CI-HW5

Problem 1 (SRE vs CRE based on Neyman’s estimator)

The estimator for SRE is

$$\hat{\tau}^{\mathrm{strat}}=q\hat{\tau}(f)+(1-q)\hat{\tau}(m)$$

Its variance is

$$\begin{align} \mathbb{V}_{sp}[\hat{\tau}^{\mathrm{strat}}] & = \mathbb{V}_{sp}[q\cdot+(1-q)\cdot] \\ & = q^{2}\mathbb{V}_{sp}[\cdot]+(1-q)^{2}\mathbb{V}_{sp}[\cdot] \\ & = \frac{q}{N}\left( \frac{\sigma_{t}^{2}(f)}{p}+\frac{\sigma_{c}^{2}(f)}{1-p} \right) + \frac{1-q}{N}\left( \frac{\sigma^{2}_{t}(m)}{p}+ \frac{\sigma^{2}_{c}(m)}{1-p}\right) \end{align}$$

The estimator for CRE is

$$\hat{\tau}^{\mathrm{dif}}=\bar{Y}_{t}^{\mathrm{obs}}-\bar{Y}_{c}^{\mathrm{obs}}$$

Its variance is

$$\begin{align} \mathbb{V}_{sp}[\hat{\tau}^{\mathrm{dif}}] & = \frac{\sigma^{2}_{t}}{N_{t}}+\frac{\sigma^{2}_{c}}{N_{c}} \\ & = \frac{q\sigma^{2}_{t}(f)+(1-q)\sigma_{t}^{2}(m)+q(1-q)[\mu_{t}(f)-\mu_{t}(m)]^{2}}{N_{t}}+\frac{\dots}{N_{c}} \\ \end{align}$$

So

$$\mathbb{V}_{sp}[\hat{\tau}^{\mathrm{dif}}]-\mathbb{V}[\hat{\tau}^{\mathrm{strat}}]=\frac{q(1-q)[\mu_{t}(f)-\mu_{t}(m)]^{2}}{Np}+\frac{q(1-q)[\mu_{c}(f)-\mu_{c}(m)]^{2}}{N(1-p)}\geq 0$$

Problem 2 (PRE vs SRE based on Neyman’s estimator)

2-1

For SRE, the estimator is

$$\hat{\tau}^{\mathrm{S}}=\frac{6}{16}\hat{\tau}_{X>16}+\frac{10}{16}\hat{\tau}_{X\leq 16}=7.8875$$

For PRE, the estimator is

$$\hat{\tau}^{\mathrm{P}}=\frac{2}{16}\sum_{j=1}^{8}\hat{\tau}^{\mathrm{pair}}(j)=13.425$$

They are not the same.

Both estimators are unbiased because treatment is randomly assigned within each group, and both are weighted averages of the within-group differences in means, with weights based on group sizes.

2-2

I think the estimator for PRE has smaller variance. Because the gain in precision comes from the prediction capability for potential outcomes of the covariates. Dividing samples into more groups can tap the prediction capability of the covariates to the full.

Problem 3

3-1

1. Valid. Because $N_{t}(f)/N(f)=N_{t}(m)/N(m)$, $\hat{\tau}^{\mathrm{dif}}=\hat{\tau}^{\mathrm{strat}}$.
2. Valid. Adding covariates doesn't break consistency.
3. Valid. It's equivalent to 1.
4. Valid. The reason is the same as 1.
5. Valid. It's equivalent to 4.
6. Valid. Adding covariates doesn't break consistency.
7. Valid. Adding covariates doesn't break consistency.

3-2

First we have

$$\mathbb{V}_{1}=\mathbb{V}_{3},\ \mathbb{V}_{4}=\mathbb{V}_{5}$$

and

$$\mathbb{V}_{2}= \mathbb{V}_{3},\ \mathbb{V}_{4}\leq \mathbb{V}_{1}, \ \mathbb{V}_{6}\leq \mathbb{V}_{4}, \ \mathbb{V}_{7}= \mathbb{V}_{4}$$

so

$$\mathbb{V}_{6}\leq\mathbb{V}_{4}=\mathbb{V}_{5}=\mathbb{V}_{7}\leq\mathbb{V}_{1}=\mathbb{V}_{2}=\mathbb{V}_{3}$$