1
$\begingroup$

I am working on DiD setting, the variable of interest is

$Y_{i,t} = \alpha_i + \beta_t + \gamma D_{i,t}$

while $i$ and $t$ are unit and time fixed effects. $\gamma$ is the coefficient of variables of interest D.

In another word, $\gamma$ tells us how the laws affect the outcome Y. In my research, the result is $\gamma$ is negative and significant at 1%. However, when I add one more control variable called $\theta$ into the regression, the significant levels of $\gamma$ reduces to 5%, and the coefficient of $\gamma$ become higher (less negative).

What should I say about the role of $\theta$ in this situation ?

$\endgroup$

1 Answer 1

2
$\begingroup$

This is because the first coefficient estimate was estimated in the presence of omitted variable bias (OVB), and the effect of omitted variable just previously loaded onto the the $\gamma$ coefficient.

OVB can drastically change the value of coefficients, if the true model is given by

$$Y_{it}=\alpha_i+β_t+\gamma D_{it} + \theta X_{it} + e_{it}$$

but you only fit

$$Y_{it}=\alpha_i+β_t+\gamma D_{it} + e_{it}$$

it can be shown that the parameter estimate will be given:

$$E[\hat{\gamma}]= \gamma + \theta \frac{COV(D,X)}{VAR(D)}$$

So unless $COV(D,X)=0$, or unless $\theta=0$, if you omit theta your estimate of gamma will biased in a unpredictable way.

For example if the true coefficient of $\gamma$ is $-0.5$ and the true coefficient of theta is $-2$ and variance of $D$ is $1$ and covariance of $D$ and $X$ is also $1$ omitting theta would result in the following biased coefficient:

$$E[\hat{\gamma}] = -0.5 -2 \cdot \frac{1}{1}=-2.5$$

of course the larger absolute value of coefficient the higher significance you get (ceteris paribus) since you are testing hypothesis of $\gamma \neq 0$. Thus once you add $\theta$ and thus remove omitted variable bias from $\gamma$, assuming $\gamma$'s standard error does not change, you should expect $\gamma$ to be less significant in the example above. This is basically what is happening to you.

$\endgroup$

Your Answer

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

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