Modelling the difference between two WTP measurements

[Stefan Duke]

I have a data set where each indivdiual reported two different willingness-to-pay (WTPs) variables (two very similar, but distinct products). Now I would like to model the difference between the two WTPs. E.g. who was willing to pay more, who less, etc. Clearly, this data has a lot of zeros (many respondends did not differentiate between these two prodcuts) and, unusally for WTPs positive and negative values.

I cannot think of an appropriate model. The standard would be the two-part model. As far as I understand it, the Gaussian link function could be appropriate in principle as this link functions allows for negative and positive values. However, the variable in question is not normally distributed (it is skewed to the right).

Any suggestions are appreciated!
Cheers!

[George Ford]

Don't let the nonnormality bother you, it is not required.

I assume you want coefficients on demographics?

Why not least squares?

Could stack the data (if not) and get different coefficients for each product, or estimate separately then suest.

Negative valuations are a bit weird--you'd not take the product for free?

[Carlo Lazzaro]

Stefan:
why not considering a simpler approach?

. set obs 10
Number of observations (_N) was 0, now 10.

. g id=_n

. g WTP_1=runiform()*100

. g WTP_2=runiform()*150

. g diff= WTP_2- WTP_1

. ttest diff == 0

One-sample t test
------------------------------------------------------------------------------
Variable |     Obs        Mean    Std. err.   Std. dev.   [95% conf. interval]
---------+--------------------------------------------------------------------
    diff |      10    44.97588    11.05775    34.96768     19.9615    69.99025
------------------------------------------------------------------------------
    mean = mean(diff)                                             t =   4.0674
H0: mean = 0                                     Degrees of freedom =        9

    Ha: mean < 0                 Ha: mean != 0                 Ha: mean > 0
 Pr(T < t) = 0.9986         Pr(|T| > |t|) = 0.0028          Pr(T > t) = 0.0014

.