# Power transformation on interactions in linear regression

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.

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. 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 $0$.) – whuber Jan 18 '12 at 3:51
• @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? – Robert Kubrick Jan 18 '12 at 3:54
• 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. – Macro Jan 18 '12 at 4:27