# How to interpret interaction in Log Log models

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))

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.