0
$\begingroup$

I want to check that I understand the notation for the average treatment effect (ATE) estimator correctly, and hopefully some of you can double check this. I often try to understand formulas through specific examples. Consider a randomized controlled trial with $n=10$, where 6 have been given treatment, so that $n_1=6$, and 4 have not so $n_0=4$.

Filling in the ATE-estimator:

$\hat\tau= \frac{1}{n_1} \sum_{n=1}^{n_1} y_i(1) - \frac{1}{n_0} \sum_{n=1}^{n_0} y_i(0)$

We get,

$\hat\tau= \frac{1}{6} \sum_{n=1}^{6} y_i(1) - \frac{1}{4} \sum_{n=1}^{4} y_i(0)$

Which for the following toy data set

i  t  y
1  0  1
2  0  2
3  0  2
4  0  1
5  1  3
6  1  4
7  1  5
8  1  5
9  1  6
10 1  3

corresponds to: $\hat\tau= 0.33 \frac{3+4+5+5+6+3}{6} - 0.25 \frac{1+2+2+1}{4}=1.43-0.38=1.05$

Which says that the treatment had a positive effect of, on average, 1.05 for the population who received the treatment.

$\endgroup$
2
$\begingroup$

There are a few problems with what you've written, some of which are minor. First, the true ATE is $$E[y(1)] - E[y(0)]$$where $y_i(a)$ is the potential outcome under treatment level $a$. Without assumptions, this quantity cannot be estimated because $y_i(1)$ and $y_i(0)$ are not observed for any units.

Randomization gives us the assumption of exchangeability, meaning that $y(a) \perp A \ \forall a$, which implies that $E[y(a)|A=a]=E[y(a)]$. Assuming consistency (i.e., that there are no unmeasured versions of treatment), we also have $y_i = \sum_a{I(A=a)y_i(a)}$, which implies that we can replace $y_i(a)$ with $y_i$ when we condition on $A=a$.

Under these assumptions, the difference in sample means is unbiased for the ATE. $\frac{1}{n_1}\sum_{i\in A=1}{y_i}$ is an unbiased estimator of $E[y|A=1]$, which under consistency is equal to $E[y(1)|A=1]$, which under exchangeability is equal to $E[y(1)]$. The same argument can be made for $A=0$, so we have that $\frac{1}{n_1}\sum_{i\in A=1}{y_i} - \frac{1}{n_0}\sum_{i\in A=0}{y_i}$ is unbiased for $E[y(1)] - E[y(0)]$, the ATE.

I don't know where you got the $.33$ and $.25$ in your formula. The correct math is $$\frac{3+4+5+5+6+3}{6}-\frac{1+2+2+1}{4}=\frac{26}{6}-\frac{6}{4}=2.833$$

$\endgroup$
  • $\begingroup$ Thanks for your helpful reply! My only follow-up question is an easy question on notation. I got .33 by substituting $1/n_{1}$ with 1/6 (typo 2/6), and .25 by $1/n_{0}$ with 1/4. By your reply, it seems like this notation is not included in the calculation. Thus, I wonder what the $1/n$ notation adds to the summation operator? Is it just a way to describe that the sum of all observations is to be divided by the number of observations? $\endgroup$ – Tarjei W. Havneraas Feb 3 at 10:32
  • 1
    $\begingroup$ The 6 and 4 in the denominators are those values. Summation by itself doesn't involve a denominator; the 6 and 4 are the $1/n$ for each group. You don't need to additionally include them. $\endgroup$ – Noah Feb 3 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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