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Re: Delta method in random parameter models

Posted: 13 Jan 2022, 08:47
by stephanehess
Hi

could you please tell us what the cost attribute is and how that is treated in your model as that affects the WTP calculations. I can't see it in your code

Regarding a polynomial specification, so what you are estimating is beta1 * x + beta2 * x ^ 2? Specifying arndom heterogeneity in that context is possible as you did (making both beta1 and beta2 random) but then you're making an assumption of independence in the random heterogeneity between the two unless you specify a multivariate distribution. Are you actually expexting increasing marginal sensitivities with higher values of x? If not, then you could just use a Box-Cox transform instead of the polynomial

Stephane

Re: Delta method in random parameter models

Posted: 13 Jan 2022, 16:28
by aellyson
Stephane,

The cost attribute is an opportunity cost, time. It was the time in minutes required for each option. Our discrete choice setting does not have a standard monetary cost or price. Sorry if this was not clear in my original post.

I see your important point about assuming independence in the random heterogeneity between the linear and quadratic terms. Our dummy-coded regressions (where each attribute was included categorically) did show increasing marginal sensitivities with higher values of x. The difference in the relative preference weights between L1 and L2 of the time attribute was 0.89 (a 0.59 change in preference weight over this range of levels), and the difference in the relative preference weights between L2 and L3 is 0.18 (a 0.09 change in preference weight over this range of levels).

Thanks for any thoughts and/or suggestions you have!

Re: Delta method in random parameter models

Posted: 14 Jan 2022, 15:42
by stephanehess
Ah, okay.

So let's look at it independent of what the attributes are.

Let's say we have attributes x_k and x_l.

If your utility is linear in attributes, then in preference space, you'd have V=beta_k * x_k + beta_l * x_l. The marginal rate of substitution (MRS) is given as the ratio of partial derivatives of the utility, so (dV/dx_k)/(dV/dx_l), where we have that dV/dx_k=beta_k and dV/dx_l=beta_l, so MRS=beta_k/beta_l

If we work in MRS space (WTP space if the attribute is cost) in relation x_l, then we would write V=beta_l*(beta_k * x_k + x_l). You can see that the two partial derivatives are now dV/dx_k=beta_k*beta_l and dV/dx_l=beta_l, so that the MRS is given by (beta_k*beta_l)/(beta_l), i.e. MRS=beta_k, the directly estimated parameter for MRS.

When your marginal utilities are no longer linear, then the MRS calculation becomes situation dependent. So if you have:

V=beta_k1 * x_k + beta_k2 * x_k^2 + beta_l1 * x_l + beta_l2 * x_l ^ 2

then the MRS becomes (beta_k1 + beta_k2 * xk ) / (beta_l1 + beta_l2 * xl )

and your MRS thus depends on what the levels for the attributes are. You can then not just easily rewrite the utility in MRS space as the marginal utility now depends on the attribute levels. But you can of course just work in preference space as long as you can ensure that (beta_l1 + beta_l2 * xl ) is stricly one-signed

Stephane

Re: Delta method in random parameter models

Posted: 07 Feb 2023, 20:11
by Simone R
Hi Stephane,

I just want to clarify that I understand correctly from your explanation that it is not possible to estimate a mixed logit model in WTP space if one attribute is specified quadratic?

I’ve run a mixed logit model in preference space with one attribute specified as quadratic and it worked very well. A likelihood ratio test also indicated that a quadratic specification for this one attribute is a better fit for my model than a continuous linear specification. However, I’ve been having issues when I try to run it in WTP space. I came across this forum entry when I searched for an explanation.

Thank you and kind regards,
Simone

Re: Delta method in random parameter models

Posted: 01 Mar 2023, 09:50
by stephanehess
Simone

depends. Are you trying to apply the non-linear treatment to cost? Or to some other variable?

Stephane