Finite populations and design probability

Since ${\mathcal{A}}_d$ is the full power set of ${\mathcal{X}}_d$, it contains every subset of ${\mathcal{X}}_d$ and is closed under complements and countable unions; hence it is a sigma-algebra by [F2].

For every $B\in {\mathcal{A}}_d$, the sum defining $P_y^d(B)$ is finite and nonnegative. It gives $P_y^d(\mathrm{\varnothing })=0$ and $P_y^d({\mathcal{X}}_d)={\sum }_{s\in {\mathcal{S}}_d}{p}_d(s)=1$.

Let $(B_k{)}_{k\ge 0}$ be pairwise disjoint in ${\mathcal{A}}_d$ and put $C_k=\{s\in {\mathcal{S}}_d:O_y(s)\in B_k\}$. The sets $C_k$ are pairwise disjoint and at most $\mathrm{\mid }{\mathcal{S}}_d\mathrm{\mid }$ of them are nonempty. Their union is $\{s:O_y(s)\in {\bigcup }_k{B}_k\}$, so rearranging these finite weighted sums gives $P_y^d({\bigcup }_k{B}_k)={\sum }_k{P}_y^d(B_k)$. By [F3], $P_y^d$ is a measure; by [F4] and step 1.2, its total mass is $1$, so it is a probability measure.

For every $i$, ${\pi }_i>0$ makes the displayed denominator nonzero, so the formula defines a real-valued map on ${\mathcal{X}}_d$. Its domain has the full power-set sigma-algebra, so the preimage of every Borel set is measurable by [F8]; its target is the total estimand from [F1]. Thus [F2] and [F3] make ${\widehat{T}}_{HT}$ an estimator. For each fixed $y$, it has finite support and finite real values, hence is integrable under $P_y^d$; division by the positive integer $N$ gives the same properties for ${\widehat{\bar{y}}}_{HT}$. Thus both satisfy the integrability clause in [F9].

For each $i\in U$, let $A_i=\{s\in {\mathcal{S}}_d:i\in s\}$. Then $I_i={\mathbf{1}}_{A_i}$ by [F1] and [F4], so [F5] and [F6] give $E_d[I_i]=P_d(A_i)={\pi }_i$.

On a sample $s$, the observed values satisfy $z_i=y_i$ for $i\in s$; hence ${\widehat{T}}_{HT}(O_y(s))={\sum }_{i\in U}{I}_i(s)y_i\mathrm{/}{\pi }_i$. Applying [F7] to this finite sum and using step 1.2 gives $E_y^d[{\widehat{T}}_{HT}]={\sum }_{i\in U}\frac{y_i}{{\pi }_i}{E}_d[I_i]={\sum }_{i\in U}{y}_i=T(y)$. Together with the integrability in step 1.1, this is unbiasedness by [F9]. No independence identity is used.

Since $N\ge 1$ is fixed, linearity of the finite weighted expectation gives $E_y^d[{\widehat{\bar{y}}}_{HT}]=E_y^d[{\widehat{T}}_{HT}]\mathrm{/}N=T(y)\mathrm{/}N={\bar{y}}_U$ by step 2.1. Together with step 1.1, [F9] makes this the required unbiasedness claim for the mean estimator as well.

Since ${\pi }_i={\sum }_{s:i\in s}{p}_d(s)=0$ and every support weight is strictly positive by [F1], no $s\in {\mathcal{S}}_d$ contains $i$.

Fix any $y\in {\mathbb{R}}^U$ and any $c\in \mathbb{R}\setminus \{0\}$. Define $y_i^{\mathrm{\prime }}=y_i+c$ and $y_j^{\mathrm{\prime }}=y_j$ for $j\mathrm{\ne }i$. By step 1.1, $O_y(s)=O_{y^{\mathrm{\prime }}}(s)$ for every $s\in {\mathcal{S}}_d$. Therefore the finite sums defining $P_y^d(B)$ and $P_{y^{\mathrm{\prime }}}^d(B)$ agree for every measurable record set $B$, so $P_y^d=P_{y^{\mathrm{\prime }}}^d$; [F2] ensures these are laws in the induced experiment. But $T(y^{\mathrm{\prime }})=T(y)+c\mathrm{\ne }T(y)$, so the total fails the defining implication for identifiability.

Let $\delta$ be any estimator from the common record space to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Since $O_y(s)=O_{y^{\mathrm{\prime }}}(s)$ for every support sample, [F8] gives $E_y^d[\delta ]={\sum }_{s\in {\mathcal{S}}_d}{p}_d(s)\delta (O_y(s))={\sum }_{s\in {\mathcal{S}}_d}{p}_d(s)\delta (O_{y^{\mathrm{\prime }}}(s))=E_{y^{\mathrm{\prime }}}^d[\delta ]$. Both expectations are finite because the design support is finite and $\delta$ is real-valued. By [F9], an estimator unbiased for every finite population would make these equal expectations equal both $T(y)$ and $T(y^{\mathrm{\prime }})$, which are unequal. Thus no such universally unbiased estimator exists.