Finite populations and design probability — Examples

Let $a$ be the number of $n$-subsets containing a fixed unit. A permutation of $U$ taking one unit to another gives a bijection of the sample family, so every unit occurs in exactly $a$ samples. Counting pairs $(s,i)$ with $i\in s$ in two ways gives $Na=nM$, hence $a\mathrm{/}M=n\mathrm{/}N$ and ${\pi }_i=n\mathrm{/}N$.

Let $b$ be the number of samples containing a fixed ordered pair of distinct units. Permutations of $U$ give the same $b$ for every such pair. Counting triples $(s,i,j)$ with $i\mathrm{\ne }j$ and $i,j\in s$ gives $N(N-1)b=n(n-1)M$, so $b\mathrm{/}M=n(n-1)\mathrm{/}(N(N-1))$ and ${\pi }_{ij}=n(n-1)\mathrm{/}(N(N-1))$. This also gives ${\pi }_{ij}=0$ when $n=1$.

Put $A_i=\{s:i\in s\}$. By [F5] and [F6], $E_d[I_i]=P_d(A_i)={\pi }_i$ and $E_d[I_i{I}_j]=P_d(A_i\cap A_j)={\pi }_{ij}$. Expanding the covariance in [F3] with linearity [F4] therefore gives ${\mathrm{Cov}}_d(I_i,I_j)={\pi }_{ij}-{\pi }_i{\pi }_j=\frac{n(n-1)}{N(N-1)}-\frac{n^2}{N^2}=-\frac{n(N-n)}{N^2(N-1)}$.

Since ${\pi }_i=n\mathrm{/}N>0$, the estimator in [F2] has ${\widehat{\bar{y}}}_{HT}(s,y{\mathrm{\mid }}_s)=\frac{1}{N}{\sum }_{i\in s}\frac{y_i}{n\mathrm{/}N}=\frac{1}{n}{\sum }_{i\in s}{y}_i$. Its expectation is ${\bar{y}}_U$ by [F2].

Under the two-sample design, unit $1$ is in both support samples, unit $2$ only in the first, and unit $3$ only in the second. Therefore ${\pi }_1=1$ and ${\pi }_2={\pi }_3=1\mathrm{/}2$. The weighted total is $a+2b$ for sample $\{1,2\}$ and $a+2c$ for sample $\{1,3\}$, so unit $1$ contributes its observed value $a$ without expansion.

The design expectation of the displayed total is $\frac{1}{2}(a+2b)+\frac{1}{2}(a+2c)=a+b+c=T(y)$ by [F3]. Since each ${\pi }_i>0$, [F2] also identifies its divided-by-$3$ form as the unbiased estimator of ${\bar{y}}_U$.

In the census design, the only supported sample is $U$ with probability $1$, so every ${\pi }_i=1$. Thus the total estimator is ${\sum }_{i\in U}{y}_i=T(y)$ on the sole outcome, and its mean version is $T(y)\mathrm{/}3={\bar{y}}_U$.

The only supported sample is $\{1\}$, so ${\pi }_2=0$. Under both $y$ and $y^{\mathrm{\prime }}$, the record is $(\{1\},0)$, but $T(y)=0$ and $T(y^{\mathrm{\prime }})=1$.

Any observed-record estimator has one value $d$ at $(\{1\},0)$, so its design expectation is $d$ under both populations. It cannot equal both totals $0$ and $1$. This explicit calculation verifies the instance asserted by [F2], using the unbiasedness definition [F4].