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  • Interpreting effect of X on Y when X has a linear and quadratic term in a mixed model

    Hello,

    I am using Stata 14. I have a panel dataset - firms and years (firm is the higher level). I am running a mixed model which includes both fixed and random effects.

    The model equation looks like: Y = (b0 + b0i) + (b1 + b1i) X1 + (b2) X1sq + (b3 + b3i) X2 for firm i
    where X1sq = X1^2 ; and I have a random intercept and a random slope for X1 and one for X2. The bi's denote the random effects (BLUPS)

    b3 gives the mean effect for X2 and if I want to get a firm-specific value wrt X2 then for each firm i in year t, the value = b3 + b3i

    How to get the same for X1? For a 'non-mixed' model, the marginal effect of X1 = b1+2(b2)(X1)
    but I have a random slope for b1 and none for b2, in which case would the firm-specific value = (b1 + b1i)+2(b2)(X1) for firm i in year t ?

    Thanks.

  • #2
    How to get the same for X1? For a 'non-mixed' model, the marginal effect of X1 = b1+2(b2)(X1)
    but I have a random slope for b1 and none for b2, in which case would the firm-specific value = (b1 + b1i)+2(b2)(X1) for firm i in year t ?
    WIthin the context of your model, that would be correct. (With the proviso that this is the marginal effect of X1 at the specified value of X1--there is no such thing as "the" marginal effect of X1 in this kind of model.)

    But I would question your model. What is your basis for constraining the quadratic coefficient to be constant, but allowing the linear coefficient to vary across panels? Is there some substantive justification of that in terms of the meanings of X1 and Y and how they work in the real world? That would be quite unusual. In the standard form of a quadratic, y = ax2 + bx + c , it isn't really possible to give an substantive interpretation to a or b alone. It is true that the sign of a determines whether the parabola is concave up or down. But beyond that, the combination of a and b jointly determine the appearance of the parabola within any particular range of values of x. It is probably most natural, for understanding a quadratic relationship, to think of the vertex form of the relationship: y = A(x-h)2+k, where (h, k) are the coordinates of the vertex, and A is a scale factor.

    In this more natural form, A = a, h = -b/2a, and k = c - b2/4. Or in your notation, and A = a. A = b2, h = -(b1+b1i)/(2b2), and k = b0+b0i - (b1+b1i)2/4. So, in your model, the randomness of the linear coefficient serves to slide the parabola along a parabolic path (with some jitter from the random intercept at the firm level) while the scale remains entirely the same. That may be perfectly reasonable, even ideal, depending on what X1 and Y are. But it may also be completely arbitrary and meaningless. So give it some thought.

    Added: To be clear, what I am questioning is not your use of a mixed-effects model with a quadratic term. I'm questioning having a random coefficient for the linear term but not for the quadratic term.
    Last edited by Clyde Schechter; 25 Feb 2017, 20:13.

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    • #3
      Thanks much Clyde for your response!

      Originally posted by Clyde Schechter View Post
      In the standard form of a quadratic, y = ax2 + bx + c , it isn't really possible to give an substantive interpretation to a or b alone. I'm questioning having a random coefficient for the linear term but not for the quadratic term.
      I fully agree with you on this. This is more of a learning exercise for me in mixed effects and non-linear models using a data-set I have to run different variations of relationships that are frequently studied in my discipline. I have been building up this model and the next step was to add a random slope for the quadratic term and then see what results I get, and then move to cubic later.
      In my example, X1 is inventory and Y is sales. There is research showing a quadratic relationship between the two but under the assumption of homogeneous firms.
      To attempt to answer your question, if I find my model (after adding a random slope for the quadratic) does not converge or I can not estimate reliably for some other reason, then I would be forced to return to a only a fixed effect for the quadratic term. Other than that, there has to be theoretical backing.




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      • #4
        To attempt to answer your question, if I find my model (after adding a random slope for the quadratic) does not converge or I can not estimate reliably for some other reason, then I would be forced to return to a only a fixed effect for the quadratic term.
        Quite so, and convergence difficulties are fairly common when you try to fit these fully unconstrained models as the likelihood gets much more complicated and multiple local maxima tend to crop up.

        Anyway, we are on the same page.

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