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blue-bus-red-bus

  • Metadata
    • Author: Mayberry 1973
    • Tags: #statistics
  • A famous example depicting the shortcomings of N-way structure for alternatives that are correlated
  • Consider a city where 50% of travelers choose car (C) and 50% choose bus (B)
    • Assume: CC=CBC_C = C_B
    • In a stroke of genius, the manager paint half the buses red (RB) and half the buses blue (BB), while maintaining the same level of performance: CRB=CBB=CCC_{RB} = C_{BB} = C_C
    • Thus, the probability of the car using logit formulation is now: PC=exp(βCC)exp(βCC)+exp(βCRB)+exp(βCBB)P_C = \frac{\exp(-\beta C_C)}{\exp(-\beta C_C) + \exp(-\beta C_{RB}) + \exp(-\beta C_{BB})}
    • Which given the above assumptions, reduces the probability of a user choosing a car to decrease of 50% to 33%
  • Applying the nested structure to this problem gives a more reasonable result: PC=11+exp(λ1(CBCC))P_C = \frac{1}{1+exp(-\lambda_1 (C_B - C_C))} where PB=1PCP_B = 1-P_C; PRB=11+exp(λ2(CBBCRB)P_{RB} = \frac{1}{1+\exp(-\lambda_2(C_{BB}-C_{RB})}; PBB=1PRBP_{BB} = 1 - P_{RB}; CB=1λ2log(exp(λ2CRB)+exp(λ2CBB))C_B = \frac{-1}{\lambda_2}\log(\exp(-\lambda_2 C_{RB}) + \exp(-\lambda_2 C_{BB})); And λ1\lambda_1 and λ2\lambda_2 are the primary and secondary split parameters
    • If CB=CCC_B = C_C, this model will correctly assign 50% to the car and 25% to each bus options