Hi I have the following regression model:

\begin{align}y=\beta_0 +\beta_1x_1+\beta_2x_2 +\beta_3 \ln(x_3)+\beta_4[\ln(x_3)\times x_1] + \mu_{ti} \end{align}

How would you interpret $\beta_4$?

My own thinking is that it is the $\frac{\beta_4}{100}$ is the ceteris paribus partial effect of a 1% increase in $x_3$ on the partial effect of $x_1$ on $y$. Is this in any way correct?


1 Answer 1


Your interpretation is correct.

Differentiating the expected value of $y$ with respect to $x_1$ yields $$\frac{\partial y}{\partial x_1} = \beta_1 + \beta_4 \cdot \ln x_3$$ This is the change in expected $y$ (in units of $y$) associated with a one-unit change in $x_1$. It's a function of $x_3$.

How does this function behave when you change its argument? Differentiate it with respect to $x_3$:

$$\frac{\partial y}{\partial x_3 \partial x_1} =\beta_4 \cdot \frac{1}{x_3}$$

Bring $x_3$ over to the left to get:

$$x_3 \cdot \frac{\partial y}{\partial x_3 \partial x_1} =\beta_4$$

You can (approximately) rewrite that as

$$\frac{\frac{\Delta y}{\Delta x_1}}{100\cdot \frac{\Delta x_3}{x_3}} =\beta_4$$

The numerator is the change in expected $y$ for 1 unit change in $x_1$. The denominator is $1\%$ change in $x_3$.

If I can be picky, this is for expected $y$, since regression models the conditional mean of $y$, not $y$ itself.

  • $\begingroup$ Amazing, thank you for your response! $\endgroup$
    – efan787
    Dec 12, 2022 at 23:40

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