In a multiple linear regression I have one predictor showing a cubic relationship with the outcome. If the same predictor $A$ is also part of an interaction with a continuous variable $B$, do I have to apply the polynomial transformation to both terms? $$ \hat{Y} = {c} + \beta_1{A} + \beta_2{A^2} + \beta_3{A^3} + \beta_4{A}{B} + \beta_5{A^2}{B} +\beta_6{A^3}{B}. $$

Logically it makes sense, but I've never seen a transformation applied to an interaction. Also I was wondering if the relationship can be simplified. I've found some information in this thread but my question is simply to understand if the transformation is correct.


1 Answer 1


I'm assuming you mean to have different coefficients for each term (they are all represented by $\beta$).

Think of regression as attempting to fit the function $f$ such that

$$ y = f(x_{1}, x_{2}, ..., x_{p}) + \varepsilon $$

The linear model for $f$ including only linear terms,

$$ f(x_{1}, x_{2}, ..., x_{p}) = \beta_{0} + \beta_{1}x_{1} + ... + \beta_{p} x_{p}, $$

is the first order Taylor approximation of $f$, where the coefficients can be expressed in terms of the derivatives of $f$. If one additionally includes the squared terms (and cross products):

$$\sum_{k} \alpha_{k} x_{k}^{2} + \sum_{k\neq j} \zeta_{kj} x_{k} x_{j} $$

then you have a second order taylor approximation of $f$, and similarly with higher order approximations.

The model you have there is reflective of an $f$ where the fourth order taylor approximation is such that all coefficients on the terms including $B^2$ are 0. This is certainly possible and only means that the higher order derivatives of $f$ involving $B$ are 0.

  • 1
    $\begingroup$ +1. I think of it slightly differently: the higher order derivatives are rolled into the error terms and considered as part of the "random" error. (They probably are not exactly $0$.) $\endgroup$
    – whuber
    Jan 18, 2012 at 3:51
  • $\begingroup$ @Macro What are alpha and zeta in the final formula? I'm trying to see if I can simplify the equation after fitting the model. Don't see how alpha/zeta match the coefficients. And by B^2 do you mean A^2? $\endgroup$ Jan 18, 2012 at 3:54
  • $\begingroup$ The $\alpha$ and $\zeta$ terms are the coefficients for the second order taylor expansion, which are formed out of a quadratic form involving the hessian matrix of $f$. I think I did mean $B^2$ there. $\endgroup$
    – Macro
    Jan 18, 2012 at 4:27

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