1
$\begingroup$

I came across this equation when I was reading the article: $$\textrm{Y} = \alpha + \textrm{X}_1 (\beta_0 + \beta_1 \textrm{X}_2 + \beta_2 \textrm{X}_3) + \varepsilon$$

At least, I can understand the latter two variables in the parenthesis: both $\textrm{X}_2$ and $\textrm{X}_3$ affect $\textrm{Y}$ with $\textrm{X}_1$. However, there are several things do not make sense to me:

  1. How can I interpret $\textrm{X}_1 * \beta_0$? Is it fine to say constant affects $\textrm{Y}$ with $\textrm{X}_1$?
  2. When we use interaction term, don't we have to include the variables with no interaction? Shouldn't be the equation like this? $$\textrm{Y} = \alpha + \textrm{X}_1 (\beta_0 + \beta_1 \textrm{X}_2 + \beta_2 \textrm{X}_3) + \beta_3 + \beta_4 \textrm{X}_2 + \beta_5 \textrm{X}_3 + \varepsilon$$
$\endgroup$
4
  • 1
    $\begingroup$ Is $X_1$ a binary variable or something else? $\endgroup$
    – whuber
    Jun 17 '15 at 22:40
  • 1
    $\begingroup$ It is an interval variable. In the article, it is defined as "effective number of elective parties". $\endgroup$
    – user51966
    Jun 18 '15 at 0:28
  • 2
    $\begingroup$ Set $X_1$ to zero & the model reduces to $Y=\alpha+\varepsilon$: there is no relationship between $Y$ & $X_2$, or between $Y$ & *$X_3$ for this special value of $X_1$. The general issues with such constraints are discussed here, here, & here; you'll have to include some context to allow anyone to say whether it makes sense in this particular case. (BTW the alternative you suggest appears to have an unnecessary $\beta_3$ on its own.) $\endgroup$ Jun 18 '15 at 9:48
  • 1
    $\begingroup$ It may help to understand the meaning and purpose of the model. Since $X_1$ is some kind of count or proportion, the model seems to propose that $Y$ is the product of two quantities (plus an additive offset and error), which might make perfectly good sense as a count weighted by something else or as a proportion of something. $\endgroup$
    – whuber
    Jun 18 '15 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.