Joint inclusion probability (${\pi }_{ij}=\mathbb{P}(i,j\in S)$)

A joint inclusion probability is the probability that two specified population units both belong to a random sample $S$:For the same unit twice, ${\pi }_{ii}={\pi }_i$, where ${\pi }_i$ is its inclusion probability. For distinct units, ${\pi }_{ij}$ is generally not ${\pi }_i{\pi }_j$: selection can make their inclusion indicators dependent. These probabilities describe the sampling procedure, rather than a relationship between the measured values of the two units.^[ref]

For simple random sampling of $n$ distinct labels from $N>1$ labels, counting subsets containing both $i$ and $j$ giveswith value zero if $n<2$. Since ${\pi }_i=n/N$,For $N=4,n=2$, each unit has inclusion probability $1/2$, but a particular pair has joint inclusion probability $1/6$. Conditional on including one member, the other is included with probability $(1/6)/(1/2)=1/3$, not $1/2$.

Equal individual inclusion probabilities do not determine these pair probabilities. A different procedure that chooses only $\{1,2\}$ or $\{3,4\}$, each with probability $1/2$, still includes each unit with probability $1/2$. Yet ${\pi }_{12}=1/2$ and ${\pi }_{13}=0$. Thus two procedures can agree on individual representation while producing different dependence and different precision for a population summary. The distinction motivates keeping pair information when calculating the variance of weighted survey estimators; the classical general treatment is due to D. G. Horvitz and D. J. Thompson.^[ref]

Pairwise inclusion positivity requires every distinct pair of population units to have positive joint inclusion probability under the sampling design:For Horvitz-Thompson variance estimator, the probabilities must also be known, and individual inclusion probabilities must be positive. Each pair can then, at least in principle, contribute an observed product to a pair-expansion estimate. Positivity is not independence: ${\pi }_{ij}$ need not equal ${\pi }_i{\pi }_j$.^[ref]

For simple random sampling of $n$ distinct labels from $N\ge 2$, ${\pi }_{ij}=n(n-1)/\{N(N-1)\}$. Consequently, $n\ge 2$ gives pairwise positivity, whereas selecting exactly one label gives positive individual inclusion probabilities but zero probability of ever observing two distinct values together. This distinction separates unbiased total estimation from unbiased estimation of its sampling variance.

An original two-unit proof shows what information is missing. Sample one of two fixed values $y_1,y_2$ uniformly. The Horvitz-Thompson estimator is $2y_1$ or $2y_2$, and its variance isAny estimator based only on the observed label and value has expected value of the form $\{g_1(y_1)+g_2(y_2)\}/2$. Its mixed difference over $(y_1,y_2)\in \{0,1{\}}^2$ is zero. The target's mixed difference is insteadNo such estimator can therefore be unbiased for this variance for every possible pair of values. Additional randomization internal to the estimator does not help: conditioning on the observed data still produces functions of this same separated form. This is an information limitation, not division by zero that can be repaired by assigning a convenient numerical convention.^[ref]

Strict positivity alone also does not guarantee useful precision. A very small ${\pi }_{ij}$ produces a large inverse-probability multiplier on a rare observed pair. Designing a usable survey therefore requires attention to both observability and the resulting sampling variance.