2
$\begingroup$

A generalized DID has the equation as below

$Y_{it}$ = $\alpha$ + $\beta$ $(Leniency Law)_{kt}$ + $\delta$$X_{ikt}$ + $\theta$$_t$ + $\gamma$$_i$ +$\epsilon$$_{it}$

where i,k, and t index firms, countries, and years respectively. $X_{ikt}$ is a vector of the different firm, country, and industry control, while $\gamma$ and $\theta$ are firm and year fixed effects.

From this answer of Ariel,

$\beta$ is the average difference between pre- and post for the treated population. Because the ATT is the effect on the treated population of the intervention. Remember, DID is using the common trends assumption to try to tell us what would have happened to the treated population had they not been treated. It then compares the realized treated outcomes to the predicted untreated counterfactual

I am wondering if $\beta$ is ATT, meaning that we did not mention control group in our explanation, so if we did not mention, why we include the control group into our regression?

$\endgroup$
2
$\begingroup$

The two-period DID is typically taken to estimate the ATT. This can easily be seen by doing out the expectations. Consider the specification at the unit $i$ level,

$$Y_i = \alpha + \gamma G_i + \theta T_i + \beta D_i + \epsilon_i$$

Here we use $G_i\in\{0,1\}$ to denote the group and $T_i \in \{0,1\}$ to denote the period. And $D_i = G_iT_i$ denotes treatment status.

Then to extract the treatment effect $\beta$ (and determine what it is) we will look at the potential outcomes when the group and period indicators are switched on. We have the potential outcome for the control as,

$$\mathbb{E}[Y_i(0)|G_i=1,T_i=1]=\alpha + \gamma + \theta$$

And for treated,

$$\mathbb{E}[Y_i(1)|G_i=1,T_i=1]=\mathbb{E}[Y_i(0)|G_i=1,T_i=1]+\beta$$

So we have,

$$\beta = \mathbb{E}[Y_i(1)-Y_i(0)|G_i=1,T_i=1]=\mathbb{E}[Y_i(1)-Y_i(0)|D_i=1]=ATT$$

In the case of a generalized DID estimator where there are multiple periods and multiple groups whose treatment status can be switched on at different times the answer is more complicated. As I wrote in this answer the generalized DID can always be broken down into a weighted average of two-period DIDs. Each of these will exhibit an ATT. But overall, you will have estimated an average of ATTs. The exact interpretation of this is a little bit uncertain. Under a constant treatment assumption (which is very strong) this will indeed be the ATT. However, usually, that is not the case.

Goodman-Bacon (2018) provides a little more study of how to interpret this. See section 2 of the attached paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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