Understanding average treatment effect estimator notation 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.
 A: 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 $$\frac{3+4+5+5+6+3}{6}-\frac{1+2+2+1}{4}=\frac{26}{6}-\frac{6}{4}=2.833$$
