. 1 u c 0. 1ub 0.3ud ineq 2
Yes, this could be resolved by the same intuition. The inequalities could both hold if the change caused by going from probability 0.8 to 0.4 is stronger than the change caused by going from 0.5 to 0.1.
2. In lecture we resolved the Allais paradox by using π(p) = p2 . We assumed that v(0) = 0 and v(2500) = 1. We found the paradox would be resolved if 0.193 < v(2400) < 0.942.
a. Given these three utility (ranges), is it possible that v is a concave function?
v(x)
0 2400 x
2500
v cannot be a concave function. The line connecting (0,0) and (2500,1) cannot be below v(2400).
We can also show this algebraically.
The equation for the line connecting 0, 0 and 2500,1 is: f x
To have a (potentially) concave function we need
v 2400 f 2400
0.96
b. Algebraically show that the Allais paradox can be resolved when π(p) = 3
.
Here’s a reminder about the paradox.
Problem 1: Choose between
A $2500, 0.33; $2400, 0.66; $0, 0.01
and B $2400,1 .
Problem 2: Choose between
C $2500, 0.33; $0, 0.67
and D $2400, 0.34; $0, 0.66 .
61% of participants chose both B and C.
If v(0) = 0 and v(2500) = 1 we must have:
Problem 1:
V A V B
0.33 3 v 2500 0.66 3 v 2400 0.01 3 v 0 1 3 v 2400 0.0359v 2500 0.000001v 0 0.7125v 2400
0.0359 0.7125v 2400 v 2400 0.05
Problem 2:
V C V D
0.33 3 v 2500 0.67 3 v 0 0.34 3 v 2400 0.66 3 v 0 0.0359v 2500 0.0133v 0 v 2400 0.0393
0.0359 0.0393v 2400 v 2400 0.914
The observed preferences are consistent with this p if
0.05 v 2400 0.914
c. When π(p) = p3, is it possible to resolve this paradox with a concave function v?
It’s not possible. The work is nearly identical to (a). The range for v(x) is a little lower.
3. Show that Prospect Theory doesn’t resolve the Allais paradox if we use a linear weighting function. (Try using π(p) = 0.7p.) Explain why linear weighting
functions won’t resolve the Allais paradox.
We’ll use the same Allais paradox as before.
Problem 1:
V A V B
0.7 0.33 v 2500 0.7 0.66 v 2400 0.7 0.01 v 0 0.7 1 v 2400 We can divide both sides by 0.7 and we're back to our original paradox.
0.33v 2500 0.66v 2400 0.01v 0 v 2400
0.33v 2500 0.01v 0 0.34v 2400
Problem 2:
V C V D
0.7 0.33 v 2500 0.7 0.67 v 0 0.7 0.34 v 2400 0.7 0.66 v 0 We can divide both sides by 0.7 and we're back to our original paradox.
0.33v 2500 0.67v 0 0.34v 2400 0.66v 0
0.33v 2500 0.01v 0 0.34v 2400
It's impossible for both inequalities to hold.
A linear weighting function won’t resolve the Allais paradox because we always get the effect caused by a given change in probability is the same no matter how close to 0 or 1 we are on our probability scale.
Expected utility weights
p p
p 0.7p
0.231
0 1
p
2022-06-06