HT variance and SRSWOR

For each $i\in U$, set $X_i=I_i{y}_i\mathrm{/}{\pi }_i$. The design support is finite by [F1], each $I_i$ is bounded, and $y_i\mathrm{/}{\pi }_i$ is finite by the positivity assumption; hence every $X_i$ is square-integrable and ${\widehat{T}}_{HT}={\sum }_{i\in U}{X}_i$.

Apply [F5] to this finite family on the common design probability space to obtain ${\mathrm{Var}}_d({\widehat{T}}_{HT})={\sum }_{i,j\in U}{\mathrm{Cov}}_d(X_i,X_j)$.

By [F6], ${\mathrm{Cov}}_d(X_i,X_j)=\frac{y_i{y}_j}{{\pi }_i{\pi }_j}{\mathrm{Cov}}_d(I_i,I_j)$. For $A_i=\{s:i\in s\}$, $I_i={\mathbf{1}}_{A_i}$ and $I_i{I}_j={\mathbf{1}}_{A_i\cap A_j}$, so [F4] and [F1] give $E_d[I_i]={\pi }_i$ and $E_d[I_i{I}_j]={\pi }_{ij}$; expanding [F3] gives ${\mathrm{Cov}}_d(I_i,I_j)={\pi }_{ij}-{\pi }_i{\pi }_j={\mathrm{\Delta }}_{ij}$ by [F2], and substitution in step 2.1 proves the variance formula.

Under the additional pair-positivity assumption every coefficient in ${\widehat{V}}_{HT}$ is defined and every term is integrable on the finite support. Linearity over the finite index set $U\times U$ follows by induction: the empty sum has expectation zero, and adjoining one term uses the two-variable rule [F7]. For each pair, $I_i{I}_j$ is the indicator of joint inclusion by [F1], so [F4] gives $E_d[I_i{I}_j]={\pi }_{ij}$; hence $E_d[{\widehat{V}}_{HT}]={\sum }_{i,j\in U}\frac{{\mathrm{\Delta }}_{ij}}{{\pi }_{ij}}\frac{y_i{y}_j}{{\pi }_i{\pi }_j}{E}_d[I_i{I}_j]={\sum }_{i,j\in U}\frac{{\mathrm{\Delta }}_{ij}{y}_i{y}_j}{{\pi }_i{\pi }_j}={\mathrm{Var}}_d({\widehat{T}}_{HT})$ by step 3.1.

If ${\pi }_{ij}=0$, the coefficient ${\mathrm{\Delta }}_{ij}\mathrm{/}{\pi }_{ij}$ in the all-pairs expression is undefined; removing that unsupported term changes the expression, and step 4.1 no longer proves unbiasedness, without ruling out another design-specific estimator.

For each i,j, $I_i{I}_j$ indicates joint inclusion and $I_i$ indicates inclusion by [F1], so [F4] gives their expectations as ${\pi }_{ij}$ and ${\pi }_i$. Since ${\sum }_j{I}_j=n$ on every sample, finite linearity by induction using [F5] gives ${\sum }_j{\pi }_{ij}=E_d[I_i{\sum }_j{I}_j]=n{\pi }_i$ and ${\sum }_j{\pi }_i{\pi }_j={\pi }_i{E}_d[{\sum }_j{I}_j]=n{\pi }_i$; subtracting by [F2] gives ${\sum }_j{\mathrm{\Delta }}_{ij}=0$.

By [F1] and [F2], ${\mathrm{\Delta }}_{ij}={\mathrm{\Delta }}_{ji}$ and the column sums also vanish; expansion gives $-\frac{1}{2}{\sum }_{i,j}{\mathrm{\Delta }}_{ij}(x_i-x_j{)}^2=-\frac{1}{2}({\sum }_i{x}_i^2{\sum }_j{\mathrm{\Delta }}_{ij}+{\sum }_j{x}_j^2{\sum }_i{\mathrm{\Delta }}_{ij}-2{\sum }_{i,j}{\mathrm{\Delta }}_{ij}{x}_i{x}_j)={\sum }_{i,j}{\mathrm{\Delta }}_{ij}{x}_i{x}_j$, which is ${\mathrm{Var}}_d({\widehat{T}}_{HT})$ by [F3] and is nonnegative by [F6].

If every ${\pi }_{ij}>0$, each coefficient in ${\widehat{V}}_{SYG}$ is defined and every term is integrable on the finite support. Finite-sum linearity follows by induction from [F5]; by [F1], each $I_i{I}_j$ indicates joint inclusion, so [F4] gives $E_d[I_i{I}_j]={\pi }_{ij}$ and $E_d[{\widehat{V}}_{SYG}]=-\frac{1}{2}{\sum }_{i,j}\frac{{\mathrm{\Delta }}_{ij}}{{\pi }_{ij}}{E}_d[I_i{I}_j](x_i-x_j{)}^2=-\frac{1}{2}{\sum }_{i,j}{\mathrm{\Delta }}_{ij}(x_i-x_j{)}^2={\mathrm{Var}}_d({\widehat{T}}_{HT})$ by step 2.1.

On a supported sample, diagonal terms vanish and symmetry gives ${\widehat{V}}_{SYG}(s)=-{\sum }_{i<j,i,j\in s}\frac{{\mathrm{\Delta }}_{ij}}{{\pi }_{ij}}(x_i-x_j{)}^2$; if every distinct-pair ${\mathrm{\Delta }}_{ij}\le 0$, every summand is nonnegative, proving the sufficient condition.

There are $\left(\genfrac{}{}{0px}{}{N}{n}\right)$ supported samples; a fixed unit occurs in $\left(\genfrac{}{}{0px}{}{N-1}{n-1}\right)$ of them, so ${\pi }_i=n\mathrm{/}N$. If $n=1$, no sample contains distinct $i,j$; if $2\le n\le N$, $\left(\genfrac{}{}{0px}{}{N-2}{n-2}\right)$ samples contain them, hence in either case ${\pi }_{ij}=n(n-1)\mathrm{/}(N(N-1))$ for $i\mathrm{\ne }j$.

By [F1], $E_d[I_i]={\pi }_i$ and $E_d[I_i{I}_j]={\sum }_{s:i,j\in s}{p}_d(s)={\pi }_{ij}$, while ${\pi }_{ii}={\pi }_i$; expanding [F5] gives ${\mathrm{Cov}}_d(I_i,I_j)=E_d[I_i{I}_j]-E_d[I_i]E_d[I_j]={\pi }_{ij}-{\pi }_i{\pi }_j={\mathrm{\Delta }}_{ij}$. Substitution from step 1.1 gives ${\mathrm{Cov}}_d(I_i,I_j)=\frac{n(n-1)}{N(N-1)}-\frac{n^2}{N^2}=-\frac{n(N-n)}{N^2(N-1)}$ for $i\mathrm{\ne }j$.

Since ${\pi }_i=n\mathrm{/}N$, $N^{-1}{\sum }_i{I}_i{y}_i\mathrm{/}{\pi }_i={\bar{y}}_S$; by [F4], ${\mathrm{Var}}_d({\bar{y}}_S)=\frac{1}{N^2}({\sum }_i\frac{{\mathrm{\Delta }}_{ii}{y}_i^2}{{\pi }_i^2}+{\sum }_{i\mathrm{\ne }j}\frac{{\mathrm{\Delta }}_{ij}{y}_i{y}_j}{{\pi }_i^2})$. From ${\pi }_{ii}={\pi }_i$, ${\mathrm{\Delta }}_{ii}\mathrm{/}{\pi }_i^2=(N-n)\mathrm{/}n$, and step 2.1 gives ${\mathrm{\Delta }}_{ij}\mathrm{/}{\pi }_i^2=-(N-n)\mathrm{/}(n(N-1))$ for $i\mathrm{\ne }j$; thus ${\mathrm{Var}}_d({\bar{y}}_S)=\frac{N-n}{N^{2}n}({\sum }_i{y}_i^2-\frac{1}{N-1}{\sum }_{i\mathrm{\ne }j}{y}_i{y}_j)=\frac{N-n}{N^{2}n}\frac{N{\sum }_i{y}_i^2-({\sum }_i{y}_i{)}^2}{N-1}=\frac{N-n}{Nn}{S}_U^2$.

When $n=N$, the sample is the population and the factor $1-n\mathrm{/}N$ is zero. When $n=1$, step 3.1 gives $(1-1\mathrm{/}N)S_U^2$; the usual sample variance denominator $n-1$ is unavailable, but the design variance is defined.