Does triple interaction need to include all main effect variables? I have a triple interaction: AxBxD, where A and B are continuous variables and D is a dummy. My regression is Y = A + B + AxB + AxD + AxBxD
In this case, do I HAVE TO include BxD also? In theory here the effect of B will not be affected by whether D=1 or D=0.
The problem is, if I omit BxD, AxBxD is significant as expected. But if I include BxD, AxBxD becomes insignificant, and BxD is also insignificant. Can anyone give some suggestions whether I have to include BxD?
Thanks very much!
 A: It's awkward to try to give direct, literal answers to "should I?" and "do I have to?" questions on this site.  It's preferable to talk about what the consequences of a certain decision are likely to be.  
If you include a second-order, AxBxD interaction term without the first-order BxD term, you are liable to mistake a BxD effect for an AxBxD effect.  After all, how would you be able to distinguish the two?  (I'm using "effect" loosely to mean a statistical connection rather than a true effect of a cause.)   
A first-order interaction necessitates that the connection between one predictor and Y is different depending on the level of a second predictor.  Similarly, a second-order interaction necessitates that the first-order interaction pattern and associated coefficient (whether zero or non-zero) is itself different depending on the level of a third predictor.  In order to test for this latter difference, one certainly needs to know just what that first-order interaction coefficient is.
Interactions will be to some degree collinear with their component main effects, and higher-order interactions, with their component lower-order interactions.  Thus the different terms will compete for shared variance and will "interfere" with the statistical significance of the others.  The usual, and my recommended, practice is to include only those interactions that show significant and/or substantial effects, whatever your criteria might be, and to ignore determinations of non-significance for all but the highest-order interactions included in any given iteration of model-building.
A: Consider your proposed model for the expectation of the response (assuming also an intercept term):
$$\newcommand{\E}{\operatorname{E}}
\E Y = \beta_0 + \beta_1 A + \beta_2 B +  \beta_3 D + \beta_4 AB + \beta_5 AD + \beta_6 ABD
$$
When $D=0$
$$
\E Y = \beta_0 + \beta_1 A + \beta_2 B + \beta_4 AB 
$$
When $D=1$
$$
\E Y = (\beta_0 +  \beta_3) + (\beta_1 +\beta_5) A + \beta_2 B + (\beta_4 + \beta_6) AB
$$
When $D=0$ & $A=0$
$$
\E Y = \beta_0  + \beta_2 B
$$
When $D=1$ & $A=0$
$$
\E Y = (\beta_0 +  \beta_3) + \beta_2 B 
$$
So you're imposing a constraint that when $A$ is zero, the slopes for $B$ are equal whatever the value of $D$. Omitting the $BD$ term while including the $ABD$ term is not, therefore, to say "the effect of $B$ will not be affected by whether $D=$1 or $D=0$", but rather "the effect of $B$ will not be affected by whether $D=$1 or $D=0$ in the special case that $A=0$". Note that it requires the scale on which you measure $A$ to have a meaningful zero point, at least for the nonce. If that's what you want, fine—but I'd guess it isn't.
Such considerations arise whenever the Marginality Principle is violated. See Venables (1998), "Exegeses on linear models", S-Plus Users' Conference, Washington DC; also Including the interaction but not the main effects in a model, & Do all interaction terms need their individual terms in a regression model?. Valid reasons for fitting a model that violates the MP include both believing in the model (say to use it for prediction) & disbelieving in it (say to test it); but there's no sense in imposing arbitrary constraints or in testing arbitrary hypotheses.
Less fundamental, & checkable, but worth noting, is that the effect of $A$ on the change in slopes of $B$ from $D=0$ to $D=1$ is also assumed linear, so forcing it through the origin can move your regression line further away from the data when that assumption is wrong. Even when the coefficient of the $BD$ term is known to be zero, including it can allow a locally better-fit model when the sample values of $A$ are far from zero.
