Interpretation of interaction term in log-linear models

Question

Given the model:

$\log(y) = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_1 x_2$

Which can be for example a Poisson or a Negative Binomial model ($y$ would be a count variable) or a logistic regression ($y$ would be odds), what is the interpretation of $\beta_3$?

I came up with the following, and I ask you (1) if it is correct (2) if there are other or simpler interpretations.

Theorem

For a given $x_1=\bar{x}_1$, when $x_2$ changes from $\bar{x}_2$ to $\bar{x}_2+1$, $y$ is multiplied by $e^{\beta_2} (e^{\beta_3})^{\bar{x}_1}$

Proof

Be $y_1 = y_1(x_1=\bar{x}_1, x_2=\bar{x}_2)$ and $y_2 = y_2(x_1=\bar{x}_1, x_2=\bar{x}_2+1)$

$\log(y_1) = \beta_0 + \beta_1 \bar{x}_1 + \beta_2 \bar{x}_2 + \beta_3 \bar{x}_1 \bar{x}_2$

\begin{align} \log(y_2) &= \beta_0 + \beta_1 \bar{x}_1 + \beta_2 (\bar{x}_2+1) + \beta_3 \bar{x}_1 (\bar{x}_2+1) &\\ &= \beta_0 + \beta_1 \bar{x}_1 + \beta_2 \bar{x}_2 + \beta_2 + \beta_3 \bar{x}_1 \bar{x}_2 + \beta_3 \bar{x}_1 & \end{align}

$ \log(y_2) - \log(y_1) = \log\left(\dfrac{y_2}{y_1}\right) = \beta_2 + \beta_3 \bar{x}_1$

$\dfrac{y_2}{y_1} = \exp(\beta_2+\beta_3 \bar{x}_1) = e^{\beta_2} (e^{\beta_3})^{\bar{x}_1}$

$y_2 = e^{\beta_2} (e^{\beta_3})^{\bar{x}_1} y_1$

Answer 1

A logarithmic scale relates to multiplication or something approximate to it. If a dependent variable like $x_1$ or $x_2$ increases then the effect on the response variable $y$ is a multiplication with a certain factor.

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The linear model on the log scale becomes a power law in the linear scale

$$y = \alpha_0 \cdot \alpha_1^{x_1} \cdot \alpha_2^{x_2} \cdot \alpha_3^{x_1x_2}$$

with $\alpha_i = \exp \beta_i$.

Your interpretation is indeed correct that you can group the interaction with a main effect following algebraic rules

$$y = \alpha_0 \cdot \alpha_1^{x_1} \cdot \left(\alpha_2 \cdot \alpha_3^{x_1}\right)^{x_2} $$

The term $\alpha_3^{x_1}$ can be seen as a modification of the power law coefficient for the $x_2$ variable. (Or the other way around $x_2$ modifies the coefficient for the $x_1$ variable. That interpretation is a bit arbitrary).

The interaction term is a linear term but the actual relationship might be some different non-linear relationship. This is potentially useful for your interpretation as well. For example, the actual relationship could be instead $y = \alpha_0 \cdot \alpha_1^{x_1} \cdot \left(\alpha_2 + {x_1}\right)^{x_2} $ or $y = \alpha_0 \cdot {x_1} \cdot \left(\alpha_2 + {x_1}\right)^{x_2} $ . With the model being an approximation.

Answer 2

$\beta_3$ is the $log$ of the ratio of $y$ at $(x_1+1, x_2+1)$ versus $(x_1, x_2)$ divided by both the ratio of change from $(x_1+1, x_2)$ to $(x_1, x_2)$ and $(x_1, x_2+1)$ to $(x_1, x_2)$. This is true whether the $x$ variable are categorical (with a base case of (0,0) ) or continuous.

Said differently, $\beta_3$ is log of the ratio of the total change in $y$ of one unit in both $x_1$ and $x_2$ versus a unit change from baseline in $x_1$ times a unit change from baseline in $x_2$

Let $y=f(x_1, x_2)$

$$log(f(x_1,x_2)) = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_1 x_2$$

$$log(f(x_1+1,x_2)) = log(f(x_1,x_2)) + \beta_1 + \beta_3 x_2$$

$$log(f(x_1,x_2+1)) = log(f(x_1,x_2)) + \beta_2 + \beta_3 x_1$$

$$log(f(x_1+1,x_2+1)) = log(f(x_1,x_2)) + \beta_1 + \beta_2 + \beta_3 x_1 + \beta_3 x_2 + \beta_3$$

$$\beta_3 = log\left( \frac{\frac{f(x_1+1,x_2+1)}{f(x_1,x_2)}}{\frac{f(x_1+1,x_2)}{f(x_1,x_2)} \frac{f(x_1,x_2+1)}{f(x_1,x_2)}} \right)$$

Staying on the log scale, we can also say that $\beta_3$ is the change in slope of $log(y)$ with $x_1$ as you move in the $x_2$ direction [1].

$$\frac{\delta^2 log(f)}{\delta x_1 \delta x_2} = \beta_3$$