0
$\begingroup$

Imagine a variable $y$ across $i$ sections, i.e. $y_i$. For every $y_i$ there exist a $x_{1,i}$, a $x_{2,i}$ and a $x_{3,i}$. However, the following relation exists:

$x_{1,i} *x_{2,i} = x_{3,i}$.

For example, $x_1$ is some number (e.g. number of accidents in the year under observation), $x_2$ is the average volume associated with the number (e.g. average costs resulting from the accidents in the year under observation) and $x_3$ is the resulting total volume (e.g. total cost volume resulting from accidents in the year under observation).

Now someone estimates the following:

$y_i = \alpha + \beta_1 x_{1,i} + \beta_2x_{2,i} + \beta_3x_{3,i}$.

I am aware of how to usually interpret interaction terms of two continuous variables. But in this case, the interaction term itself is a standalone variable. I am confused - can I just interpret $\beta_3$ as being a normal coefficient of some variable, i.e. $y$ changes by $\beta_3$ if $x_3$ changes by 1? Or is that not possible due to the relation between the variable and the two variables before? Since a change of $x_3$ by 1 would necessarily imply a change of the two variables before aswell.

$\endgroup$
2
$\begingroup$

When you write:

$y = \alpha + \beta_1 x_{1} + \beta_2x_{2} + \beta_3x_{3}$

with your definition of $x_{3}$, it's the same as

$y = \alpha + \beta_1 x_{1} + \beta_2x_{2} + \beta_3 x_{1}x_{2}$

which is the customary form for an interaction term between $ x_{1}$ and $x_{2}$. In that sense $x_3$ isn't really "standalone"; you've just effectively used the symbol $x_3$ to represent the interaction.

So you think about the $\beta_3$ coefficient just as you always have thought about interaction coefficients involving continuous predictors. Yes, "a change of $𝑥_3$ by 1 would necessarily imply a change of the two variables before as well," but so would a change of the standard interaction term $x_{1}x_{2}$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Exactly as mentioned above, technically, $\beta_3$ is a normal coefficient just as it would, if $x_3$ were a completely independent variable. However, you need to be careful in interpreting the significance statistics due to the potentially high levels of multi-colinearity in such a model; expect your standard errors to skyrocket vs a model without $x_3$ in. You are more likely to get meaningful coefficients and t-stats if you manage to combine most of the useful information in one or two variables.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.