In randomized studies, we have that the difference-in-means estimator is given, for treatment/control as:

$$ \hat{\tau}_{DM} = \frac{1}{n_1}\sum_{Z_i = 1}Y_i - \frac{1}{n_0}\sum_{Z_i = 0}Y_i $$

where $n_z = |\{i:Z_i = z\}|$.

The variance of the estimator can be written as

$$ Var\left(\hat{\tau}_{DM}\right|n_0,n_1) = \frac{1}{n_0}Var\left(Y_i(0)\right) + \frac{1}{n_1}Var\left(Y_i(1)\right) $$

I know a central limit theorem should look like

$$ \sqrt{n}\left(\hat{\tau}_{DM} - \tau\right) \overset{D}\to \mathcal{N}\left(0, V_{DM}\right) $$


$$ V_{DM} = \frac{Var\left(Y_i(0)\right)}{P(Z_i = 0)} + \frac{Var\left(Y_i(1)\right)}{P(Z_i = 1)} $$

I am unsure how to derive $V_{DM}$, although it seems it should be that intuitively. How can I derive the variance?

  • $\begingroup$ Could you explain your notation? What do "$Y_i(0)$" and "$Y_i(1)$" mean and how are they related to "$Y_i$"? $\endgroup$
    – whuber
    Jun 2, 2021 at 13:48
  • 2
    $\begingroup$ $(Y_i(1),Y_i(0))$ are the potential outcomes for a unit $i$, and $Y_i$ is the realized or observed outcome, represented by $Y_i = Z_iY_i(1) + (1-Z_i)Y_i(0)$. Thanks! $\endgroup$
    – user321627
    Jun 2, 2021 at 15:38

1 Answer 1


First, the finite sample variance that you have above is called the Neyman conservative variance. This result requires a little trick about how we write the estimator and an assumption of a random sample from an infinite population.

Define, $K=\frac{n_1}{n}$. Now consider,

$$\hat\tau = \bar{Y_i(1)}-\bar{Y_i(0)}=\frac{1}{n_1}\sum_{i=1}Z_iY_i - \frac{1}{n_0}\sum_i (1-Z_i)Y_i \\= \frac{1}{n}\sum_{i=1}\frac{Z_iY_i}{K} - \frac{1}{n}\sum_i \frac{(1-Z_i)Y_i}{1-K}\\=\frac{1}{n}\sum_i(\frac{Z_iY_i}{K}-\frac{(1-Z_i)Y_i}{1-K})=\frac{1}{n}\sum_i(\frac{Y_i(1)}{K}-\frac{Y_i(0)}{1-K})$$

Define the population average treatment effect or PATE by,

$$\tau = \mathbb{E}[Y_i(1)-Y_i(0)]$$

Then, since we have that these are iid sequences by the WLLN,

$$\hat\tau \overset{p}{\to} \tau$$

Notice that $K\overset{p}{\to}\Pr[Z_i=1]$. So by Slutzky's theorem and the CLT we have,

$$Avar(\hat\tau)= \frac{Var(Y_i(1))}{\Pr[Z_i=1]}+\frac{Var(Y_i(0))}{\Pr[Z_i=0]}$$

where $Avar(\cdot)$ refers to the asymptotic variance or the variance of the limiting distribution of the estimator.


$$\sqrt{n}(\hat\tau - \tau)\overset{d}{\to}N(0,\frac{Var(Y_i(1))}{\Pr[Z_i=1]}+\frac{Var(Y_i(0))}{\Pr[Z_i=0]})$$

As desired.


I glossed over this in the main answer but on second thought I think it is useful to note. It is easy to see that the variance of the parameter would be given by,

$$Var(\hat\tau) = \frac{n}{n_1} Var(Y_i(1)) + \frac{n}{n_0} Var(Y_i(0)) - Cov(Y_i(1),Y_i(0))$$

Of course, this is not identified within the sample. However under asymptotic inference, given true random sampling from the population, the covariance term will be $0$ because control and treatment are sampled independently. Thus, yielding the asymptotic variance that we see above.

For anyone interested in results under more general assumptions, I would recommend this paper by Li et al..

  • $\begingroup$ Could you explain what "$Avar$" refers to? I suppose you mean some version of variance used to handle the fact that the variance of $\hat \tau$ is undefined -- but it would be interesting to know the details. $\endgroup$
    – whuber
    Jun 3, 2021 at 12:23
  • $\begingroup$ Good point! I just edited the post to make it clear that I am referring to the asymptotic variance. $\endgroup$
    – Ariel
    Jun 3, 2021 at 13:26
  • $\begingroup$ Okay--but your characterization of it is confusing, because the population parameter ($\tau$) is not a random variable and therefore is considered either to have no variance or to have a variance of zero. $\endgroup$
    – whuber
    Jun 3, 2021 at 15:16
  • $\begingroup$ Ahh you are of course correct! I corrected my description of it. $\endgroup$
    – Ariel
    Jun 3, 2021 at 15:22

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