Variance of ATE (Average Treatment Effect) from log-linked gamma model 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?
 A: 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 $$ \nabla g^\top = \left(\frac{\partial g}{\partial \beta_0},\frac{\partial g}{\partial \beta_1},\frac{\partial g}{\partial \beta_2},\frac{\partial g}{\partial \beta_3} \right).$$.
