Jinsil Kim
2024.08.28
A population consisting of $N$ units.
For each unit there exist two potential outcomes, $Y_i(0)$ and $Y_i(1)$, corresponding to the outcome under control and treatment
The vector of treatment assignments, $\mathbf{W}$, with $i^{\text {th }}$ element $W_i$
Let $N_{\mathrm{c}}=\sum_{i=1}^N\left(1-W_i\right)$ and $N_{\mathrm{t}}=\sum_{i=1}^N W_i$ be the number of units assigned to the control and active treatment respectively, with $N_c+N_t=N$.
The population average treatment effect $$ \tau_{\mathrm{fs}}=\frac{1}{N} \sum_{i=1}^N\left(Y_i(1)-Y_i(0)\right)=\bar{Y}(1)-\bar{Y}(0), $$
$\bar{Y}(0)$ and $\bar{Y}(1)$ are the averages of the potential control and treated outcomes respectively: $$ \bar{Y}(0)=\frac{1}{N} \sum_{i=1}^N Y_i(0), \quad \text { and } \quad \bar{Y}(1)=\frac{1}{N} \sum_{i=1}^N Y_i(1) $$
Theorem 2. The sampling variance of $\hat{\tau}^{\mathrm{dif}}=\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}$ is
$$ \mathbb{V}W\left(\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)=\frac{S_c^2}{N{\mathrm{c}}}+\frac{S_t^2}{N_{\mathrm{t}}}-\frac{S_{t c}^2}{N} $$
$S_c^2$ and $S_t^2$ are the variances of $Y_i(0)$ and $Y_i(1)$ in the sample, defined as:
$$ S_c^2=\frac{1}{N-1} \sum_{i=1}^N\left(Y_i(0)-\bar{Y}(0)\right)^2, \quad \text { and } \quad S_t^2=\frac{1}{N-1} \sum_{i=1}^N\left(Y_i(1)-\bar{Y}(1)\right)^2, $$
$S_{t c}^2$ is the sample variance of the unit-level treatment effects, defined as: $$ \begin{aligned} S_{t c}^2 & =\frac{1}{N-1} \sum_{i=1}^N\left(Y_i(1)-Y_i(0)-(\bar{Y}(1)-\bar{Y}(0))\right)^2 \ & =\frac{1}{N-1} \sum_{i=1}^N\left(Y_i(1)-Y_i(0)-\tau_{\mathrm{fs}}\right)^2 . \end{aligned} $$
Proof of Theorem 2 -> Appendix
$$ \hat{\mathbb{V}}^{\text {neyman }}=\frac{s_c^2}{N_{\mathrm{c}}}+\frac{s_t^2}{N_{\mathrm{t}}} $$
Proof of Theorem 3 (brief version)
An unbiased estimator for $S_c^2$ is $$ s_c^2=\frac{1}{N_{\mathrm{c}}-1} \sum_{i: W_i=0}\left(Y_i(0)-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)^2=\frac{1}{N{\mathrm{c}}-1} \sum_{i: W_i=0}\left(Y_i^{\mathrm{obs}}-\bar{Y}_{\mathrm{c}}^{\mathrm{obs}}\right)^2 . $$
An unbiaed estimator for $S_t^2$, the population variance of $Y_i(1)$, $$ s_t^2=\frac{1}{N_{\mathrm{t}}-1} \sum_{i: W_i=1}\left(Y_i(1)-\bar{Y}{\mathrm{t}}^{\mathrm{obs}}\right)^2=\frac{1}{N{\mathrm{t}}-1} \sum_{i: W_i=1}\left(Y_i^{\mathrm{obs}}-\bar{Y}_{\mathrm{t}}^{\mathrm{obs}}\right)^2 $$
$S_{\text {tc }}^2$ (the population variance of the unit-level treatment effects), is generally impossible to estimate empirically
because we never observe both $Y_i(1)$ and $Y_i(0)$ for the same unit.
If the treatment effects are constant and additive $\left(Y_i(1)-Y_i(0)=\tau_{\mathrm{fs}}\right.$ for all units), then $S_{\text {tc }}^2=0$
A detailed proof -> Appendix.
First note that the average treatment effect is:
$$ \tau_{\mathrm{fs}}=\bar{Y}(1)-\bar{Y}(0)=\frac{1}{N} \sum_{i=1}^{N}\left(Y_{i}(1)-Y_{i}(0)\right) $$
And the standard estimator of $\tau_{\mathrm{fs}}$ is:
$$ \begin{aligned} \hat{\tau}^{\mathrm{dif}} & =\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}=\frac{1}{N_{\mathrm{t}}} \sum_{i=1}^{N} W_{i} \cdot Y_{i}^{\mathrm{obs}}-\frac{1}{N_{\mathrm{c}}} \sum_{i=1}^{N}\left(1-W_{i}\right) \cdot Y_{i}^{\mathrm{obs}} \ & =\frac{1}{N} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{t}}} \cdot W_{i} \cdot Y_{i}(1)-\frac{N}{N_{\mathrm{c}}} \cdot\left(1-W_{i}\right) \cdot Y_{i}(0)\right) \end{aligned} $$
For the variance calculations, define a centered treatment indicator $D_{i}$ as
$$ D_{i}=W_{i}-\frac{N_{\mathrm{t}}}{N}= \begin{cases}\frac{N_{\mathrm{c}}}{N} & \text { if } W_{i}=1 \ -\frac{\mathrm{N}{\mathrm{t}}}{N} & \text { if } W{i}=0\end{cases} $$
$\mathbb{E}\left[D_{i}\right]=0$, and $\mathbb{V}\left(D_{i}\right)=\mathbb{E}\left[D_{i}^{2}\right]=N_{\mathrm{c}} N_{\mathrm{t}} / N^{2}$.
Later we also need its cross moment, $\mathbb{E}\left[D_{i} \cdot D_{j}\right]$. For $i \neq j$ the distribution of this cross product is
$$ \operatorname{Pr}{W}\left(D{i} \cdot D_{j}=d\right)= \begin{cases}\frac{N_{\mathrm{t}} \cdot\left(N_{\mathrm{t}}-1\right)}{N \cdot(N-1)} & \text { if } d=N_{\mathrm{c}}^{2} / N^{2} \text{ (both treated}) \ 2 \cdot \frac{N_{\mathrm{t}} \cdot N_{\mathrm{c}}}{N \cdot(N-1)} & \text { if } d=-N_{\mathrm{t}} N_{\mathrm{c}} / N^{2} \text{ (only one treated)} \ \frac{N_{\mathrm{c}} \cdot\left(N_{\mathrm{c}}-1\right)}{N \cdot(N-1)} & \text { if } d=N_{\mathrm{t}}^{2} / N^{2} \text{ (both not treated}) \ 0 & \text { otherwise }\end{cases} $$
thereby leading to
$$ \mathbb{E}{W}\left[D{i} \cdot D_{j}\right]= \begin{cases}\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{2}} & \text { if } i=j \ -\frac{N_{\mathrm{t}} \cdot N_{\mathrm{c}}}{N^{2} \cdot(N-1)} & \text { if } i \neq j\end{cases} $$
In terms of $D_{i}$, our estimate of the average treatment effect is:
$$ \begin{align*} \bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}} & =\frac{1}{N} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{t}}} \cdot\left(D_{i}+\frac{N_{\mathrm{t}}}{N}\right) \cdot Y_{i}(1)-\frac{N}{N_{\mathrm{c}}} \cdot\left(\frac{N_{\mathrm{c}}}{N}-D_{i}\right) \cdot Y_{i}(0)\right) \ & =\frac{1}{N} \sum_{i=1}^{N}\left(Y_{i}(1)-Y_{i}(0)\right)+\frac{1}{N} \sum_{i=1}^{N} D_{i} \cdot\left(\frac{N}{N_{\mathrm{t}}} \cdot Y_{i}(1)+\frac{N}{N_{\mathrm{c}}} \cdot Y_{i}(0)\right) \ & =\tau_{f s}+\frac{1}{N} \sum_{i=1}^{N} D_{i} \cdot\left(\frac{N}{N_{\mathrm{t}}} \cdot Y_{i}(1)+\frac{N}{N_{\mathrm{c}}} \cdot Y_{i}(0)\right) \tag{A.1} \end{align*} $$
Because $\mathbb{E}{W}\left[D{i}\right]=0$ and all potential outcomes are fixed, the estimator $\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}$ is unbiased for the average treatment effect, $\tau_{\mathrm{fs}}=\bar{Y}(1)-\bar{Y}(0)$. Next, because the only random element in Equation (A.1) is $D_{i}$, the variance of $\hat{\tau}=$ $\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}$ is equal to the variance of the second term in Equation (A.1). Defining $Y_{i}^{+}=\left(N / N_{\mathrm{t}}\right) Y_{i}(1)+\left(N / N_{\mathrm{c}}\right) Y_{i}(0)$, the latter is equal to:
$$ \begin{equation*} \mathbb{V}{W}\left(\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)=\mathbb{V}{W}\left(\frac{1}{N} \sum_{i=1}^{N} D_{i} \cdot Y_{i}^{+}\right)=\frac{1}{N^{2}} \cdot \mathbb{E}{W}\left[\left(\sum{i=1}^{N} D_{i} \cdot Y_{i}^{+}\right)^{2}\right] \tag{A.2} \end{equation*} $$
Expanding Equation (A.2), we get:
$$ \begin{aligned} & \mathbb{V}{W}\left(\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)=\mathbb{E}{W}\left[\frac{1}{N^{2}} \sum_{i=1}^{N} \sum_{j=1}^{N} D_{i} D_{j} Y_{i}^{+} Y_{j}^{+}\right] \ & =\frac{1}{N^{2}} \sum_{i=1}^{N}\left(Y_{i}^{+}\right)^{2} \cdot \mathbb{E}{W}\left[D{i}^{2}\right]+\frac{1}{N^{2}} \sum_{i=1}^{N} \sum_{j \neq i} \mathbb{E}{W}\left[D{i} \cdot D_{j}\right] \cdot Y_{i}^{+} \cdot Y_{j}^{+} \ & =\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{4}} \sum_{i=1}^{N}\left(Y_{i}^{+}\right)^{2}-\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{4} \cdot(N-1)} \sum_{i=1}^{N} \sum_{j \neq i} Y_{i}^{+} \cdot Y_{j}^{+} \ & =\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(Y_{i}^{+}\right)^{2}-\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{4} \cdot(N-1)} \sum_{i=1}^{N} \sum_{j=1}^{N} Y_{i}^{+} \cdot Y_{j}^{+} \ & =\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(Y_{i}^{+}-\overline{Y^{+}}\right)^{2} \ & =\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{t}}} \cdot Y_{i}(1)+\frac{N}{N_{\mathrm{c}}} \cdot Y_{i}(0)-\left(\frac{N}{N_{\mathrm{t}}} \cdot \bar{Y}(1)+\frac{N}{N_{\mathrm{c}}} \cdot \bar{Y}(0)\right)\right)^{2}\ & = \frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{t}}} \cdot Y_{i}(1)-\frac{N}{N_{\mathrm{t}}} \cdot \bar{Y}(1)\right)^{2} \ &~~~ +\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{c}}} \cdot Y_{i}(0)-\frac{N}{N_{\mathrm{c}}} \cdot \bar{Y}(0)\right)^{2} \ &~~~ +\frac{2 \cdot N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{3} \cdot(N-1)} \sum_{i=1}^{N}\left(\frac{N}{N_{\mathrm{t}}} \cdot Y_{i}(1)-\frac{N}{N_{\mathrm{t}}} \cdot \bar{Y}(1)\right) \cdot\left(\frac{N}{N_{\mathrm{c}}} \cdot Y_{i}(0)-\frac{N}{N_{\mathrm{c}}} \cdot \bar{Y}(0)\right) \ &= \frac{N_{\mathrm{c}}}{N \cdot N_{\mathrm{t}} \cdot(N-1)} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right)^{2}+\frac{N_{\mathrm{t}}}{N \cdot N_{\mathrm{c}} \cdot(N-1)} \sum_{i=1}^{N}\left(Y_{i}(0)-\bar{Y}(0)\right)^{2} \ &~~~ +\frac{2}{N \cdot(N-1)} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right) \cdot\left(Y_{i}(0)-\bar{Y}(0)\right) \tag{A.3} \end{aligned} $$
Recall the definition of $S_{t c}^{2}$, which implies that
$$ \begin{aligned} & S_{t c}^{2}= \frac{1}{N-1} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)-\left(Y_{i}(0)-\bar{Y}(0)\right)\right)^{2} \ &= \frac{1}{N-1} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right)^{2}+\frac{1}{N-1} \sum_{i=1}^{N}\left(Y_{i}(0)-\bar{Y}(0)\right)^{2} \ & -\frac{2}{N-1} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right) \cdot\left(Y_{i}(0)-\bar{Y}(0)\right) \ &= S_{t}^{2}+S_{c}^{2}-\frac{2}{N-1} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right) \cdot\left(Y_{i}(0)-\bar{Y}(0)\right) \end{aligned} $$
Hence, the expression in (A.3) is equal to
$$ \begin{aligned} \mathbb{V}{W}\left(\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)= & \frac{N{\mathrm{c}}}{N \cdot N_{\mathrm{t}}} \cdot S_{t}^{2}+\frac{N_{\mathrm{t}}}{N \cdot N_{\mathrm{c}}} \cdot S_{c}^{2} \ & +\frac{1}{N} \cdot\left(S_{t}^{2}+S_{c}^{2}-S_{t c}^{2}\right)=\frac{S_{t}^{2}}{N_{\mathrm{t}}}+\frac{S_{c}^{2}}{N_{\mathrm{c}}}-\frac{S_{t c}^{2}}{N} \end{aligned} $$
Now we investigate the bias of the Neyman estimator for the sampling variance, $\mathbb{V}{\text {neyman }}$, under the assumption of a constant treatment effect. Assuming a constant treatment effect, $S{t c}^{2}$ is equal to zero, so we need only find unbiased estimators for $S_{c}^{2}$ and $S_{t}^{2}$ to provide an unbiased estimator of the variance of $\bar{Y}{\mathrm{t}}^{\mathrm{obs}}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}$. Consider the estimator
$$ s_{t}^{2}=\frac{1}{N_{\mathrm{t}}-1} \sum_{i: W_{i}=1}\left(Y_{i}^{\mathrm{obs}}-\bar{Y}_{\mathrm{t}}^{\mathrm{obs}}\right)^{2} $$
The goal is to show that the expectation of $s_{t}^{2}$ is equal to
$$ S_{t}^{2}=\frac{1}{N-1} \sum_{i=1}^{N}\left(Y_{i}(1)-\bar{Y}(1)\right)^{2}=\frac{N}{N-1}\left(\overline{Y^{2}}(1)-(\bar{Y}(1))^{2}\right) $$
First,
$$ \begin{align*} s_{t}^{2} & =\frac{1}{N_{\mathrm{t}}-1} \sum_{i=1}^{N} 1\left{W_{i}=1\right} \cdot\left(Y_{i}^{\mathrm{obs}}-\bar{Y}{\mathrm{t}}^{\mathrm{obs}}\right)^{2} \ & =\frac{1}{N{\mathrm{t}}-1} \sum_{i=1}^{N} 1\left{W_{i}=1\right} \cdot\left(Y_{i}(1)-\bar{Y}{\mathrm{t}}^{\mathrm{obs}}\right)^{2} \ & =\frac{1}{N{\mathrm{t}}-1} \sum_{i=1}^{N} 1\left{W_{i}=1\right} \cdot Y_{i}^{2}(1)-\frac{N_{\mathrm{t}}}{N_{\mathrm{t}}-1}\left(\bar{Y}_{\mathrm{t}}^{\mathrm{obs}}\right)^{2} \tag{A.4} \end{align*} $$
Consider the expectation of the two terms in (A.4) in turn. Using again $D_{i}=\mathbf{1}{W{i}=1}-$ $N_{\mathrm{t}} / N$, with $\mathbb{E}\left[D_{i}\right]=0$, we have
$$ \begin{aligned} \mathbb{E}\left[\frac{1}{N_{\mathrm{t}}-1} \sum_{i=1}^{N} \mathbf{1}{W{i}=1} \cdot Y_{i}^{2}(1)\right] & =\frac{1}{N_{\mathrm{t}}-1} \sum_{i=1}^{N} \mathbb{E}\left[\left(D_{i}+\frac{N_{\mathrm{t}}}{N}\right) \cdot Y_{i}^{2}(1)\right] \ & =\frac{N_{\mathrm{t}}}{N_{\mathrm{t}}-1} \cdot \overline{Y^{2}}(1) \end{aligned} $$
Next, the expectation of the second factor in the second term in (A.4):
$$ \begin{aligned} \mathbb{E}{W} & {\left[\left(\bar{Y}{\mathrm{t}}^{\mathrm{obs}}\right)^{2}\right]=\mathbb{E}{W}\left[\frac{1}{N{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j=1}^{N} W_{i} \cdot W_{j} \cdot Y_{i}^{\mathrm{obs}} \cdot Y_{j}^{\mathrm{obs}}\right] } \ = & \mathbb{E}{W}\left[\frac{1}{N{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j=1}^{N} W_{i} \cdot W_{j} \cdot Y_{i}(1) \cdot Y_{j}(1)\right] \ = & \frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j=1}^{N} \mathbb{E}{W}\left[\left(D{i}+\frac{N_{t}}{N}\right) \cdot\left(D_{j}+\frac{N_{\mathrm{t}}}{N}\right) \cdot Y_{i}(1) \cdot Y_{j}(1)\right] \ = & \frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j=1}^{N} Y_{i}(1) \cdot Y_{j}(1) \cdot\left(\mathbb{E}\left[D_{i} \cdot D_{j}\right]+\frac{N_{\mathrm{t}}^{2}}{N^{2}}\right) \ = & \frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} Y_{i}^{2}(1) \cdot\left(\mathbb{E}{W}\left[D{i}^{2}\right]+\frac{N_{\mathrm{t}}^{2}}{N^{2}}\right) \ & +\frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j \neq i} Y_{i}(1) \cdot Y_{j}(1) \cdot\left(\mathbb{E}{W}\left[D{i} \cdot D_{j}\right]+\frac{N_{t}^{2}}{N^{2}}\right)\ = & \frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} Y_{i}^{2}(1) \cdot\left(\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{2}}+\frac{N_{\mathrm{t}}^{2}}{N^{2}}\right) \ & +\frac{1}{N_{\mathrm{t}}^{2}} \sum_{i=1}^{N} \sum_{j \neq i} Y_{i}(1) \cdot Y_{j}(1) \cdot\left(-\frac{N_{\mathrm{c}} \cdot N_{\mathrm{t}}}{N^{2} \cdot(N-1)}+\frac{N_{\mathrm{t}}^{2}}{N^{2}}\right) \ = & \frac{1}{N_{\mathrm{t}}} \cdot \overline{Y^{2}}(1)+\frac{N_{\mathrm{t}}-1}{N \cdot(N-1) \cdot N_{\mathrm{t}}} \sum_{i=1}^{N} \sum_{j \neq i} Y_{i}(1) \cdot Y_{j}(1) \ = & \frac{1}{N_{\mathrm{t}}} \cdot \overline{Y^{2}}(1)-\frac{N_{\mathrm{t}}-1}{N \cdot(N-1) \cdot N_{\mathrm{t}}} \sum_{i=1}^{N} Y_{i}^{2}(1)+\frac{N_{\mathrm{t}}-1}{N \cdot(N-1) \cdot N_{\mathrm{t}}} \sum_{i=1}^{N} \sum_{j=1}^{N} Y_{i}(1) \cdot Y_{j}(1) \ = & \frac{1}{N_{\mathrm{t}}} \cdot \overline{Y^{2}}(1)-\frac{N_{\mathrm{t}}-1}{(N-1) \cdot N_{\mathrm{t}}} \cdot \overline{Y^{2}}(1)+\frac{\left(N_{\mathrm{t}}-1\right) \cdot N}{(N-1) \cdot N_{\mathrm{t}}}(\bar{Y}(1))^{2} \ = & \frac{N_{\mathrm{c}}}{N_{\mathrm{t}} \cdot(N-1)} \cdot \overline{Y^{2}}(1)+\frac{\left(N_{\mathrm{t}}-1\right) \cdot N}{(N-1) \cdot N_{\mathrm{t}}}(\bar{Y}(1))^{2} \cdot \end{aligned} $$
Hence, the expectation of the second term in (A.4) equals
$$ -\frac{N_{\mathrm{c}}}{\left(N_{\mathrm{t}}-1\right) \cdot(N-1)} \cdot \overline{Y^{2}}(1)+\frac{N}{(N-1)} \cdot(\bar{Y}(1))^{2} $$
and adding up the expectations of both terms in in (A.4) leads to
$$ \begin{aligned} \mathbb{E}{W}\left[s{t}^{2}\right] & =\frac{N_{\mathrm{t}}}{N_{\mathrm{t}}-1} \cdot \overline{Y^{2}}(1)-\frac{N_{\mathrm{c}}}{\left(N_{\mathrm{t}}-1\right) \cdot(N-1)} \cdot \overline{Y^{2}}(1)-\frac{N}{(N-1)} \cdot(\bar{Y}(1))^{2} \ & =\frac{N}{N-1} \cdot \overline{Y^{2}}(1)-\frac{N}{(N-1)} \cdot(\bar{Y}(1))^{2}=S_{t}^{2} \end{aligned} $$
Following the same argument,
$$ \mathbb{E}{W}\left[s{c}^{2}\right]=\frac{1}{N_{c}-1} \cdot \mathbb{E}{W}\left[\sum{i=1}^{N}\left(1-W_{i}\right) \cdot\left(Y_{i}^{\text {obs }}-\bar{Y}{\mathrm{c}}^{\mathrm{obs}}\right)^{2}\right]=S{c}^{2} $$
Hence, the estimators $s_{c}^{2}$ and $s_{t}^{2}$ are unbiased for $S_{c}^{2}$ and $S_{t}^{2}$, and can be used to create an unbiased estimator for the variance of $\bar{Y}{\mathrm{t}}^{\text {obs }}-\bar{Y}{\mathrm{c}}^{\text {obs }}$, our estimator of the average treatment effect under the constant treatment effect assumption.