2
$\begingroup$

I have matched my sample using propensity score matching such that each individual has an estimated propensity score of being assigned to a treatment group. Let $T_i$={0,1} be the actual treatment group of each individual. And $score_i$ be the estimated propensity score for each individual. I ran a log-linked gamma model with the following model specification: $$log(\mu)=\beta_0+\beta_1*score+\beta_2*T+\beta_3*score*T$$

Average Treatment Effect (ATE) is esimated using: $$ATE=exp(\hat{\beta_0}+\hat{\beta_1}*\overline{score}+\hat{\beta_2}*1+\hat{\beta_3}*\overline{score}*1)-exp(\hat{\beta_0}+\hat{\beta_1}*\overline{score}+\hat{\beta_2}*0+\hat{\beta_3}*\overline{score}*0)$$

where $\overline{score}$ is the average propensity score. How do I calculate the variance of the ATE?

$\endgroup$
2
  • $\begingroup$ A simple way is with bootstrapping, which is common and valid in this type of analysis. $\endgroup$
    – Noah
    Dec 18, 2018 at 21:25
  • $\begingroup$ I agree with using the bootstrap. I also think you can do better than the log-linked gamma specification. It seems like a restrictive and unnecessary assumption. You can easily do a kernel weighted estimator in stata or other packages without imposing any assumptions on the form of the outcome equation. $\endgroup$ Nov 1, 2021 at 4:12

1 Answer 1

0
$\begingroup$

You can use the delta rule. The essentials of the delta rule is that if you have an estimator $\hat \beta$ which $$\sqrt n(\hat \beta - \beta) \rightarrow \mathcal N(\mathbf 0,\mathbf S)$$

and you want to consider some function of the estimate $\hat \theta = g(\hat \beta)$ then

$$\sqrt n(g(\hat \beta) - g(\beta)) \rightarrow \mathcal N(\mathbf 0,\nabla g^\top \mathbf S \nabla g),$$

which is also sometimes written in the form $$g(\hat \beta) \sim \mathcal N\left(g(\beta),\frac{1}{n}\nabla g^\top \mathbf S \nabla g\right).$$

This suggest the approach (1) define the function $g$, (2) find the relevant derivative and (3) compute $\frac{1}{n}\nabla g^\top \mathbf S \nabla g$

In your case $\beta=(\beta_0,\beta_1,\beta_2,\beta_3) ^\top$ and $g(\beta) = ATE(\beta)$. According to the way you have defined ATE, I calculate the derivatives to be

$$ \frac{\partial g}{\partial \beta_0} = g(\beta) \\ \frac{\partial g}{\partial \beta_1} = g(\beta)\bar s \\ \frac{\partial g}{\partial \beta_2} = \exp(\beta_0 + \beta_1 \bar s + \beta_2 + \beta_3 \bar s) \\ \frac{\partial g}{\partial \beta_3} = \exp(\beta_0 + \beta_1 \bar s + \beta_2 + \beta_3 \bar s)\bar s$$

and $$ \nabla g^\top = \left(\frac{\partial g}{\partial \beta_0},\frac{\partial g}{\partial \beta_1},\frac{\partial g}{\partial \beta_2},\frac{\partial g}{\partial \beta_3} \right).$$.

$\endgroup$
2
  • $\begingroup$ Is S the covariance matrix of Beta hat? $\endgroup$
    – tatami
    Feb 19, 2019 at 3:23
  • $\begingroup$ This is approach seems like the right one if the pscore is known. However, it ignores the uncertainty in the estimation of the propensity score itself. A bootstrap that recalculates the propensity score and the betas simultaneously at each replication would fix this problem. $\endgroup$ Nov 1, 2021 at 4:09

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.