Can I use the indicator variable and the count/continuous variable of the same thing in the same linear regression model?

Question

For example, if I want to determine the effect of rain on daily traffic volume. I would include a binary variable whether or not it rained, and a continuous variable for the amount of rain. Would this be redundant?

Answer 1

You could do this, however it is not without it's problems. If you do, then the coefficient for the indicator will tell you if rainy days differ from dry days, and the continuous variable will tell you if the amount of rain makes a difference if it has rained. On the other hand the indicator will also be correlated with amount of rain so it could lead to an inflated standard error. We can demonstrate this with a simple simulation in R:

set.seed(1)
n_sim <- 1000
simvec_est_1 <- numeric(n_sim)
simvec_SE_1 <- numeric(n_sim)
simvec_est_2 <- numeric(n_sim)
simvec_SE_2 <- numeric(n_sim)
N <- 100
for(i in 1:n_sim) {
x1 <- rnorm(N, 0, 1)
  x1 <- x1 * (x1 > 0)  # non-rainy days
x2 <- rep(1,N)
  x2 <- x2 * (x1 > 0) # non-rainy days
y <- 10 + x1*3 + rnorm(N,0,5)  # fixed effect of 3 for x1
simvec_est_1[i] <- summary(lm(y ~ x1))$coef[2,1]
  simvec_SE_1[i] <- summary(lm(y ~ x1))$coef[2,2]
simvec_est_2[i] <- summary(lm(y ~ x1 + x2))$coef[2,1]
  simvec_SE_2[i] <- summary(lm(y ~ x1 + x2))$coef[2,2]
}
mean(simvec_est_1); mean(simvec_SE_1[i])
mean(simvec_est_2); mean(simvec_SE_2[i])

which results in:

[1] 3.011803
[1] 0.8578615
[1] 3.042935
[1] 1.159548

As we can see, the estimates are unbiased, but the standard error is inflated.

Answer 2

As long as you include the variables together I do not see the problem here. The coefficient for the binary variable will tell you if rainy days differ from dry days and the continuous variable will tell you, conditional on it having rained, does how much it rains make a difference.

The important thing here is to focus on the joint effect of the two variables by fitting a model with both of them (plus other covariates) and then the same model but without the two variables. That gives you the overall effect of your composite variable. Since you know that the two variables are related it seems pointless to test for the variance inflation factor as you already know the answer.

Answer 3

If you're asking on a general level, if you can include the continuous variable and its indicator in the same model, I don't think anyone here can tell you if you can or can't have both variables. It's not that the statistical modelling police is going to arrest you or your program would crash...

If you ask specifically for the variables in your example, I also can't say anything for sure, but I'm afraid there's a high chance for seeing weird things in your model. You can, for example, see that the indicator variable has coefficient zero, wrongfully inferring that there's no link between traffic jams and an indicator for rain. On the other hand, including both variables may lower your error.

Some guiding questions:

1. What is the purpose of the model? Is it a descriptive model? Is it a predictive one?
2. What other variables do you have? Is there an intercept? If there isn't an intercept your model would predict zero traffic volume for non-rainy days. How are you with that?

In addition, I don't think that your variables must have a high Pearson correlation coefficient.

Let us denote the indicator variable by $X$ and the continuous by $Y$. Assuming both nonnegative and not constant. $$ Corr(X,Y) = \frac{E(XY)-E(X)E(Y)}{\sqrt{Var(X)Var(Y)}}. $$

Denote by $p$ the proportion of rainy days in your data. I Assume $1 > p > 0$.

Since $XY$=$Y$, $E(XY)=E(Y)$. $X$ is a binary rv, hence $E(X) = p,\:Var(X)=p(1-p)$.

So your correlation coefficient will be: $$ Corr(X,Y) = \frac{E(Y)-pE(Y)}{\sqrt{p(1-p)Var(Y)}} = \sqrt{\frac{1-p}{p}}\frac{E(Y)}{\sqrt{Var(Y)}}. $$

$Corr(X,Y)$ can have any value in $(0,1]$, depending on $p$ and $Y$.