 Horvitz-Thompson variance estimator

The Horvitz-Thompson variance estimator estimates the sampling variance of a Horvitz-Thompson estimator using observed values and known individual and joint inclusion probabilities. It is a random estimate of a fixed design-dependent variance, rather than the variance itself.

Let $I_i$ be an inclusion indicator, ${\pi }_i={\mathbb{E}}_p{I}_i>0$, and ${\pi }_{ij}={\mathbb{E}}_p(I_i{I}_j)$, with ${\pi }_{ii}={\pi }_i$. For fixed population values $y_i$, write $a_i=y_i/{\pi }_i$. Bilinearity of covariance givesUnder pairwise inclusion positivity, a sample $S$ provides the estimatorThe diagonal terms are included. Taking expected values replaces $I_i{I}_j/{\pi }_{ij}$ by 1, proving ${\mathbb{E}}_p{\hat{V}}_{\mathrm{HT}}=V_p$. This is the pairwise counterpart of inverse-probability expansion for the finite population total.^[ref]

An original exact example uses fixed values $(3,0,1)$. Select $\{1,2\},\{1,3\},\{2,3\}$ with probabilities $1/2,1/4,1/4$, respectively. Then $({\pi }_1,{\pi }_2,{\pi }_3)=(3/4,3/4,1/2)$, and the possible total estimates are $4,6,2$. Their expected value is the true total 4, with varianceFor those same samples, the variance estimates are $4,-2,2$. Their expected value is also 2. In particular, for $S=\{1,3\}$, the two diagonal contributions are 4 and 2, while the combined off-diagonal contribution is $-8$, giving $-2$.

Thus this unbiased estimator of variance can be negative in a realized sample. Replacing a negative result by zero changes its expected value; in this example that modification raises the mean estimate to $5/2$. A positive square root is therefore not always available as a sample standard error from this particular formula. If some ${\pi }_{ij}=0$, dropping an unobservable cross-product does not preserve the proof; the missing term needs separate analysis.