Analysis of Probability Proportional to Size (PPS) Sampling Statements

Evaluating Statement A: Unbiased Estimator

The question concerns Probability Proportional to Size (PPS) sampling with replacement for a sample size $n=2$. A standard estimator for the population total $Y = \sum_{i=1}^{N} y_i$ in this design is:

$ \hat{Y} = \frac{1}{n} \sum_{i \in s} \frac{y_i}{p_i} $

Where $s$ is the sample, $y_i$ is the value of the study variable for unit $i$, and $p_i$ is the probability of selecting unit $i$ in a single draw. The estimator $T$ proposed is:

$ T = \left(\frac{1}{2}\right) \sum_{i \in s} \frac{y_i}{p_i} $

With $n=2$, this matches the form $\hat{Y}$. This estimator is known to be unbiased for the population total $Y$ under PPS sampling with replacement.

Conclusion: Statement A is correct.

Evaluating Statement B: Inclusion Probabilities

The inclusion probability $\pi_i$ for unit $i$ in a sample of size $n=2$ drawn with replacement is the probability that unit $i$ is selected in at least one of the two draws. It is calculated as:

$ \pi_i = 1 - P(\text{unit } i \text{ not selected in draw 1}) \times P(\text{unit } i \text{ not selected in draw 2}) $

$ \pi_i = 1 - (1 - p_i) \times (1 - p_i) = 1 - (1 - p_i)^2 $

Using the given probabilities $p_1 = 0.2$ and $p_2 = 0.3$:

• For unit 1: $\pi_1 = 1 - (1 - 0.2)^2 = 1 - (0.8)^2 = 1 - 0.64 = 0.36$.
• For unit 2: $\pi_2 = 1 - (1 - 0.3)^2 = 1 - (0.7)^2 = 1 - 0.49 = 0.51$.

These calculated values match the statement.

Conclusion: Statement B is correct.

Evaluating Statement C: Joint Inclusion Probability

The joint inclusion probability $\pi_{ij}$ is the probability that both unit $i$ and unit $j$ ($i \ne j$) are included in the sample of size $n=2$ drawn with replacement. This occurs if unit $i$ is drawn first and unit $j$ second, OR if unit $j$ is drawn first and unit $i$ second.

$ \pi_{ij} = P(\text{draw 1}=i, \text{draw 2}=j) + P(\text{draw 1}=j, \text{draw 2}=i) $

Due to independence in sampling with replacement:

$ \pi_{ij} = p_i p_j + p_j p_i = 2 p_i p_j $

Using $p_1 = 0.2$ and $p_2 = 0.3$:

$ \pi_{12} = 2 \times 0.2 \times 0.3 = 2 \times 0.06 = 0.12 $

This calculated value matches the statement.

Conclusion: Statement C is correct.

Evaluating Statement D: Sum of Inclusion Probabilities

We need to calculate the inclusion probabilities for units 3 and 4 using the formula $\pi_i = 1 - (1 - p_i)^2$. Given $p_3 = 0.1$ and $p_4 = 0.4$:

• $\pi_3 = 1 - (1 - 0.1)^2 = 1 - (0.9)^2 = 1 - 0.81 = 0.19$.
• $\pi_4 = 1 - (1 - 0.4)^2 = 1 - (0.6)^2 = 1 - 0.36 = 0.64$.

The sum of all inclusion probabilities is:

$ \sum_{i=1}^4 \pi_i = \pi_1 + \pi_2 + \pi_3 + \pi_4 = 0.36 + 0.51 + 0.19 + 0.64 = 1.70 $

The statement claims the sum is $2$. Since $1.70 \ne 2$, the statement is false.

Conclusion: Statement D is incorrect.

Overall Conclusion

Statements A, B, and C were found to be correct based on the principles of PPS sampling with replacement.