Harmonic choice model

Abstract

For decades, discrete choice modelling was practically dominated by only two models: multinomial probit and logit. This paper presents a novel alternative—harmonic choice model. It is qualitatively similar to multinomial probit and logit: if one choice alternative greatly exceeds all (falls below at least one of) other alternatives in terms of utility then it is chosen with probability close to one (zero). Compared to probit and logit, the new model has relatively flat tails and it is steeper in the neighborhood of zero (when all alternatives yield the same utility and the decision maker chooses among them at random).

Appendix

Proof of Proposition 1

If the choice set contains at least three elements, then we can select three alternatives A, B, C ∊ Ω. If the independence from irrelevant alternatives (6) holds, then we must have

$$O\left(A|\Omega \right)-O\left(B|\Omega \right)=O\left(A,B\right)-O\left(B,A\right)$$

$$O\left(B|\Omega \right)-O\left(C|\Omega \right)=O\left(B,C\right)-O\left(C,B\right)$$

$$O\left(C|\Omega \right)-O\left(A|\Omega \right)=O\left(C,A\right)-O\left(A,C\right)$$

Adding these three equations together yields

$$O\left(A,B\right)+O\left(B,C\right)+O\left(C,A\right)=O\left(A,C\right)+O\left(C,B\right)+O\left(B,A\right)$$

Using the definition of choice odds (5) we can rewrite this equation as

$$\frac{P\left(B,A\right)}{P\left(A,B\right)}+\frac{P\left(C,B\right)}{P\left(B,C\right)}+\frac{P\left(A,C\right)}{P\left(C,A\right)}=\frac{P\left(C,A\right)}{P\left(A,C\right)}+\frac{P\left(B,C\right)}{P\left(C,B\right)}+\frac{P\left(A,B\right)}{P\left(B,A\right)}$$

Finally, using probabilistic completeness, we can rewrite this equation as

$$\frac{1-P\left(A,B\right)}{P\left(A,B\right)}+\frac{1-P\left(B,C\right)}{P\left(B,C\right)}+\frac{1-P\left(C,A\right)}{P\left(C,A\right)}=\frac{1-P\left(A,C\right)}{P\left(A,C\right)}+\frac{1-P\left(C,B\right)}{P\left(C,B\right)}+\frac{1-P\left(B,A\right)}{P\left(B,A\right)}$$

Simplifying and rearranging then yields (3). $\square$

Proof of Proposition 2

Consider first the case when binary choice probability function $P:\Omega \times \Omega \to \left(\mathrm{0,1}\right)$ satisfies (3) for any A,B,C ∊ Ω. Using probabilistic completeness (4), we can rewrite (3) as follows:

$$\frac{1}{P\left(A,B\right)}+\frac{1}{P\left(B,C\right)}+\frac{1}{1-P\left(A,C\right)}=\frac{1}{1-P\left(A,B\right)}+\frac{1}{1-P\left(B,C\right)}+\frac{1}{P\left(A,C\right)}$$

Rearranging this equation yields

$$\frac{1}{P\left(A,B\right)}-\frac{1}{1-P\left(A,B\right)}=\frac{1}{1-P\left(B,C\right)}-\frac{1}{P\left(B,C\right)}-\left[\frac{1}{P\left(A,C\right)}+\frac{1}{1-P\left(A,C\right)}\right]$$

The left-hand side of this equation does not depend on C. Hence, the right-hand side must also not depend on C. Let us then fix C and define a real-valued function

$$u\left(.\right)\equiv \frac{1}{1-P\left(.,C\right)}-\frac{1}{P\left(.,C\right)}$$

We obtain then

$$\frac{1}{P\left(A,B\right)}-\frac{1}{1-P\left(A,B\right)}=u\left(B\right)-u\left(A\right)$$

Rearranging yields quadratic equation

$${P}^{2}\left(A,B\right)\left[u\left(B\right)-u\left(A\right)\right]+P\left(A,B\right)\left[u\left(A\right)-u\left(B\right)-2\right]+1=0$$

If $u\left(A\right)=u\left(B\right)$ then we have an immediate solution $P\left(A,B\right)=1/2$. Otherwise, the solution to this quadratic equation is given by

$$P\left(A,B\right)=\frac{1}{2}+\frac{\sqrt{1+{\left[u\left(A\right)-u\left(B\right)\right]}^{2}/4}-1}{u\left(A\right)-u\left(B\right)}$$

Note that utility function $u\left(.\right)$ is unique up to addition of a constant. If we fix the third alternative to be C’, then this corresponds to a different real-valued utility function:

$$u^{\prime}\left(.\right)\equiv \frac{1}{1-P\left(.,C^{\prime}\right)}-\frac{1}{P\left(.,C^{\prime}\right)}=u\left(A\right)-u\left(C^{\prime}\right)$$

Reversely, if there is utility function $u:\Omega \to {\mathbb{R}}$ such that binary choice probability is given by (7) for any A,B ∊ Ω, then a relatively straightforward algebra yields

$$\frac{1}{P\left(A,B\right)}=\frac{1}{P\left(B,A\right)}+u\left(B\right)-u\left(A\right)$$

$$\frac{1}{P\left(B,C\right)}=\frac{1}{P\left(C,B\right)}+u\left(C\right)-u\left(B\right)$$

$$\frac{1}{P\left(C,A\right)}=\frac{1}{P\left(A,C\right)}+u\left(A\right)-u\left(C\right)$$

Adding these three equations together and rearranging then yields (3). $\square$

Proof of Proposition 3

Let us consider four choice alternatives A,B,C,D ∊ Ω such that P(A,B) ≥ P(C,D).

If condition (3) holds for A,B,C ∊ Ω, then we have

$$\frac{1}{P\left(A,B\right)}+\frac{1}{P\left(B,C\right)}+\frac{1}{P\left(C,A\right)}=\frac{1}{P\left(A,C\right)}+\frac{1}{P\left(C,B\right)}+\frac{1}{P\left(B,A\right)}$$

If condition (3) holds for B,C,D ∊ Ω, then we have

$$\frac{1}{P\left(D,B\right)}+\frac{1}{P\left(B,C\right)}+\frac{1}{P\left(C,D\right)}=\frac{1}{P\left(D,C\right)}+\frac{1}{P\left(C,B\right)}+\frac{1}{P\left(B,D\right)}$$

Subtracting one of these equalities from another then yields

$$\frac{1}{P\left(A,B\right)}-\frac{1}{P\left(C,D\right)}+\frac{1}{P\left(D,C\right)}-\frac{1}{P\left(B,A\right)}=\frac{1}{P\left(A,C\right)}-\frac{1}{P\left(B,D\right)}+\frac{1}{P\left(D,B\right)}-\frac{1}{P\left(C,A\right)}$$

If P(A,B) ≥ P(C,D), then $\frac{1}{P\left(A,B\right)}-\frac{1}{P\left(C,D\right)}\le 0$. Moreover, if probabilistic completeness holds, then we also have $\frac{1}{P\left(D,C\right)}-\frac{1}{P\left(B,A\right)}\le 0$. Therefore, we must have

$$\frac{1}{P\left(A,C\right)}-\frac{1}{P\left(B,D\right)}+\frac{1}{P\left(D,B\right)}-\frac{1}{P\left(C,A\right)}\le 0$$

If probabilistic completeness holds, then this inequality can be rearranged as

$$\frac{1}{P\left(A,C\right)}-\frac{1}{1-P\left(A,C\right)}\le \frac{1}{P\left(B,D\right)}-\frac{1}{1-P\left(B,D\right)}$$

Since function 1/x-1/(1-x) is strictly decreasing in x, we must then have P(A,C) ≥ P(B,D).$\square$

Proof of Proposition 4

We first prove the sufficiency part. If the independence from irrelevant alternatives (6) holds then

$$O\left(A|\Omega \right)-O\left(B|\Omega \right)=O\left(A,B\right)-O\left(B,A\right)$$

Using probabilistic completeness, this equation can be rewritten as

$$\frac{1}{P\left(A|\Omega \right)}-\frac{1}{P\left(B|\Omega \right)}=\frac{1}{P\left(A,B\right)}-\frac{1}{1-P\left(A,B\right)}$$

By proposition 1 and 2 there is utility function $u:\Omega \to {\mathbb{R}}$ such that the right-hand side of this equation is equal to $u\left(B\right)-u\left(A\right)$. Rearranging then yields

$$\frac{1}{P\left(A|\Omega \right)}+u\left(A\right)=\frac{1}{P\left(B|\Omega \right)}+u\left(B\right)$$

A similar argument implies that $\frac{1}{P\left(A|\Omega \right)}+u\left(A\right)$ is constant for any other choice alternative. Let us denote this constant by x(Ω). Then choice probability is given by $P\left(A|\Omega \right)=1/\left[x\left(\Omega \right)-u\left(A\right)\right]$. Summing over all choice alternatives A ∊ Ω and using (4) then yields Eq. (8) that implicitly defines constant x(Ω). In general, Eq. (8) has n real roots but only the highest root is such that all choice probabilities are strictly positive.

For the “necessity” part, if alternative A ∊ Ω is chosen with probability $P\left(A|\Omega \right)=1/\left[x\left(\Omega \right)-u\left(A\right)\right]$ then

$$\frac{1}{P\left(A|\Omega \right)}-\frac{1}{P\left(B|\Omega \right)}=u\left(B\right)-u\left(A\right)=\frac{1}{P\left(A,B\right)}-\frac{1}{1-P\left(A,B\right)}$$

Using probabilistic completeness, the equality between the left-most and the right-most part can be rewritten as the independence from irrelevant alternatives (6). $\square$