Horvitz–Thompson estimator

Remarks

The estimator is unbiased: write $\hat{Y}_{HT} = \sum_{i=1}^N \frac{y_i}{\pi_i} \mathbf{1}[i \in S]$, and take the expected value, using $E(\mathbf{1}[i \in S]) = \pi_i$, to get $E(\hat{Y}_{HT}) = Y$.

Example: in a simple random sample of size $n$ drawn without replacement, every unit has $\pi_i = n/N$. Then $\hat{Y}_{HT} = \frac{N}{n}\sum_{i \in S} y_i = N \bar{y}$, where $\bar{y}$ is the sample mean.

Non-example: let some unit $j$ have $\pi_j = 0$. Then it never appears in the sample, so no estimator built from the sample can be unbiased for every choice of $y_j$. The condition $\pi_i > 0$ fails, and the definition does not apply.

With the joint inclusion probabilities $\pi_{ij} = P(i \in S, j \in S)$ and $\pi_{ii} = \pi_i$, the variance is $$\operatorname{Var}(\hat{Y}_{HT}) = \sum_{i=1}^N \sum_{j=1}^N (\pi_{ij} - \pi_i \pi_j) \frac{y_i}{\pi_i} \frac{y_j}{\pi_j}.$$