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?

  • $\begingroup$ A simple way is with bootstrapping, which is common and valid in this type of analysis. $\endgroup$
    – Noah
    Commented 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. $\endgroup$ Commented Nov 1, 2021 at 4:12

1 Answer 1


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

  • $\begingroup$ Is S the covariance matrix of Beta hat? $\endgroup$
    – tatami
    Commented 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. $\endgroup$ Commented Nov 1, 2021 at 4:09

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