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Some New Approaches to Old Problems: Behavioral
Models of Preference
Michael H. BirnbaumCalifornia State University,
Fullerton
Testing Algebraic Models with Error-Filled Data
• Algebraic models assume or imply formal properties such as stochastic dominance, coalescing, transitivity, gain-loss separability, etc.
• But these properties will not hold if data contain “error.”
Some Proposed Solutions
• Neo-Bayesian approach (Myung, Karabatsos, & Iverson.
• Cognitive process approach (Busemeyer)
• “Error” Theory (“Error Story”) approach (Thurstone, Luce) combined with algebraic models.
Variations of Error Models
• Thurstone, Luce: errors related to separation between subjective values. Case V: SST (scalability).
• Harless & Camerer: errors assumed to be equal for certain choices.
• Today: Allow each choice to have a different rate of error.
• Advantage: we desire error theory that is both descriptive and neutral.
Basic Assumptions
• Each choice in an experiment has a true choice probability, p, and an error rate, e.
• The error rate is estimated from (and is the “reason” given for) inconsistency of response to the same choice by same person over repetitions
One Choice, Two Repetitions
A B
A
B€
pe2
+ ( 1 − p )( 1 − e )2
p ( 1 − e ) e + ( 1 − p )( 1 − e ) e
p ( 1 − e ) e + ( 1 − p )( 1 − e ) e
€
p ( 1 − e )2
+ ( 1 − p ) e2
Solution for e
• The proportion of preference reversals between repetitions allows an estimate of e.
• Both off-diagonal entries should be equal, and are equal to:
( 1 − e ) e
Estimating eProbability of Reversals in Repeated Choice
0
0.1
0.2
0.3
0.4
0.5
0 0.1 0.2 0.3 0.4 0.5
Error Rate (e)
Estimating p
Observed = P(1 - e)(1 - e)+(1 - P)ee
0.00
0.20
0.40
0.60
0.80
1.00
0.00 0.20 0.40 0.60 0.80 1.00
True Choice Probabiity, P
Error Rate = 0
Error Rate = .02
Error Rate = .04
Error Rate = .06
Error Rate = .08
Error Rate = .10
Error Rate = .12
Error Rate = .14
Error Rate = .16
Error Rate = .18
Error Rate = .20
Error Rate = .22
Error Rate = .24
Error Rate = .26
Error Rate = .28
Error Rate = .30
Error Rate = .32
Error Rate = .34
Error Rate = .36
Error Rate = .38
Error Rate = .40
Error Rate = .42
Error Rate = .44
Error Rate = .46
Error Rate = .48
Error Rate = .50
Testing if p = 0
Test if P = 0
0
0.1
0.2
0 0.1 0.2 0.3 0.4 0.5
Probability of Reversals 2e(1 - e)
Ex: Stochastic Dominance
: 05 tickets to win $12
05 tickets to win $14
90 tickets to win $96
B: 10 tickets to win $12
05 tickets to win $90
85 tickets to win $96
122 Undergrads: 59% repeated viols (BB) 28% Preference Reversals (AB or BA) Estimates: e = 0.19; p = 0.85170 Experts: 35% repeated violations 31% Reversals Estimates: e = 0.196; p = 0.50 Chi-Squared test reject H0: p < 0.4
Testing 2, 3, 4-Choice Properties
• Extending this model to properties using 2, 3, or 4 choices is straightforward.
• Allow a different error rate on each choice.
• Allow a true probability for each choice pattern.
Response CombinationsNotation (A, B) (B, C) (C, A)
000 A B C *
001 A B A
010 A C C
011 A C A
100 B B C
101 B B A
110 B C C
111 B C A *
Weak Stochastic Transitivity
€
P ( A f B ) = P ( 000 ) + P ( 001 ) + P ( 010 ) + P ( 011 )
P ( B f C ) = P ( 000 ) + P ( 001 ) + P ( 100 ) + P ( 101 )
P ( C f A ) = P ( 000 ) + P ( 010 ) + P ( 100 ) + P ( 110 )
WST Can be Violated even when Everyone is Perfectly
Transitive
€
P ( 001 ) = P ( 010 ) = P ( 100 ) =1
3
€
P ( A f B ) = P ( B f C ) = P ( C f A ) =2
3
Model for Transitivity
€
P ( 000 ) = p000
( 1 − e1
)( 1 − e2
)( 1 − e3
) + p001
( 1 − e1
)( 1 − e2
) e3
+
+ p010
( 1 − e1
) e2
( 1 − e3
) + p011
( 1 − e1
) e2e
3+
+ p100
e1
( 1 − e2
)( 1 − e3
) + p101
e1
( 1 − e2
) e3
+
+ p110
e1e
2( 1 − e
3) + p
111e
1e
2e
3
A similar expression is written for the other seven probabilities. These can in turn be expanded to predict the probabilities of showing each pattern repeatedly.
Expand and Simplify• There are 8 X 8 data patterns in an
experiment with 2 repetitions.• However, most of these have very small
probabilities.• Examine probabilities of each of 8
repeated patterns.• Probability of showing each of 8
patterns in one replicate OR the other, but NOT both. Mutually exclusive, exhaustive partition.
New Studies of Transitivity
• Work currently under way testing transitivity under same conditions as used in tests of other decision properties.
• Participants view choices via the WWW, click button beside the gamble they would prefer to play.
Some Recipes being Tested
• Tversky’s (1969) 5 gambles.• LS: Preds of Priority Heuristic• Starmer’s recipe• Additive Difference Model• Birnbaum, Patton, & Lott (1999) recipe.• New tests: Recipes based on Schmidt
changing utility models.
Priority Heuristic
• Brandstaetter, Gigerenzer, & Hertwig (in press) model assumes people do NOT weight or integrate information.
• Each decision based on one reason only. Reasons tested one at a time in fixed order.
Choices between 2-branch gambles
• First, consider minimal gains. If the difference exceeds 1/10 the maximal gain, choose best minimal gain.
• If minimal gains not decisive, consider probability; if difference exceeds 1/10, choose best probability.
• Otherwise, choose gamble with the best highest consequence.
Priority Heuristic Preds.
A: .5 to win $100 .5 to win $0
B: $40 for sureReason: lowest consequence.
C: .02 to win $100 .98 to win $0Reason: highest consequence.
D: $4 for sure
Priority Heuristic Implies• Violations of Transitivity• Satisfies New Property: Priority
Dominance. Decision based on dimension with priority cannot be overrulled by changes on other dimensions.
• Satisfies Independence Properties: Decision cannot be altered by any dimension that is the same in both gambles.
Fit of PH to Data
• Brandstaetter, et al argue that PH fits the data of Kahneman and Tversky (1979) and Tversky and Kahneman (1992) and other data better than does CPT or TAX.
• It also fits Tversky’s (1969) violations of transitivity.
Tversky Gambles
• Some Sample Data, using Tversky’s 5 gambles, but formatted with tickets instead of pie charts.
• Data as of May 17, 2005, n = 251.• No pre-selection of participants.• Participants served in other studies,
prior to testing (~1 hr).
Three of Tversky’s (1969) Gambles
• A = ($5.00, 0.29; $0, 0.79)• C = ($4.50, 0.38; $0, 0.62)• E = ($4.00, 0.46; $0, 0.54)Priority Heurisitc Predicts:A preferred to C; C preferred to E, And E preferred to A. Intransitive.
Results-ACEpattern Rep 1 Rep 2 Both
000 10 21 5
001 11 13 9
010 14 23 1
011 7 1 0
100 16 19 4
101 4 3 1
110 176 154 133
111 13 17 3
sum 251 251 156
Test of WSTA B C D E
A 0.712 0.762 0.771 0.852
B 0.339 0.696 0.798 0.786
C 0.174 0.287 0.696 0.770
D 0.101 0.194 0.244 0.593
E 0.148 0.182 0.171 0.349
Comments• Results are surprisingly transitive.• Differences: no pre-test, selection;• Probability represented by # of tickets
(100 per urn);• Participants have practice with variety
of gambles, & choices;• Tested via Computer.
Response Patterns
Choice ( 0 = first; 1 = second)
LPH
LHP
PLH
PHL
HLP
HPL
TAX
($26,.1;$0) ($25,.1;$20) 1 1 1 1 1 1 1
($100,.1;$0) ($25,.1;$20) 1 1 1 0 0 0 1
($26,.99;$0) ($25,.99;$20) 1 1 1 1 1 1 1
($100,.99;$0) ($25,.99;$20) 1 1 1 0 0 0 0
Data Patterns (n = 260)Frequency Both Rep 1 or 2 not both Est. Prob
0000 1 2.5 0.03
0001 0 4.5 0.02
0010 0 3.5 0.01
0011 0 1 0
0100 0 8.5 0
0101 4 16 0.02
0110 6 22 0.04
0111 98 42.5 0.80
1000 1 2.5 0.01
1001 0 0 0
1010 0 1 0
1011 0 .5 0
1100 0 .5 0
1101 0 6.5 0
1110 0 5 0
1111 9 24.5 0.06
Summary• True & Error model with different error rates
seems a reasonable “null” hypothesis for testing transitivity and other properties.
• Requires data with replications so that we can use each person’s self-agreement or reversals to estimate whether response patterns are “real” or due to “error.”
• Priority Heuristic model’s predicted violations of transitivity do not occur and its prediction of priority dominance is violated.