How to interpret interaction in Log Log models

Question

I have the following model, and I am not sure how to interpret the elasticity of the interaction term (log-log coefficients):

Log(member) = 3.61 + 0.52 Log(Poor) - 0.26 Log(Sick) + 0.04 (Log(poor) * Log(sick))

Answer 1

The original question is missing what member, poor and sick mean, so I will answer this question more abstractly (assuming all three are continuous variables that can be logged).

The expected value of $\ln y$ given $x_1$ and $x_2$ in your model is

$$ \ln y =\alpha + \beta_1 \ln x_1 + \beta_2 \ln x_2 + \beta_3 \ln x_1 \cdot \ln x_2$$

Taking the derivative of that with respect to $x_1$ (using the chain rule), you get

$$\frac{1}{y}\cdot\frac{\partial y}{\partial x_1} = \frac{\beta_1}{x_1}+ \frac{\beta_3}{x_1} \cdot \ln x_2$$

Multiplying both sides by $x_1$ and rearranging, you get

$$\frac{\partial y}{\partial x_1} \cdot \frac{x_1}{y} = \beta_1+ \beta_3 \cdot \ln x_2$$

The left-hand-side is the very definition of an elasticity of $y$ with respect to $x_1$. This means that you can interpret $\beta_1+ \beta_3 \ln x_2$ as the percent change in $y$ from a 1% increase in $x_1$. This is a function that depends on $x_2$. $\beta_3$ is positive, as is $\beta_1$, which means $x_2$ will generally make the elasticity grow larger.

The elasticity here is $0.52 + 0.04 \cdot \ln x_2$, and it is probably easiest to just graph this (though I have no idea if the range for $x_2$ is realistic here):

The means the over the range for $x_2$ above, a 1% increase in $x_1$ is associated with between 0.5% to 0.7% change in $y$. This is still fairly inelastic.

And you can actually continue to do this. The elasticity is

$$\varepsilon = \beta_1+ \beta_3 \cdot \ln x_2$$

Take the derivative of that with respect to $x_2$ and rearrange:

$$\frac{\partial \varepsilon}{\partial x_2} \cdot x_2 = \beta_3 $$

This means that $\frac{\beta_3}{100}$ gives you the change in the elasticity with respect to $x_1$ from a 1% increase in $x_2$, which is .0004 according to your model.