CI-HW3

HW3.

Problem 1(Regression approach)

a.

$\hat{\tau}=15.0658$

$[3.5060532, 26.625451]$.

b.

$\hat{\tau}=15.0658$

$[2.1225326, 28.008972]$

c.

Compare with (d): the estimate is different and the confidence interval is a bit wider. The reason is that they compute different things. (d) is $Y-X$.

Compare with (f): Adding covariates changed the estimation of $\tau$, as well as the variance of the estimator.

Problem 2 (Model-based Analysis)

a.

$\tau_{fs}:$

1. $\tau_{fs}=\tau(\boldsymbol{Y}^{obs},\boldsymbol{Y}^{mis},\boldsymbol{W})=h(\boldsymbol{Y}(1),\boldsymbol{Y}(0),\boldsymbol{W})$.
2. $f(\tau\mid\boldsymbol{Y}^{obs},\boldsymbol{W})$.
3. $f(\boldsymbol{Y}^{miss}\mid \boldsymbol{Y}^{obs},\boldsymbol{W})=\int _{\theta}f(\boldsymbol{Y}^{miss}\mid \boldsymbol{Y}^{obs},\boldsymbol{W},\theta)p(\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W})d\theta$.
4. $f(\boldsymbol{Y}^{mis},\boldsymbol{Y}^{obs}\mid \boldsymbol{W},\theta),f(\boldsymbol{Y}^{mis},\boldsymbol{Y}^{obs},\boldsymbol{W}\mid \theta)$.

$\tau_{sp}:$

1. $\tau_{sp}=\mathbb{E}[\dots]=\tau(\theta)$.
2. $f(\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W})= \dfrac{p(\theta)f(\boldsymbol{Y}^{obs},\boldsymbol{W}\mid \theta)}{f(\boldsymbol{Y}^{obs},\boldsymbol{W})}$.

b.

$\tau_{sp}$:

$$\begin{align} f(\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W}) & \propto f(\boldsymbol{Y}^{obs}\mid \boldsymbol{W},\theta)p(\theta) \\ & \propto f(Y_{1}(0),Y_2(0),Y_3(1),Y_4(1)\mid \theta) \quad (p(\theta)\propto 1)\\ & = f(Y_1(0)\mid \mu_c)\dots f(Y_4(1)\mid \mu_t) \\ & \propto \exp (\frac{-(\mu_{c}-2)^{2}-(\mu_{c}-6)^{2}}{2})\cdot \exp(\frac{-(\mu_{t}-7)^{2}-(\mu_{t}-5)^{2}}{2}) \\ & = \exp\left( -(\mu_{c}-4)^{2}+* \right)\cdot \exp\left( -(\mu_{t}-6)^{2}+* \right) \\ & \propto \exp\left( -(\mu_{c}-4)^{2} \right)\cdot \exp\left( -(\mu_{t}-6)^{2} \right) \end{align}$$

So

$$\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W}\sim N \left( \begin{pmatrix} 4 \\ 6 \end{pmatrix}, \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \right)$$

Therefore

$$\mathbb{E}[\tau_{sp}\mid\boldsymbol{Y}^{obs},\boldsymbol{W}]=\mathbb{E}[\mu_{t}-\mu_{c}\mid\boldsymbol{Y}^{obs},\boldsymbol{W}]=6-4=2$$

---

$\tau_{fs}$:

$$\begin{align} f(\boldsymbol{Y}^{miss}\mid \boldsymbol{Y}^{obs},\boldsymbol{W}) & =\int _{\theta}f(\boldsymbol{Y}^{miss}\mid \boldsymbol{Y}^{obs},\boldsymbol{W},\theta)p(\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W})d\theta \\ & = \int _{\theta} f(\boldsymbol{Y}^{mis}\mid \theta)p(\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W})d\theta\quad (\text{independence}) \\ \end{align}$$

Note that $\boldsymbol{Y}^{mis}=(Y_1(1),Y_2(1),\dots,Y_4(0))$, so

$$\boldsymbol{Y}^{mis}\mid \theta \sim N \left( \begin{pmatrix} \mu_{t} \\ \mu_{t} \\ \mu_{c} \\ \mu_{c} \end{pmatrix}, \begin{pmatrix} 1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{pmatrix} \right)$$

and we know from above

$$\theta\mid \boldsymbol{Y}^{obs},\boldsymbol{W}\sim N \left( \begin{pmatrix} 4 \\ 6 \end{pmatrix}, \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \right)$$

With the knowledge of multivariate normal distribution (I don't have that knowledge though), we can get

$$Y^{mis} \mid Y^{obs}, W \sim N_4 \left( \begin{pmatrix} 6 \\ 6 \\ 4 \\ 4 \end{pmatrix}, \begin{pmatrix} 3/2 & 1/2 & 0 & 0 \\ 1/2 & 3/2 & 0 & 0 \\ 0 & 0 & 3/2 & 1/2 \\ 0 & 0 & 1/2 & 3/2 \end{pmatrix} \right)$$

So

$$\mathbb{E}[\tau_{fs}\mid Y^{obs}, W ] = \dots = 2$$

Problem 3 (Treatment Effect Heterogeneity)

1.

$S^{2}(\tau)=0$ means $Y_i(1)-Y_i(0)$ is constant, which leads to $S^{2}(1)=S^{2}(0)$.

2.

$Y_i(1)$	$Y_i(0)$
1	2
2	1
3	3

3.

$\tau_{i}=Y_{i}(1)-Y_{i}(0)\implies Y_{i}(1)=Y_{i}(0)+\tau_{i}$.

Take the variance of both sides, we have

$$S^{2}(1)=S^{2}(0)+S^{2}(\tau)+2S(Y(0),\tau)\implies 2S(Y(0),\tau)=S^{2}(1)-S^{2}(0)-S^{2}(\tau)$$

So if $S^{2}(1)<S^{2}(0)$, $S(Y(0),\tau)<0$.

4.

$Y_{i}(1)$	$Y_{i}(0)$	$\tau_{i}$
-100	1000	-1100
1	1000	-999
0	999	-999