I would like to know more on the relation between non linearity and interaction effect. For example, if we have a linear model of the form

$$ y = \beta_0 + \beta_1x_1 + \beta_2x_2 + \epsilon $$

we know that when we want to test for interaction between explanatory variable we could add a multiplicative term giving us the following non linear model :

$$ y = \beta_0 + \beta_1x_1 + \beta_2x_2 + \beta_3x_1x_2 +\epsilon = \beta_0 + (\beta_1 + \beta_3x_2)x_1 + \beta_2x_2 + \epsilon = \beta_0 + \tilde{\beta}_1x_1 + \beta_2x_2 + \epsilon $$

and the idea is to say that if we want to differentiate with respect to $x_1$, in order to keep the interpretation of the derivatives from a linear function, we need to fix $x_2$ in $\tilde{\beta_1}$ and that corresponds to an interaction between $x_1$ and $x_2$. Given this procedure, interaction effect implies non linearity.

Now if we take the reverse and consider a non linear model of the form

$$ y = \beta_0 + x_1^{\beta_1} + x_2^{\beta_2} + \epsilon $$

the model is clearly non linear, but there is at my knowledge no interaction between these variables. Thus it seems to me that there is no implication if we start from a non linear model concerning the interaction. Am I right ?

Thank you !

  • 1
    $\begingroup$ If $\beta_1=2$, the model could be viewed as interacting $x_1$ with itself. $\endgroup$
    – Dave
    Commented Jun 6, 2023 at 10:23
  • $\begingroup$ @Dave You make a good point since the interaction of a variable with itself is a mystery for me. I started to look at interaction with this exampl e : the effect of two explanatory variables (marketing) in which the same amount has been invested is greater than the effect of investing only in X_1 and $x_2$. $\endgroup$
    – G2MWF
    Commented Jun 6, 2023 at 13:14


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.