Adv Behav Econ pdf


N O N E X P E C T E D - U T I L I T Y T H E O R Y



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113
N O N E X P E C T E D - U T I L I T Y T H E O R Y
Consider the class of prospects defined over three outcomes 
x
1

x
2

x
3
such that
x
1
s
x
2
s
x
3
. Since any such prospects can be described as a vector of probabilities
(
p
1
, 1
2
p
1
2
p
3
,
p
3
) we can also locate them, graphically, in two-dimensional
probability space. Figure 4.1a is a probability triangle that does this for the four
prospects {
s
1

r
1

s
2

r
2
} from the original Allais paradox problems. By conven-
tion, the horizontal axis measures the probability of the worst consequence ($0)
increasing from left to right; the vertical axis measures the probability of the best
consequence ($5M) increasing from bottom to top. Hence 
s
1
, which results in the
intermediate consequence of $1M for sure, is located at the bottom left corner of
the triangle; 
s
2
and 
r
2
, which each assign positive probability to only two of the
three possible consequences, are located on the triangle boundaries; while 
r
1
,
which assigns positive probability to all three consequences, lies on the interior of
the triangle. Two lines have been drawn in the triangle joining the pairs of
prospects involved in the two choices. It is easy to establish that these two lines
are parallel.
Taking into consideration ordering plus continuity, we can see that preferences
over prospects in any given triangle can be represented by a set of indifference
curves. Hence, every conventional theory implies the existence of a set of indif-
ference curves in this space though the precise form of indifference curves varies
between them.
The addition of the independence axiom of EU restricts the set of indifference
curves to being 
upward sloping
(left to right), 
linear
, and 
parallel
. One such set of
indifference curves is illustrated in figure 4.1b (preferences are increasing moving
north-west). Independence is a strong restriction that leaves only one feature of
0.1
0.01
p
3
p
1
0.89 0.9
x
3
= $5M
x
2
= $1M
x
1
= 0
r
1
r
2
s
1
s
2
Figure 
4.1a The Allais paradox problems in a probability triangle.



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