Incidence Rate Ratio Equal to One

Question

I am a student and I have a statistics question. I ran a generalized linear model using quasipossion as the distribution family ( y variable was continuous). The result I obtained for one of my variables ,the three way interaction term, was significant p = 0.02 but the incidence rate ratio was equal to 1. I think this means that even though the p-value is significant, the incidence rate ratio being equal to 1 basically discredits the model ( in terms of this term) , as it doesn't show any decrease or increase in risk. Is this a correct interpretation? Thanks.

Answer 1

Interactions in non-linear models are tricky. I would argue that the exponentiated interaction coefficient is not informative and does not "discredit" the model.

Here's the intuition for a two-way interaction that you can adopt to the three-way case. Your QP model for the expected value of $y$ conditional on $x$ and $z$ is $$E[y \vert x,z] = \exp\{\alpha + \beta \cdot x + \gamma \cdot z + \delta \cdot x \cdot z\}.$$

Now suppose $x$ increased by one. The ratio of the new to old expected values is

$$\frac{E[y \vert x+1,z]}{E[y \vert x,z]} = \frac{\exp\{\alpha + \beta \cdot (x+1) + \gamma \cdot z + \delta \cdot (x+1) \cdot z\}}{\exp\{\alpha + \beta \cdot x + \gamma \cdot z + \delta \cdot x \cdot z\}}= \exp\{\beta\} \cdot \exp\{\delta \cdot z\}.$$

So the fact that $\exp\{\delta\} = 1$ is not very informative about how $z$ modifies the effect of $x$ on $y$ since the answer depends on the value of $z$. So if $\delta = 0.005$, then $\exp \{.005\} \approx 1$, but if $z$ is large enough, the $\exp\{.005 \cdot z\}$ term will still be greater than one.

I personally find it easier to think about $\delta$ in terms of changing the relationship of $y$ and $x$. This means looking at the un-exponentiated coefficient. Differentiating the expected value of $y$ with respect to $x$, we get

$$\frac{ \partial E[y \vert x,z]}{\partial x} = \exp \{\alpha + \beta \cdot x + \gamma \cdot z + \delta \cdot x \cdot z\} \cdot (\beta+ \delta \cdot z)=E[y \vert x,z]\cdot (\beta+ \delta \cdot z).$$

This uses the chain rule and the fact that $\frac{\partial\exp \{x\}}{\partial x} = \exp \{x\}$.

We can re-write the derivative as

$$\frac{ \partial E[y \vert x,z]}{\partial x} \cdot \frac{1}{E[y \vert x,z]}= \beta+ \delta \cdot z.$$

This is a semi-elasticity, which tells us the change in $y$ (in percent) from a one-unit change in $x$. People often multiply by 100 here. Note that it is a function of $z$.

We can then ask how that semi-elasticity itself depends on $z$, so we take the derivative with respect to $z$ this time to get $\delta$ all alone.

So if $\delta$ is 0.005, I would interpret that as a one-unit increase in $z$ is associated with a $100 \cdot 0.005 = 1/2\%$ bigger change in $y$ from a one-unit change $x$.