When I run a standard difference in differences specification with a log-transformed dependent variable like:

$$\log(Outcome_{it}) = \beta_1 + \beta_2Treat_i +\beta_3Post_t +\beta_4(Treat\times Post)_{it} +\varepsilon_{it}$$

How do I interpret the coefficient $\beta4$?

Normally in log-level models, I would use the approximation $\%\Delta y= 100\beta_4\Delta x$, but this approximation is only valid for small changes in $x$ (and small $\beta$). In my case $x$ is a dummy variable and as such either $0$ or $1$ ($Treat\times Post$). Is this change considered to be 'small' or do I have too use $\%\Delta y=e^{\beta-1}$ for interpreting the coefficient?


2 Answers 2


You can should treat the interaction variable as a dummy and follow this advice from David Giles:

If $Treat\cdot Post$ switches from 0 to 1, the % impact on $Y$ is $100 \cdot (\exp(\beta_4 - \frac{1}{2} \hat \sigma_{\beta_4}^2)-1).$


What's required is that $\beta \cdot \Delta x$ be small. If you know that $\Delta x$ is 1, then that means that $\beta$ has to be small. How small? The true proportionate change in $Outcome$ when the dummy rises by $1$ is $\exp(\beta)-1$. The approximate change is $\beta$. The error from the approximation is:

\begin{equation} \textrm{Error} = \exp(\beta)-1-\beta \end{equation}

For small $|\beta|$, this is pretty small. For example, for $\beta=0.1$ (approximate 10% change), the true percent change in $Outcome$ when the dummy turns on is 10.5%. Given the usual standard errors in empirical work, I'm happy to ignore this. By the time you get to a $\beta$ of 0.2 (approximate 20%), the true percent change is 22%. Willing to ignore this much approximation error? Again, I am, but you may not be. This is now a 10% approximation error. By the time you get to $\beta=0.3$, the true percent change in outcome is 35% rather than 30%, and I am not happy to ignore this any more.

So, my rule of thumb is to ignore this approximation error for $|\beta|<0.2$ and worry about it for $\beta$ bigger than that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.