# Linear Price - Zero center on Average of Tested prices

Working on a study where the price shown is calculated by adding each feature level price and applying a random shock (-50% to +50%). With this setup, we'd have many unique prices across all the respondents.We are considering to solve price as a linear attribute. We understand that HB would zero-center the prices in estimating the utilities and that we need to use zero-centered price value to when simulating a specific price point in an excel simulator.

While using the average of the tested prices to zero-center makes sense, we are wondering or trying to understand the affect of using 'average of tested prices' to zero-center when we want to simulate a non-tested price (i.e., an interpolated price). For example, if the prices are \$2, \$4, \$6 & \$8, the zero-centered values would be -3,-1,1,3. Now,if we want to simulate a price of \$3, we'd use a zero-centered value of -1 to multiply the price utility. But,if we assume that \$3 was also a price originally tested (shown to repsondents), then the zero-centered value would be -1.6 for \$3.Thoughts please?
asked Oct 25, 2018

## 1 Answer

0 votes
\$3 is indeed half-way between \$2 and \$4, which on your zero-centered scale (if you are zero-centering based on the increments of \$2, \$4, \$6, and \$8) is now half-way between -3 and -1, or -2.  So, you would use -2 to multiply by the coefficient to compute the part-worth utility for \$3.

But, if prior to running the utility estimation you scanned all the prices shown to respondents, including the \$3, to create your zero-centering transform, then that array would have been the prices (\$2, \$3, \$4, \$6, and \$8).  The zero-centered prices would have been coded for utility estimation as -2.6, -1.6, -0.6, 1.4, and 3.4.  If you had coded your design matrix for utility estimation using that transform, then now to calculate the price at \$3 you should multiply -1.6 times the coefficient.

Just be consistent with whatever transformation you used during utility estimation.
answered Oct 25, 2018 by Platinum (186,465 points)
Thank you, Bryan for your reply! Sorry about the typo - I meant to say -2  for \$3, when \$3 is not part of the tested prices.
I guess I didn't put out my question properly in my earlier posting - so, the inclusion or exclusion of prices with in the price range changes the factor (in this example: -2 vs -1.6) that we need to multiply  the coefficient with to get to the utility.  How much does this difference in the factor affect the part-worth utility estimate for \$3? I think, to start with, we can expect difference in the coefficient estimates because we are including an additional price point (\$3) - but what role does the factor play in the final part-worth utility? In other words, is using zero-centering based on scanned values affecting results?
The transformation you choose for your continuous variable doesn't matter at all if conducting aggregate logit.  But, for HB estimation (due to default priors in our HB settings for our software), transforming your continuous independent variables to be zero-centered and with a range that doesn't exceed about 10 in total magnitude (or on the other hand doesn't have a range smaller than about 0.2 in magnitude from lowest to highest value) will lead to faster and better convergence.