CI-HW4

HW4.

Problem 1

1-1

$Y_{i}(0)$ is the potential outcome of unit $i$ if it doesn't receive treatment.

$Y_{i}^{obs}$ is the observed outcome of unit $i$. Whether it is $Y_{i}(1)$ or $Y_{i}(0)$ depends on the assignment.

1-2

It represents the expected outcome for individuals who received treatment, had they not received the treatment.

1-3

$\mathbb{E}[Y_{i}(0)]$ is the expected potential outcome of unit $i$ under no treatment.

While $\mathbb{E}[Y_{i}^{obs}\mid W_{i}=0]=\mathbb{E}[Y_{i}(0)\mid W_{i}=0]$ is the expected potential outcome of unit $i$ under no treatment, conditional on the unit actually being assigned to the control group.

1-4

$\mathbb{E}[Y_{i}(0)\mid W_{i}=1]$: the expected potential outcome of $i$ under no treatment, conditioning on it being assigned to the treatment group.

$\mathbb{E}[Y_{i}(0)\mid W_{i}=0]$: the expected potential outcome of $i$ under no treatment, conditioning on it being assigned to the control group.

1-5

$\mathbb{E}[Y_{i}^{obs}\mid W_{i}=0]$ and $\mathbb{E}[Y_{i}(0)\mid W_{i}=0]$.

1-6

This assignment guarantees that

$$\overrightarrow{W}\perp (\boldsymbol{Y}(0),\boldsymbol{Y}(0))$$

Therefore

$$p_{i}(\boldsymbol{Y}(0),\boldsymbol{Y}(1))=P(W_{i}=1)=.5$$

and

$$P(\overrightarrow{W}\mid \boldsymbol{Y}(0),\boldsymbol{Y}(1))=P(\overrightarrow{W})=\begin{cases} 1/C_{n}^{n/2} & \text{if}\ \overrightarrow{1}\cdot\overrightarrow{W}=\frac{n}{2} \\ 0 & \text{o.w.} \end{cases}$$

So unconfounded assumption is satisfied.

And because $1/C_{n}^{n/2}$ can be expressed as

$$c\cdot \prod_{i=1}^{n}0.5 = c\cdot \prod_{i=1}^{n} (.5)^{W_{i}}(1-.5)^{1-W_{i}}$$

So $\overrightarrow{W}$ satisfies the individualistic assumption.

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All quantities in (5) can be identified.

Problem 2

2-1

I prefer a small puncture procedure. Because the successful rate is higher.

2-2

I would choose the open surgical procedure, because it has a higher success rate in both stone-size groups

The paradox arises because the proportion of small-stone cases differs between the two procedures. Small stones are easier to treat and have higher success rates, and they make up a much larger share of the puncture group than of the open-surgery group. As a result, small-stone cases dominate the overall success rate of the puncture procedure and inflate it, even though open surgery performs better within each stone-size group.

2-3

It is possible. Because $W_{i}$ can still be influenced by $X_{i}$. Patients with larger stone are more likely to be assigned to open surgical procedure.

$W_{i}\perp (Y_{i}(0),Y_{i}(1))$ is enough to justify

$$\hat{\tau}^{\mathrm{dif}}=\bar{Y}^{obs}(1)-\bar{Y}^{obs}(0)$$

is valid for estimating $\tau_{fs}$ and $\tau_{sp}$.

The result is $\hat{\tau}^{\mathrm{dif}}=-0.0457$.

It doesn't conflict with my results in (2), because (2) is saying that

$$\mathbb{E}[Y_{i}^{obs}\mid W_{i}=1,X=x]-\mathbb{E}[Y_{i}^{obs}\mid W_{i}=0,X=x]>0$$

for $x=1,0$.

Problem 3

• It is from a random experiment.
• It is reasonable
• Because it is a complete randomization. $\boldsymbol{W}$ is independent of all of the covariates and potential outcomes.