Models of Choice

Agenda Administrivia –Readings –Programming –Auditing –Late HW –Saturated –HW 1 Models of Choice –Thurstonian scaling –Luce choice theory –Restle choice theory Quantitative vs. qualitative tests of models. Rumelhart & Greeno (1971) Conditioning… Next assignment

Choice The same choice is not always made in the “same” situation. Main assumption: Choice alternatives have choice probabilities.

Overview of 3 Models Thurstone & Luce –Responses have an associated ‘strength’. –Choice probability results from the strengths of the choice alternatives. Restle –The factors in the probability of a choice cannot be combined into a simple strength, but must be assessed individually.

Thurstone Scaling Assumptions –The strongest of a set of alternatives will be selected. –All alternatives gives rise to a probabilistic distribution (discriminal dispersions) of strengths.

Thurstone Scaling Let x j denote the discriminal process produced by stimulus j. The probability that Object k is preferred to Stimulus j is given by –P(x k > x j ) = P(x k - x j > 0)

Thurstone Scaling Assume x j & x k are normally distributed with means  j &  k, variances  j &  k, and correlation r jk. Then the distribution of x k - x j is normal with –mean  k -  j –variance  j 2 +  k r jk  j  k =  jk 2

Thurstone Scaling

Special cases: –Case III: r = 0 If n stimuli, n means, n variances, 2n parameters. –Case V: r = 0,  j 2 =  k 2 If n stimuli, n means, n parameters.

Luce’s Choice Theory Classical strength theory explains variability in choices by assuming that response strengths oscillate. Luce assumed that response strengths are constant, but that there is variability in the process of choosing. –The probability of each response is proportional to the strength of that response.

A Problem with Thurstone Scaling Works well for 2 alternatives, not more.

Luce’s Choice Theory For Thurstone with 3 or more alternatives, it can be difficult to predict how often B will be selected over A. The probabilities of choice may depend on what other alternatives are available. Luce is based on the assumption that the relative frequency of choices of B over C should not change with the mere availability of other choices.

Luce’s Choice Axiom Mathematical probability theory cannot extend from one set of alternatives to another. For example, it might be possible for: –T1 = {ice cream, sausages} P(ice cream) > P(sausage) –T2 = {ice cream, sausages, sauerkraut} P(sausage) > P(ice cream) Need a psychological theory.

Luce’s Choice Axiom Assumption: The relative probabilities of any two alternatives would remain unchanged as other alternatives are introduced. –Menu: 20% choose beef, 30% choose chicken. –New menu with only beef & chicken: 40% choose beef, 60% choose chicken.

Luce’s Choice Axiom P T (S) is the probability of choosing any element of S given a choice from T. –P {chicken, beef, pork, veggies} (chicken, pork)

Luce’s Choice Axiom Let T be a finite subset of U such that, for every S  T, Ps is defined, Then: –(i) If P(x, y)  0, 1 for all x, y  T, then for R  S  T, P T (R) = P S (R) P T (S) –(ii) If P(x, y) = 0 for some x, y in T, then for every S  T, P T (S) = P T-{x} (S-{x})

Luce’s Choice Axiom S R T (i) If P(x, y)  0, 1 for all x, y  T, then for R  S  T, P T (R) = P S (R) P T (S)

Luce’s Choice Axiom (ii) If P(x, y) = 0 for some x, y in T, then for every S  T, P T (S) = P T-{x} (S-{x}) Why? If x is dominated by any element in T, it is dominated by all elements. Causes division problems. S T X

Luce’s Choice Theorem Theorem: There exists a positive real- valued function v on T, which is unique up to multiplication by a positive constant, such that for every S  T,

Luce’s Choice Theorem Proof: Define v(x) = kP T (x), for k > 0. Then, by the choice axiom (proof of uniqueness left to reader),

Thurstone & Luce Thurstone's Case V model becomes equivalent to the Choice Axiom if its discriminal processes are assumed to be independent double exponential random variables –This is true for 2 and 3 choice situations. –For 2 choice situations, other discriminal processes will work.

Restle A choice between 2 complex and overlapping choices depends not on their common elements, but on their differential elements. –$10 + an apple –$10 XXX X XXX P($10+A, $10) = (4 - 3)/( ) = 1

Quantitative vs. Qualitative Tests Dimensions StimulusLegsEyeHeadBody A11110 A21010 A31011 A41101 A50111 B11100 B20110 B30001 B40000

Quantitative vs. Qualitative Tests Dimensions StimulusLegsEyeHeadBody A11110 A21010 A31011 A41101 A50111 B11100 B20110 B30001 B40000 Prototype vs. Exemplar Theories

Quantitative Test P(Correct) StimulusDataPrototypeExemplar A A A A A B B B B GOF Made-up #s

Qualitative Test Dimensions StimulusLegsEyeHeadBody A11110 A21010 A31011 A41101 A50111 B11100 B20110 B30001 B40000 <- More ‘protypical’ <- Less ‘prototypcial’

Qualitative Test Dimensions StimulusLegsEyeHeadBody A11110 A21010 A31011 A41101 A50111 B11100 B20110 B30001 B40000 <- Similar to A1, A3 <- Similar to A2, B6, B7 Prototype: A1>A2 Exemplar: A2>A1

Quantitative Test P(Correct) StimulusDataPrototypeExemplar A A A A A B B B B GOF Made-up #s

Quantitative vs. Qualitative Tests You ALWAYS have to figure out how to split up your data. –Batchelder & Riefer, 1980 used E1, E2, etc instead of raw outputs. –Rumelhart & Greeno, 1971 looked at particular triples.

Caveat Qualitative tests are much more compelling and, if used properly, telling, but –qualitative tests can be viewed as specialized quantitative tests, i.e., on a subset of the data. –“qualitative” tests often rely on quantitative comparisons.