BENEFITS, COSTS, AND PERCEPTIONS OF A FUNGUS-RESISTANT BANANA 131


where Xij is a vector of attributes associated with banana bunch alterna- tive j of a choice set C and consumer i, and βs is a segment-specific vector of taste parameters. The differences in βs vectors enable this approach to cap- ture the heterogeneity in banana bunch attribute preferences across segments. Assuming that the error terms are identically and independently distributed


(IID) and follow a Type I distribution, the probability Pij/s of alternative j being chosen by the ith individual in segment s is then given by Pij/s = exp(βsXij) exp(βsXih)


^ C


. h=1 A membership likelihood function M* is introduced to classify the con-


sumer into one of the S finite number of latent segments with some probabil- ity, Pis. The membership likelihood function for consumer i and segment s is given by M*is = λsZi + ξis, where Z represents the observed characteristics of the household, λk(k = 1, 2, . . ., S) is the segment-specific parameters to be esti- mated, and ξis is the error term. Assuming that the error terms in the consumer membership likelihood function are IID across consumers and segments and


follow a Type 1 distribution, the probability that consumer i belongs to seg- ment s can be expressed as


Pis = exp(λsXi)


^ S


k=1 exp(λkXi)


The segment-specific parameters λk denote the contributions of the various consumer characteristics to the probability of segment membership, Pis. A pos- itive (negative) and significant λ implies that the associated consumer charac-


teristic, Zi, increases (decreases) the probability that the consumer i belongs to segment s. Pis sums to one across the S latent segments, where 0 ≤ Pis ≤ 1. By bringing equations 4C.2 and 4C.3 together, we can construct a mixed-


logit model that simultaneously accounts for banana bunch choice and seg- ment membership. The joint unconditional probability of individual i belonging to segment s and choosing banana bunch alternative j can be given by


Pijs = (Pij/s) × (Pis) = 3


^ C


h=1


exp(βsXij) exp(βsXih)


43 exp(λsXi)


^ S


k=1 exp(λkXi) 4 . (4C.4) . (4C.3) (4C.2)


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