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.