0
$\begingroup$

I am trying to understand the connection between sample average treatment effect (SATE) and average treatment effect (Population)-PATE or ATE

$SATE=\frac{1}{n} \sum^n_i (Y_i(1)-Y_i(0))$

$ATE=\frac{1}{N} \sum^N_i (Y_i(1)-Y_i(0))=E(Y_i(1)-Y_i(0))$

The estimator:

$\hat{\Delta}=\frac{1}{n} \sum^n_i (Y_i*(T_i=1)-Y_i(T_i=0))$, $T_i$ is the treatment status and we assume treatment is randomly assigned.

$\hat{\Delta}$ is sample mean difference between treated and untreated subjects.

So $\hat{\Delta}$ is the estimate for both $SATE$ and $ATE$, right?

One more question: Why should we be interested in this $SATE$? this $SATE$ may vary from sample to sample. Why should we care whether treatment works or not in one particular sample?

$\endgroup$

1 Answer 1

0
$\begingroup$

The main difference can be seen in the sum indices. The sample ATE uses a sample $1\dots n$, whereas the population ATE includes the entire population $1\dots N$. The estimator $\hat{\Delta}$ is still based on a sample. You'd be correct in saying that it gives you an estimate of the sample ATE, which is also an estimate of the population ATE if the data is from an RCT, but it is not the same as the population ATE.

$\endgroup$
2
  • $\begingroup$ Thank you! I am not sure I fully understand your comments. Do you mean the estimator $\hat{\Delta}$ estimates both $SATE$ and $PATE$, right? The estimator $\hat{\Delta}$ is not the same as $SATE$ but it is an estimate of $SATE$, right?$SATE$ is a causal parameter involving one unobservable potential outcome. $SATE$ is not same as $PATE$. I wonder why we should be interested in this $SATE$ if the results from a trial is not generalizable? $\endgroup$
    – Vincent
    Jul 14, 2023 at 13:59
  • $\begingroup$ So, you're right in saying that if the data is from an RCT, you'll get an estimate of the SATE and the PATE. For the SATE it's an estimate because for each individual you can only observe a single outcome. For the PATE it's the same thing but additionally you have the sample uncertainty. If the sample is drawn randomly, the estimate should be generalizable to the population and the bigger the sample, the more so. $\endgroup$
    – Scriddie
    Jul 14, 2023 at 16:56

Your Answer

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

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