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I am sure this has been asked before (similar here but no answer). But I have not found an answer yet.

To give you a short frame: I am researching firm level data and I am ivestigating several contigenicies between continuous variables and a categorical variable indicating firm type in a binary format.

Given the function

$Y_a=\alpha+X_1(\beta_0+\beta_1X_2+\beta_3X_3)+\epsilon$

or even

$Y_ b=\alpha+X_1(\beta_0+\beta_1X_2+\beta_3X_3...+\beta_nX_n)+\epsilon$

ang given $X_1$ is the firm type dummy coded $(0;1)$, how do you interpret the simple effect $\beta_0$ on $Y$?

For $X_1=0$ , $Y=\alpha$ - so is it merely the intercept? Or is it the simple effect of $X_1$ on $Y$ for $X_1=0$?

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Rachel
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1 Answers1

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I think I found an answer myself.

First, the above fails to take account of the individual effect. The indicated formula should read the following instead:

$Y=\alpha+X_1(\beta_0+\beta_1X_2+\beta_3X_3)+\beta_4X_2+\beta_5X_3+\epsilon$

For $X_1=0$ we find $Y=\alpha+\beta_4X_2+\beta_5X_3+\epsilon$

For $X_1=1$ we find $Y=\alpha+\beta_0X_1+X_1(\beta_1X_2+\beta_3X_3)+\beta_4X_2+\beta_5X_3+\epsilon$

Thus $\beta_0$ indicates the simple effect of $X_1$ on $Y$ - that is it indicates the change in $Y$ when $X_1$ moves from $0$ to $1$.

$\beta_4$ and $\beta_5$ in turn indicate the effect of $X_2$ and $X_3$ on $Y$ for $X_1=0$ and $\beta_1$ and $\beta_3$ show the effect of $X_2$ and $X_3$ on $Y$ for $X_1=1$.

Rachel
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