[Jeff Pitblado (StataCorp)]

Give the example code

regress y lc.x##lc.w##lc.z l.y, vce(cluster ccode)
margins , dydx(l.x) at(lc.z=(-.4(0.1).4) lc.w=(-.5 0 .5))
marginsplot

For each fixed value of lc.w, the plotted marginal effects should be
linear in lc.z. Contrasting between the same two levels of lc.w within each
level of lb.z should yield the same test statistic.

[Steve Johnson]

The example code is

regress y lc.x##lc.w##lc.z l.y, vce(cluster ccode)
margins , dydx(l.x) at(lc.z=(-.4 0 .4) lc.w=(-.5 0 .5)) post
marginsplot
contrast {_at -1 0 0 1} {_at 0 -1 0 0 1} {_at 0 0 -1 0 0 1} {_at 0 0 0 -1 0 0 1} {_at 0 0 0 0 -1 0 0 1} {_at 0 0 0 0 0 -1 0 0 1} {_at -1 0 0 0 0 0 1} {_at 0 -1 0 0 0 0 0 1} {_at 0 0 -1 0 0 0 0 0 1}, pveffects

However, as you argued, since marginsplot in the example dataset indicates that the three plots are parallel, it should be not surprising that the differences between plot w= -0.5 and w=0 are similarly significant at all values of Z. The same holds true for comparing w= -0.5 and w=0.5 and w= 0 and w=0.5.

But in my dataset marginsplot shows that the 3 plots are not parallel.


However the test statistics indicate that the differences are similarly (in)significant for w= -0.5, w=0 and w= 0.5, which can be seen in the output of my last post. Although obviously the differences between the 3 plots at low levels of Z are much bigger than at high levels of Z.

[Jeff Pitblado (StataCorp)]

Sorry, I meant to take the parallel comment back, and replace it with
the argument that your results are reasonable since the marginal effects
are a linear function of lc.z when holding lc.w fixed and
vice versa.

Here is a mathematical argument to support my claim.

For a given pair of values for lc.w and lc.z, the
contrasted marginal effects look something like

C = _b[lc.x#lc.w]*dw + _b[lc.x#lc.w#lc.z]*dw*lc.z

where dw is the difference between two of your levels for
lc.w.

If we let L be

L = _b[lc.x#lc.w] + _b[lc.x#lc.w#lc.z]*lc.z

then we have

C = dw*L

The standard error of C is then

se(C) = dw*se(L)

Now since the test statistics for each individual contrast is

t = C/se(C)

we get

t = L/se(L)

which is no longer a function of dw.

[Steve Johnson]

Ok, great! This is convincing!
Once again, thanks a lot for your effort.
Best, Steve