0
$\begingroup$

Suppose I have a multiple regression model:

$y = \beta_0 + \beta_1x_1 + \beta_2x_2 + \beta_3x_1x_2 + \epsilon$

where

$y$ is continuous

$x_1$ is dichotomous (0 or 1)

$x_2$ is continuous

If $x_1 = 0$, then this model becomes:

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

which has the format y = intercept + slope*x_2 + epsilon

and if $x_1 = 1$, then this model becomes:

$y = \beta_0 + \beta_1(1) + \beta_2x_2 + \beta_3(1)x_2 + \epsilon$

$y = \beta_0 + \beta_1 + \beta_2x_2 + \beta_3x_2 + \epsilon$

$y = (\beta_0 + \beta_1) + (\beta_2 + \beta_3)x_2 + \epsilon$

which has the format y = intercept + slope*x_2 + epsilon, just like when $x_1 = 0$.

So the format of the final fitted regression equation will be the same regardless of the value of the dichotomous variable, and the result is that you will never see that an interaction term was included in the model.

Putting numbers to this, suppose when $x_1 = 0$, the fitted regression equation is:

$y = 10 + 20x_2 + \epsilon$

and when $x_1 = 1$, the fitted regression equation is:

$y = 11 + 25x_2 + \epsilon$

My question is: How would we interpret the coefficients of the fitted model in both cases? Would there be some hidden explanation of the interaction term involved?

Here's my best guess:

First equation:

Interpreting the fitted intercept: When $x_2 = 0$, the estimated value of the response variable is 11 on average.

Interpreting the fitted slope: When $x_1 = 0$ (i.e., for the first level of the dichotomous variable), a unit increase in the continuous variable $x_2$ results in a 20 unit increase in the response variable on average.

Second equation:

Interpreting the fitted intercept: When $x_2 = 0$, the estimated value of the response variable is 15 on average.

Interpreting the fitted slope: When $x_1 = 1$, a unit increase in the continuous variable $x_2$ results in a 25 unit increase in the response variable on average.

Regarding the interaction:

There is a 5-unit increase in the fitted slope for $x_2$ when $x_1 = 0$ vs. when $x_1 = 1$. There is a 1-unit increase in the fitted intercept...not sure how to finish this.

$\endgroup$

1 Answer 1

0
$\begingroup$

Talking about "fitted slopes" and "fitted intercepts" for two separate equations based on the value of your binary predictor $x_1$ will get you into trouble in general. You might be able to get away with that in this simple case, but applying it to more complicated situations can lead you astray.

I find it's least confusing always to start with the entire equation:

$$y = \beta_0 + \beta_1x_1 + \beta_2x_2 + \beta_3x_1x_2 + \epsilon$$

and recognize that each of the coefficients represents a change from what you would have otherwise predicted. I think that your approach might already have led to some confusion.

First equation:

Interpreting the fitted intercept: When $x_2 = 0$, the estimated value of the response variable is 11 on average.

Interpreting the fitted slope: When $x_1 = 0$ (i.e., for the first level of the dichotomous variable), a unit increase in the continuous variable $x_2$ results in a 20 unit increase in the response variable on average.

I think that the "11" is a typo and you meant "10" for the intercept. The interpretation of the slope is OK.

Second equation:

Interpreting the fitted intercept: When $x_2 = 0$, the estimated value of the response variable is 15 on average.

Interpreting the fitted slope: When $x_1 = 1$, a unit increase in the continuous variable $x_2$ results in a 25 unit increase in the response variable on average.

I don't know where the value of 15 comes from. If $x_2 = 0$ and $x_1 = 1$, then $y = 11$. Again, the slope interpretation is OK.

Regarding the interaction:

There is a 5-unit increase in the fitted slope for $x_2$ when $x_1 = 0$ vs. when $x_1 = 1$.

That's OK as far as it goes.

The interaction term, however, has nothing to do with the intercept (even in your re-formulation into 2 separate equations). The interaction term, $\beta_3 x_1 x_2$, only comes into play when both $x_1$ and $x_2$ are non-zero. The intercept $\beta_0$ is the case when both of them are 0; your attempt to re-define an intercept as $\beta_0 + \beta_1$ when $x_1 = 1$ is the value only when $x_2=0$ and thus then interaction term involving $\beta_3$ is necessarily 0.

Now, think about the other way to interpret $\beta_3 = 5$, as implied in your example. If $x_2=0$, then flipping $x_1$ from 0 to 1 increases $y$ by 1 unit. Otherwise, flipping $x_1$ from 0 to 1 increases $y$ by $1 +5x_2$.

That result comes directly when you start with the entire equation and recognize that each term represents a change over the intercept and the other previous terms.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer. Yes, I had some typos there - thanks for letting me know! I'm confused about this section: "The intercept 𝛽0 is the case...𝛽3 is necessarily 0." I agree, but I'm not sure why this matters. I remember seeing this re-grouping in an undergraduate statistics course, so I think it's valid, but I'm having trouble with interpretation. $\endgroup$ May 13 at 17:01

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.