4
$\begingroup$

Considering the following question about the Independence of Irrelevant Alternatives assumption:

Alternatives to multinomial logistic regression

It seems as if IIA is only a problem when using a multinomial logit model, but, as the answers seems to imply, this assumption is not necessary for nested probit models and/or mixed multinomial logit.

Why is this the case? What about nested logit models? It seems as if the IIA assumption is only a problem for logit but is not necessary for probit models. Is this true?

$\endgroup$
4
$\begingroup$

A violation of the IIA assumption is basically a case of correlation between the residuals for the equations predicting each of the $k-1$ categories (excluding the baseline) of the response variable. A classic example is when you have two similar modes of transport like a blue bus and a red bus: people unmeasured characteristics that make them more likely to choose blue bus may also make them likely to choose red bus because both are similar alternatives (e.g. maybe among the unmeasured characteristics is a preference for public transportation that affects both blue and red bus choice).

Multinomial logit and independent multinomial probit assume IIA because they assume that the residuals in each equation are uncorrelated. Nested logit and probit partially relax this assumption by dividing alternatives into subsets that are similar among themselves (e.g. placing red and blue bus within a subset of public transportation) but independent from the ones in the other subsets. Multilevel multinomial logit tries to deal with IIA by modelling the unobservable component: IIA is not an issue if the correlated unobservable heterogeneity is accounted for by the model. Mixed logit deals with IIA by allowing coefficients to vary between decision-makers, reflecting individual observed differences in preferences. Lastly, alternative specific multinomial probit models the full correlation matrix of the residuals, allowing for them to be correlated and doing away with IIA completely.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.