Odds ratios with probit latent regres... PreviousNext
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 Mr Mo DANG-ARNOUX posted on Tuesday, July 17, 2012 - 7:26 am
Dear Pr Muthen,

I wish to calculate the odds ratios of responses to four ordinal
3-level items, u1, u2, u3, u4, depending on the values of two covariates, x and z,
with the hypothesis that the effect is completely mediated through a latent
construct eta. eta is linked to u1-u4 via probit functions.

We have

eta BY u1*-u4* (load1-load4);
eta@1; ! scale eta residual variance

eta ON
x (bx)
z (bz);


I would like to compute an odds ratio for u1=2 vs u1=0 or 1, depending on
the covariate x, such as:

OR(x=1/x=0) =
[ P(u1=2 | x=1) / P(u1=0 or 1| x=1) ] /
[ P(u1=2 | x=0) / P(u1=0 or 1| x=0) ]

A. If the link between eta and u1-u4 was logit, do we have
OR(x=1/x=0) = load1 . bx ?

B. If the link is probit, the probabilities depend on the value of the
other covariate z. Does it make sense to express an OR(x=1/x=0) with z fixed to
some reference value, say 0 (or its mean if z is continuous)?

Are there any studies expressing results in this way ?
Odds ratios are more commonly reported (and compact) than probabilities,
but I also need probit to benefit from WLSMV estimation
(faster, and simpler to get indirect effects).

Thank you very much for any guidance,
 Bengt O. Muthen posted on Tuesday, July 17, 2012 - 9:00 am
A.

OR(x=1/x=0) = exp(load1 . b)

B.

With probit you have to actually express the probabilities that go into the OR. You can't just work with exp(coeff). But if you express probabilities, you can certainly condition on various z values.
 Mr Mo DANG-ARNOUX posted on Wednesday, July 18, 2012 - 1:06 am
Thank you very much for your fast answer, Pr Muthen!

A. Yes, I had indeed forgotten exp(.) (or ln OR), thank you for correcting my error.

B. This is actually what I had in mind: computing the probabilities in the OR, conditioned on a given value of z. If z is binary, with z=0 the "normal" reference value (e.g. "no pain"):

OR(x=1/x=0)_{z=0} =
[ P(u1=2 | x=1, z=0) / P(u1=0 or 1| x=1, z=0) ] /
[ P(u1=2 | x=0, z=0) / P(u1=0 or 1| x=0, z=0) ]

Since the resulting OR for x depends on the chosen value for z, does it still make sense to give the result in terms of one odds ratio?

Or do you mean that several odds ratios should be given for several values of z ? As OR_{"no pain"} and OR_{"pain present"}?
 Linda K. Muthen posted on Wednesday, July 18, 2012 - 11:56 am
He means that several odds ratios should be given for several values of z.
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